On extremal leaf status and internal status of trees
OOn extremal leaf status and internal status oftrees
Haiyan Guo ∗ , Bo Zhou † School of Mathematical Sciences, South China Normal UniversityGuangzhou 510631, P.R. China
Abstract
For a vertex u of a tree T , the leaf (internal, respectively) status of u is the sum of the distances from u to all leaves (internal vertices, respec-tively) of T . The minimum (maximum, respectively) leaf status of a tree T is the minimum (maximum, respectively) leaf statuses of all vertices of T . The minimum (maximum, respectively) internal status of a tree T isthe minimum (maximum, respectively) internal statuses of all vertices of T . We give the smallest and largest values for the minimum leaf status,maximum leaf status, minimum internal status, and maximum internalstatus of a tree and characterize the extremal cases. We also discuss theseparameters of a tree with given diameter or maximum degree. Mathematics Subject Classifications:
Keywords and phrases: minimum leaf status, maximum leaf status,minimum internal status, maximum internal status, tree, diameter, maxi-mum degree
Let G be a connected graph of order n ≥ V ( G ). For u, v ∈ V ( G ),the distance between u and v in G , denoted by d G ( u, v ), is the length of a shortestpath connecting u and v in G . Let ∅ (cid:54) = A ⊆ V ( G ). For u ∈ V ( G ), the A -statusof u in G is defined as s G ( u, A ) = (cid:88) v ∈ A d G ( u, v ) . The minimum A -status of G is s A ( G ) = min { s G ( u, A ) : u ∈ V ( G ) } , while themaximum A -status of G is S A ( G ) = max { s G ( u, A ) : u ∈ V ( G ) } . The A -centroid(or A -median) of G is defined as { u ∈ V ( G ) : s G ( u, A ) = s A ( G ) } . ∗ E-mail: [email protected] † Corresponding author. E-mail: [email protected] a r X i v : . [ c s . D M ] A ug et s G ( u ) = s G ( u, V ( G )) for u ∈ V ( G ), s ( G ) = s V ( G ) ( G ) and S ( G ) = S V ( G ) ( G ). Then s G ( u ) is the status (or transmission) of u in G [4,8,18], s ( G ) is theminimum status of G , and S ( G ) is the maximum status of G . Both minimum andmaximum statuses have been studied extensively, and it should be noted that theminimum (maximum, respectively) status appeared also in its normalized formdivided by n − T be a tree. For u ∈ V ( T ), denote by N T ( u ) the set of vertices adjacentto u in T and the cardinality of N T ( u ) is the degree of u in T , denoted by δ T ( u ).A vertex of degree one in a tree is called a leaf and a vertex of degree at leasttwo in a tree is called an internal vertex. Let L ( T ) and I ( T ) be the set of leavesand the set of internal vertices of T , respectively.Slater [16] studied structure of the A -centroid of a tree T with ∅ (cid:54) = A ⊆ V ( T ).For example, it was shown in [16, Theorem 5] that the A -centroid induces a pathin a tree for any subset A . The another related concept is called A -center, whichis defined to be the set { u ∈ V ( T ) : e A ( u, T ) } , where e A ( u, T ) = max { d T ( u, v ) : v ∈ A } . It was shown in [16] that L ( T )-center and L ( T )-centroid have quitedifferent properties. Wang [19] characterized the trees with maximum distancebetween the L ( T )-centroid and the centroid or the I ( T )-centroid, and maximumdistance between the I ( T )-centroid and the centroid, respectively. Here for twosubset A and B of vertices of a connected graph G , the distance between A and B is smallest distance between a vertex from A and a vertex from B in G .The minimum leaf status (internal status, respectively) of T is defined to bethe minimum L ( T )-status ( I ( T )-status, respectively), denoted by ls ( T ) ( is ( T ),respectively). That is, ls ( T ) = s L ( T ) ( T ) and is ( T ) = s I ( T ) ( T ) . The maximum leaf status (internal status, respectively) of T is defined to be themaximum L ( T )-status ( I ( T )-status, respectively), denoted by LS ( T ) ( IS ( T ),respectively). That is, LS ( T ) = S L ( T ) ( T ) and IS ( T ) = S I ( T ) ( T ) . In this paper, we study the extremal properties of these four parameters of trees.We give the smallest and largest values for the minimum leaf status, maximumleaf status, minimum internal status, and maximum internal status of a tree andcharacterize the extremal cases. We also discuss these parameters of trees withgiven diameter or maximum degree.We note related work of Dimitrov et al. [7], where, if restricted to trees, theystudied the extremal properties of (cid:80) u,v ∈ I ( T ) d T ( u, v ) and (cid:80) u,v ∈ V ( T ) { u,v }∩ L ( T ) (cid:54) = ∅ d T ( u, v ) fortrees T . 2 Preliminaries
The diameter of a connected graph G is the maximum distance between twovertices. Denote by S n and P n the star and the path of order n , respectively.A double star is a tree with diameter 3, which is obtainable by adding an edgebetween the centers of two nontrivial stars.For a vertex u of a nontrivial tree T , the components of T − u are called thebranches of T at u . For A ⊆ V ( T ), the A -branch-weight of u in T , denoted by bw T ( u, A ), is defined to bemax {| A ∩ V ( B ) | : B is a branch of T at u } . For a tree T , a vertex in the A -centroid is called an A -centroid vertex. Thefollowing lemma is a restatement of [16, Theorem 8]. Lemma 2.1.
Let T be a tree of order n ≥ . Then u is an A -centroid vertex ifand only if bw T ( u, A ) ≤ bw T ( v, A ) for any v ∈ V ( T ) . For u, v ∈ V ( T ), denote by n T ( u, v | A ) the number of vertices in A closerto u than to v . Let T be a tree with u ∈ V ( T ). For A = L ( T ) , I ( T ), Wang[19, Proposition 3.1] stated that u is an A -centroid vertex of T if and only if n T ( u, v | A ) ≥ n T ( v, u | A ) for any v ∈ N T ( u ).We give a somewhat easy necessary and sufficient condition for a vertex of atree T being an A -centroid vertex for A = L ( T ) , I ( T ). Lemma 2.2.
Let T be a tree of order n ≥ with u ∈ V ( T ) . For A = L ( T ) , I ( T ) , u is an A -centroid vertex of T if and only if bw T ( u, A ) ≤ | A | .Proof. Let r = δ T ( u ) and N T ( u ) = { u , . . . , u r } . For i = 1 , . . . , r , let B i bethe branch of T at u containing u i and let a i = | A ∩ V ( B i ) | . Assume that a ≥ · · · ≥ a r . Then, by definition, bw T ( u, A ) = a .Suppose that bw T ( u, A ) ≤ | A | , i.e., a ≤ | A | . For any v ∈ V ( T ) \ ( V ( B ) ∪{ u } ),say v ∈ V ( B i ) with 2 ≤ i ≤ r , as T − V ( B i ) is a subtree of a branch at v , wehave bw T ( v, A ) ≥ | A ∩ ( V ( T ) \ V ( B i )) | ≥ (cid:80) rj =1 a j − a i ≥ a = bw T ( u, A ). If A = L ( T ), then (cid:80) rj =1 a j = | A | , so (cid:80) rj =2 a j ≥ | A | , and for any v ∈ V ( B ),we have bw T ( v, A ) ≥ | A ∩ ( V ( T ) \ V ( B )) | = (cid:80) rj =2 a j ≥ | A | ≥ bw T ( u, A ). If A = I ( T ), then (cid:80) rj =1 a j = | A | −
1, so (cid:80) rj =2 a j ≥ | A | −
1, and for any v ∈ V ( B ),we have bw T ( v, A ) ≥ | A ∩ ( V ( T ) \ V ( B )) | = 1 + (cid:80) rj =2 a j ≥ | A | ≥ bw T ( u, A ).Therefore bw T ( u, A ) ≤ bw T ( v, A ) for any v ∈ V ( T ), which implies that u is an A -centroid vertex of T by Lemma 2.1.Conversely, suppose that u is an A -centroid vertex of T . Case 1. A = L ( T ).If | A | = 2, then T ∼ = P n and u may be any internal vertex. Then bw T ( u, A ) =1 ≤ | A | . Suppose that | A | ≥
3. If bw T ( u, A ) > | A | , i.e., a > | A | , then (cid:80) ri =2 a i < | A | , so s T ( u, L ( T )) − s T ( u , L ( T )) = a − (cid:80) ri =2 a i >
0, implying that s T ( u, L ( T )) >s T ( u , L ( T )), a contradiction. It follows that bw T ( u, A ) ≤ | A | .3 ase 2. A = I ( T ).If | A | = 1, then T ∼ = S n and u is the center. If | A | = 2, then T is a doublestar and u may be either internal vertices. So we have bw T ( u, A ) = 0 , ≤ | A | if | A | = 1 ,
2. Suppose that | A | ≥
3. If bw T ( u, A ) > | A | , i.e., a > | A | , then (cid:80) ri =2 a i < | A | −
1, so s T ( u, I ( T )) − s T ( u , I ( T )) = a − − (cid:80) ri =2 a i >
0, acontradiction. It follows that bw T ( u, A ) ≤ | A | .By combining Cases 1 and 2, we have bw T ( u, A ) ≤ | A | .A leaf peripherian vertex of a tree T on n vertices is a vertex of T withmaximum leaf status. Note that every vertex of P n is a leaf peripherian vertexand LS ( P n ) = n −
1. In the following lemma, we show that a leaf peripherianvertex of a tree that is not a path must be a leaf.
Lemma 2.3.
Let T be a tree that is not a path. Let u ∈ V ( T ) . If u is a leafperipherian vertex of T , then u ∈ L ( T ) .Proof. We prove the lemma by contradiction. Suppose that u is a leaf peripherianvertex of T but u / ∈ L ( T ). Then δ T ( u ) ≥
2. Let r = d T ( u ) and N T ( u ) = { u , . . . , u r } , where r ≥
2. For i = 1 , . . . , r , let B i be the branch of T at u containing u i , L i = L ( T ) ∩ V ( B i ) and a i = | L i | . Assume that a ≤ · · · ≤ a r .Suppose first that r = 2 and a = a . As T is not a path, we have a = a ≥ v ∈ L and let z be the unique vertex adjacent to v in T . Then s T ( u, L ( T )) = (cid:88) w ∈ L \{ v } d T ( u, w ) + d T ( u, v ) + (cid:88) w ∈ L d T ( u, w ) ,s T ( v, L ( T )) = (cid:88) w ∈ L \{ v } d T ( v, w ) + (cid:88) w ∈ L ( d T ( v, u ) + d T ( u, w )) , and so s T ( v, L ( T )) − s T ( u, L ( T ))= (cid:88) w ∈ L \{ v } ( d T ( v, w ) − d T ( u, w )) − d T ( u, v ) + (cid:88) w ∈ L d T ( v, u ) > (cid:88) w ∈ L \{ v } ( d T ( z, w ) − d T ( u, w )) − d T ( u, v ) + (cid:88) w ∈ L d T ( v, u ) ≥ − (cid:88) w ∈ L \{ v } d T ( z, u ) − d T ( u, v ) + (cid:88) w ∈ L d T ( v, u )= − (cid:88) w ∈ L \{ v } ( d T ( v, u ) − − d T ( u, v ) + (cid:88) w ∈ L d T ( v, u )= (cid:88) w ∈ L \{ v } a − > . s T ( v, L ( T )) > s T ( u, L ( T )). This implies that u can not be a leaf peripherianvertex of T , a contradiction.Suppose next that r = 2 and a > a , or r ≥
3. Then s T ( u, L ( T )) = (cid:88) v ∈ L d T ( u, v ) + r (cid:88) j =2 (cid:88) v ∈ L j d T ( u, v ) ,s T ( u , L ( T )) = (cid:88) v ∈ L ( d T ( u, v ) −
1) + r (cid:88) j =2 (cid:88) v ∈ L j ( d T ( u, v ) + 1) , and so s T ( u , L ( T )) − s T ( u, L ( T )) = (cid:88) v ∈ L ( −
1) + r (cid:88) j =2 (cid:88) v ∈ L j r (cid:88) i =2 a i − a > . Thus s T ( u , L ( T )) > s T ( u, L ( T )). This implies that u is not a leaf peripherianvertex of T , also a contradiction.Therefore, u ∈ L ( T ), as desired.A internal peripherian vertex of a tree T is a vertex of T with maximuminternal status. Lemma 2.4.
Let T be a tree. Suppose that u is a internal peripherian vertex of T . Then u ∈ L ( T ) .Proof. We prove the lemma by contradiction. Suppose that u / ∈ L ( T ). Then δ T ( u ) ≥ v ∈ N T ( u ) with δ T ( v ) = 1, then it is obvious that s T ( v, I ( T )) > s T ( u, I ( T )), a contradiction. So δ T ( v ) ≥ v ∈ N T ( u ).Let r = δ T ( u ) and N T ( u ) = { u , . . . , u r } , where r ≥
2. For i = 1 , . . . , r , let B i bethe branch of T at u containing u i , I i = I ( T ) ∩ V ( B i ) and a i = | I i | . Assume that a ≤ · · · ≤ a r . Let z ∈ I . Then s T ( u, I ( T )) = (cid:88) w ∈ I \{ z } d T ( u, w ) + d T ( u, z ) + r (cid:88) i =2 (cid:88) w ∈ I i d T ( u, w ) ,s T ( z, I ( T )) = (cid:88) w ∈ I \{ z } d T ( z, w ) + d T ( z, u ) + r (cid:88) i =2 (cid:88) w ∈ I i ( d T ( z, u ) + d T ( u, w )) , and so s T ( z, I ( T )) − s T ( u, I ( T ))5 (cid:88) w ∈ I \{ v } ( d T ( z, w ) − d T ( u, w )) + r (cid:88) i =2 (cid:88) w ∈ I i d T ( z, u ) ≥ − d T ( z, u )( a −
1) + d T ( z, u ) r (cid:88) i =2 a i = d T ( z, u ) (cid:32) − a + 1 + r (cid:88) i =2 a i (cid:33) > . Thus s T ( z, I ( T )) > s T ( u, I ( T )), a contradiction.A tree is called starlike if it has at most one vertex of degree greater than 2.So, a star and a path are both particular starlike trees.A diametric path of a tree is a longest path in this tree (whose length equals thediameter). Evidently, the terminal vertices of a diametric path of any nontrivialtree are leaves.A caterpillar is a tree such that the deletion of all leaves outside a diametricpath (if any exists) yields a path.A leaf edge in a tree is an edge incident with a leaf.For a tree T with uw ∈ E ( T ) and vw (cid:54)∈ E ( T ), if T (cid:48) = T − uw + vw is a tree,then we also say that T (cid:48) is obtained from T by moving the edge uw from u to v .A hanging path at a vertex u of a tree T is a path uu . . . u (cid:96) with δ T ( u ) ≥ δ T ( u (cid:96) ) = 1 and if (cid:96) ≥ δ T ( u i ) = 2 for i = 1 , . . . , (cid:96) − P be a path in a tree T . For v ∈ V ( T ) \ V ( P ), the distance between v and P is defined to be d T ( v, P ) = min { d T ( v, w ) : w ∈ V ( P ) } . Theorem 3.1.
Let T be a tree of order n . Then ls ( T ) ≥ n − with equality if and only if T is starlike.Proof. Let T be a tree of order n that minimizes the minimum leaf status. Let u be an L ( T )-centroid vertex. Suppose that there is a vertex v , different from u ,with degree at least 3. Denote by v , v , . . . , v r − all the neighbors of v , where v lies on the unique path connecting u and v in T , and r = δ T ( v ). Let T (cid:48) = T − { vv i : i = 1 , . . . , r − } + { uv i : i = 1 , . . . , r − } . Let V (cid:48) be the set of leaves in the branch of T at u containing v , and V (cid:48)(cid:48) the setof leaves of all the branches of T at v containing one of v , v , . . . , v r − . Then s T ( u, L ( T )) − s T (cid:48) ( u, L ( T (cid:48) )) = s T ( u, V (cid:48) ) − s T (cid:48) ( u, V (cid:48) )6 (cid:88) w ∈ V (cid:48)(cid:48) ( d T ( u, w ) − d T (cid:48) ( u, w ))= (cid:88) w ∈ V (cid:48)(cid:48) d T ( u, v )= | V (cid:48)(cid:48) | d T ( u, v ) > , and so ls ( T ) = s T ( u, L ( T )) > s T (cid:48) ( u, L ( T (cid:48) )) ≥ ls ( T (cid:48) ), a contradiction. Thus,all vertices different from u are of degree 1 or 2. That is, there is at most onevertex of degree at least 3, or T is starlike. By Lemma 2.2, in a starlike tree T of order n , the vertex of maximum degree is an L ( T )-centroid vertex, and thus ls ( T ) = | E ( T ) | = n − n , a and b with 2 ≤ a, b ≤ n − , T n ; a,b be the tree of order n obtained from two stars S a +1 and S b +1 by connecting their centers by a path oflength n − a − b −
1. For convenience, let T n,a = T n ; a,a . Theorem 3.2.
Let T be a tree of order n ≥ . Then ls ( T ) ≤ (cid:22) ( n + 1) (cid:23) with equality if and only if T ∼ = T n, (cid:100) n (cid:101) if n is even or n ≡ , and T ∼ = T n, n − , T n, n +34 if n ≡ .Proof. Let T be a tree of order n that maximizes the minimum leaf status.Let x be an L ( T )-centroid vertex. Let r = δ T ( x ) and N T ( x ) = { x , . . . , x r } .For i = 1 , . . . , r , let B i be the branch of T at x containing x i and let a i = | L ( T ) ∩ V ( B i ) | . Assume that a ≥ · · · ≥ a r .By Theorem 3.1, T can not be a starlike tree and thus there are at least twovertices of degree at least three, and it is obvious that one such vertex lies insome branch of T at x . Claim 1.
If a branch of T at x contains a vertex of degree at least three in T ,then there is exactly one such vertex in this branch, and its neighbors in T areall leaves except one lying on the unique path connecting to x .Assume that B is a branch of T at x with a vertex of degree at least three in T . Let P = v v . . . v p be a longest path in T from a leaf of T in B to x , where v p = x , v p − = x and p ≥
2. Then δ T ( v i ) ≥ i = 1 , . . . , p −
1. Supposethat i >
1. Then p ≥
3. Let T (cid:48) be the tree obtained from T by moving all edgesoutside P incident with v i from v i to v . By Lemma 2.2, x is an L ( T (cid:48) )-centroidvertex. It is evident that ls ( T (cid:48) ) > ls ( T ), a contradiction. Thus, v is the onlyvertex in B with degree at least three in T and all its neighbors are leaves of T except the one lying on the path connecting to x . This proves Claim 1. Claim 2. T is obtainable by connecting the centers of two copies of S a +1 by apath, where a ≥ r = 2. Then, by Lemma 2.2, both B and B contain avertex of degree at least three in T , and by Claim 1, there is exactly one vertex7f degree at least three in each of B and B , and its neighbors in T are all leavesexcept one lying on the path connecting to x . So, T is obtainable by connectingthe centers of two stars, say S a +1 and S a +1 , by a path, where a ≥ a ≥ a > a . Let z be the vertex in B with degree a + 1 in T . Then bw T ( x, L ( T )) = a > a = bw T ( z, L ( T )). So x is not an L ( T )-centroid vertex byLemma 2.1, a contradiction. Thus a = a .Suppose next that r ≥
3. Assume that B is a branch of T at x containing avertex of degree at least three in T . By Claim 1, there is exactly one such vertex v , and the neighbors of v in T are all leaves except the one lying on the pathbetween v and x . Let z ∈ L ( T ) ∩ V ( B r ), k = d T ( x, z ) and (cid:96) = d T ( x, v ). ByLemma 2.1, a = bw T ( x, L ( T )) ≤ bw T ( v , L ( T )) = (cid:80) ri =2 a i .Suppose that a < (cid:80) ri =2 a i . Suppose first that a r = 1. Let T (cid:48) = T − xx + x z. Note that (cid:80) ri =1 a i = | L ( T ) | . As a < (cid:80) ri =2 a i , we have 2 a < | L ( T ) | , so a ≤ | L ( T ) |− = | L ( T (cid:48) ) | . Then, by Lemma 2.2, x is an L ( T (cid:48) )-centroid vertex. Thus ls ( T (cid:48) ) − ls ( T ) = s T (cid:48) ( x, L ( T (cid:48) )) − s T ( x, L ( T ))= ( k + (cid:96) + 1) a − k − ( (cid:96) + 1) a = ( a − k> . That is, ls ( T (cid:48) ) > ls ( T ), a contradiction. Suppose next that a r ≥
2. Let z , . . . , z a r be the leaves of T in B r , where z = z . By Claim 1, these leaves are adjacent to acommon vertex, say v r . We consider a ≤ (cid:80) r − i =2 a i and a > (cid:80) r − i =2 a i separately.In the former case, let T (cid:48) = T − xx − { v r z i : i = 2 , . . . , a r } + { z i z i +1 : i = 1 , . . . , a r − } + x z a r . As a ≤ | L ( T ) |− a r = | L ( T ) (cid:48) | , x is an L ( T (cid:48) )-centroid vertex by Lemma 2.2, so ls ( T (cid:48) ) − ls ( T ) = ( k + a r − (cid:96) + 1) a − ka r − ( (cid:96) + 1) a = ( k + a r − a − ka r > , i.e., ls ( T (cid:48) ) > ls ( T ), a contradiction. In the latter case, let T (cid:48) = T − { xx i : i = 2 , . . . , r − } + { x r x i : i = 2 , . . . , r − } . By Lemma 2.2, x r is an L ( T (cid:48) )-centroid vertex. Then ls ( T (cid:48) ) − ls ( T ) = a − a r > a = (cid:80) ri =2 a i .If | V ( B i ) | > ≤ i ≤ r , then for the tree T (cid:48)(cid:48) = T − { xx j : j = 2 , . . . , r with j (cid:54) = i } + { x i x j : j = 2 , . . . , r with j (cid:54) = i } , as x is an L ( T (cid:48)(cid:48) )-centroid vertex by Lemma 2.2, we have ls ( T (cid:48)(cid:48) ) > ls ( T ). Thiscontradiction shows that | V ( B ) | = · · · = | V ( B r ) | = 1. This proves Claim 2.8y Claim 2, T ∼ = T n,a . Then the diameter of T is d = n + 1 − a , and anyinternal vertex is an L ( T )-centroid vertex. Let v be the neighbor of some leaf.So ls ( T ) = s T ( v, L ( T )) = a + a ( d −
1) = a d = a ( n + 1 − a ) := f ( a ) , which is strictly increasing when a ≤ n +14 and strictly decreasing when a ≥ n +14 .If n ≡ f ( a ) ≤ f (cid:0) n +14 (cid:1) = ( n +1) = (cid:106) ( n +1) (cid:107) with equality if andonly if a = n +14 . If n ≡ f ( a ) ≤ f (cid:0) n +24 (cid:1) = n ( n +2)8 = (cid:106) ( n +1) (cid:107) with equality if and only if a = n +24 . If n ≡ f ( a ) ≤ f (cid:0) n (cid:1) = n ( n +2)8 = (cid:106) ( n +1) (cid:107) with equality if and only if a = n . If n ≡ f ( a ) ≤ f (cid:0) n − (cid:1) = f (cid:0) n +34 (cid:1) = ( n − n +3)8 = (cid:106) ( n +1) (cid:107) with equality if and only if a = n − or n +34 . So, we have T ∼ = T n,a , where a = (cid:100) n (cid:101) if n is even or n ≡ n − , n +34 if n ≡ ls ( T ) = (cid:106) ( n +1) (cid:107) . Theorem 3.3.
Let T be a tree of order n with diameter d , where ≤ d ≤ n − .Then ls ( T ) ≤ ( n − d +1) d if n − d is odd ( n − d ) d + 1 if n − d is evenwith equality if and only if T ∼ = T n, n − d +12 when n − d is odd, and T ∼ = T n ; n − d , n − d +22 ,or d ≥ and T is isomorphic to a tree obtained from T n − , n − d by adding a leafedge at a vertex of degree two when n − d is even.Proof. If d = n −
1, then the result is trivial as n − d = 1 is odd, T ∼ = P n and ls ( T ) = n − ( n − d +1) d .Suppose that d ≤ n −
2. Let T be a tree of order n with diameter d thatmaximizes the minimum leaf status. Let x be an L ( T )-centroid vertex. Let r = δ T ( x ) and N T ( x ) = { y , . . . , y r } . For i = 1 , . . . , r , let B i be the branch of T at x containing y i and let a i = | L ( T ) ∩ V ( B i ) | .Suppose that r = 2. By Lemma 2.2, a , a ≤ | L ( T ) | . So a = a = | L ( T ) | .Choose a vertex w ∈ V ( B ) with δ T ( w ) ≥ d T ( x, w ) is as small aspossible. Then bw T ( w, L ( T )) = a = | L ( T ) | . So w is an L ( T )-centroid vertex byLemma 2.2. Therefore, we may assume that r ≥
3. Let P := x x . . . x d be anarbitrary diametric path of T . Claim 1. x is in some diametric path.Suppose this is not true. That is, x lies outside any diametric path. Then, forsome i with 3 ≤ i ≤ d −
3, one has d T ( x, P ) = d T ( x, x i ) ≤ min { i − , d − i − } .9ssume that x i ∈ B r . Suppose that a j > a r for some j with 1 ≤ j ≤ r −
1. Let T (cid:48) be the tree obtained from T by moving the edges xy k with 1 ≤ k ≤ r − k (cid:54) = j from x to y r . Note that the diameter of T (cid:48) is d and L ( T (cid:48) ) = L ( T ).As a j > a r , we have bw T (cid:48) ( y r ) = max { a , . . . , a r − } = bw T ( x ). By Lemma 2.2, y r is an L ( T (cid:48) )-centroid vertex. So ls ( T ) = s T ( x, L ( T )) < s T ( x, L ( T )) + a j − a r = s T (cid:48) ( y r , L ( T (cid:48) )) = ls ( T (cid:48) ). This contradiction shows that a j ≤ a r for 1 ≤ j ≤ r − ≤ j ≤ r − a j < a r , and either δ T ( y j ) ≥ δ T ( y j ) = 1 and a r < | L ( T ) | .Let T (cid:48)(cid:48) be the tree obtained from T by moving the edges xy k with 1 ≤ k ≤ r − k (cid:54) = j from x to y j . Evidently, the diameter of T (cid:48)(cid:48) is d . Let T be the maximalsubtree of T − xy j containing y j . Note that the branches of T (cid:48)(cid:48) − y j are B k with1 ≤ k ≤ r − k (cid:54) = j , T [ { x } ∪ V ( B r )], and if δ T ( y j ) ≥
2, the branches of T at y j . As a k ≤ a r for 1 ≤ k ≤ r − bw T (cid:48)(cid:48) ( y j ) = max { a k : k = 1 , . . . , r, k (cid:54) = j } = a r .If δ T ( y j ) ≥
2, then bw T (cid:48)(cid:48) ( y j ) = a r = bw T ( x ) ≤ | L ( T ) | = | L ( T (cid:48)(cid:48) ) | by Lemma 2.2.If δ T ( y j ) = 1 and a r < | L ( T ) | , then bw T (cid:48)(cid:48) ( y j ) = a r ≤ | L ( T ) |− = | L ( T (cid:48)(cid:48) ) | . In eithercase, y j is an L ( T (cid:48)(cid:48) )-centroid vertex by Lemma 2.2. So ls ( T ) = s T ( x, L ( T )) s T ( w, L ( T )), implying that ls ( T (cid:48) ) = s T (cid:48) ( w (cid:48) , L ( T (cid:48) )) > s T ( w, L ( T )) = ls ( T ), a contradiction. It thus followsthat w = x i . Let T (cid:48)(cid:48) = T − xy + x y . As bw T (cid:48)(cid:48) ( x i ) ≤ | L ( T (cid:48)(cid:48) ) | , x i is an L ( T (cid:48)(cid:48) )-centroid vertex by Lemma 2.2. Then ls ( T (cid:48)(cid:48) ) = s T (cid:48)(cid:48) ( x i , L ( T (cid:48)(cid:48) )) > s T ( x i , L ( T )) = s T ( x, L ( T )) = ls ( T ), a contradiction. So Case 1 can not occur. Case 2. a = · · · = a r .Let w be a vertex in B r with δ T ( w ) ≥ d T ( x, w ) is as small aspossible. Let T (cid:48) = T − { xy j : 2 ≤ j ≤ r − } + { wy j : 2 ≤ j ≤ r − } .Note that bw T (cid:48) ( w ) = bw T ( x ) = a < | L ( T ) | = | L ( T (cid:48) ) | . So w is an L ( T (cid:48) )-centroidvertex by Lemma 2.2. Thus ls ( T (cid:48) ) − ls ( T ) = s T (cid:48) ( w, L ( T (cid:48) )) − s T ( x, L ( T )) = a d T ( x, w ) − a r d T ( x, w ) = 0, i.e., ls ( T (cid:48) ) = ls ( T ).Suppose that w (cid:54) = x i . Let t = δ T (cid:48) ( w ) and N T (cid:48) ( w ) = { w , . . . , w t } . For j =1 , . . . , t , let B (cid:48) j be the branch of T at w containing w j and let b j = | L ( T (cid:48) ) ∩ V ( B (cid:48) j ) | .Assume that x ∈ V ( B (cid:48) ) and x i ∈ V ( B (cid:48) t ). Let T (cid:48)(cid:48) = T (cid:48) − { ww i : 2 ≤ j ≤ t − } + { w t w i : 2 ≤ j ≤ t − } . Obviously, bw T (cid:48)(cid:48) ( w t ) = b = a , so w t is an L ( T (cid:48)(cid:48) )-centroid vertex by Lemma 2.2. Note also that b = a = a r > b t . Then ls ( T (cid:48)(cid:48) ) = s T (cid:48)(cid:48) ( w t , L ( T (cid:48)(cid:48) )) = s T (cid:48) ( w, L ( T (cid:48) )) + b − b t > s T (cid:48) ( w, L ( T (cid:48) )) = ls ( T (cid:48) ) =10 s ( T ), a contradiction. Thus w = x i . Let n ( n , respectively) be the number ofleaves in the branch of T (cid:48) at x i containing x i − ( x i +1 , respectively). Let z bethe neighbor of x i in the path connecting x i and x in T (cid:48) . Assume that n ≥ n .Suppose that n > n . Denote by T the tree obtained from T (cid:48) by moving theedges x i z , x i y , . . . , x i y r − from x i to x i +1 . By Lemma 2.2, bw T ( x i +1 ) = a = bw T (cid:48) ( x i ) ≤ | L ( T (cid:48) ) | = | L ( T ) | , and so x i +1 is an L ( T )-centroid vertex. Thus ls ( T ) = s T ( x i +1 , L ( T )) ≥ s T (cid:48) ( x i , L ( T (cid:48) ))+ n − n > s T (cid:48) ( x i , L ( T (cid:48) )) = ls ( T (cid:48) ) = ls ( T ). Thiscontradiction shows that n = n . Let T = T (cid:48) − { x i z , x i y } + { x i − z , x i +1 y } .Note that there are n + a leaves in the branch of T at x i containing x and n + a (= n + a ) leaves in the branch of T at x i containing x d . Thus bw T ( x i ) = a + n ≤ | L ( T ) | , implying that x i is an L ( T )-centroid vertex by Lemma 2.2. It followsthat ls ( T ) = s T ( x i , L ( T )) = s T (cid:48) ( x i , L ( T (cid:48) )) + 2 a > s T (cid:48) ( x i , L ( T (cid:48) )) = ls ( T (cid:48) ), alsoa contradiction. So Case 2 can not occur.Now Claim 1 follows by combining the above two cases.By Claim 1, we may assume that x i = x , where 1 ≤ i ≤ d −
1. Denote by a the number of leaves in the branch of T at x i containing x i − . Claim 2. T is a caterpillar.Suppose that T is not a caterpillar. Then d ( w, P ) ≥ w of T outside P . Let u be the unique neighbor vertex of w . Assume that d ( w, x j ) = d ( w, P ) and j ≤ i . We want to show that j = i . Suppose that j < i . Choose j such that i − j is as small as possible. If j < i − δ T ( x k ) ≥ k with j < k < i , then by moving the leaf edges at x k to x , we get a tree T (cid:48) forwhich x is still an L ( T (cid:48) )-centroid by Lemma 2.2, so ls ( T (cid:48) ) > ls ( T ), which is acontradiction. Thus δ T ( x j +1 ) = · · · = δ T ( x i − ) = 2 if j < i −
1. Denote by T (cid:48) thetree obtained from T by moving all the leaf edges at u from u to x . If x i is alsoan L ( T (cid:48) )-centroid vertex, then, as d T ( x , x i ) ≥ d T ( u, x i ) and u ∈ L ( T (cid:48) ), we have ls ( T (cid:48) ) = s T (cid:48) ( x i , L ( T (cid:48) )) > s T ( x i , L ( T )) = ls ( T ), a contradiction. Thus x i is not an L ( T (cid:48) )-centroid vertex. Note that the branches of T (cid:48) at x i containing no x j are justthe branches of T at x i not containing x j , and that x i is an L ( T )-centroid vertex.By Lemma 2.2, a + 1 = bw T (cid:48) ( x ) > | L ( T (cid:48) ) | and a ≤ | L ( T ) | . So | L ( T ) | +12 = | L ( T (cid:48) ) | s T ( x j , L ( T )) = ls ( T ), alsoa contradiction. This shows that j = i .By the choice of T , each internal vertex of T on P different from x , x d − , x i has degree two. Otherwise, as above, by moving the leaf edges from these verticesto x or x d − would result in a tree with larger minimum leaf status. Note that thenumber of leaves in the branch of T at x i containing x i − is a . Let p = δ T ( u ) − k be the number of leaves in the branch of T at x i containing w . Let n bethe maximum number of leaves in a branch of T at x i containing no x i − and w .Suppose that a + p ≤ | L ( T ) | +12 . Let T (cid:48) be the tree obtained from T by moving theleaf edges at u from u to x . By Lemma 2.2, bw T (cid:48) ( x i ) = max { a + p, k − p +1 , n } ≤ max (cid:110) a + p, | L ( T ) | (cid:111) ≤ | L ( T ) | +12 , and so x i is an L ( T (cid:48) )-centroid vertex. Then we11ave ls ( T (cid:48) ) = s T (cid:48) ( x i , L ( T (cid:48) )) > s T ( x i , L ( T )) = ls ( T ), a contradiction. It thusfollows that a + p > | L ( T ) | +12 , i.e., p > | L ( T ) | +12 − a . We form a tree T (cid:48)(cid:48) by moving (cid:106) | L ( T ) | +12 (cid:107) − a leaf edges at u from u to x and the remaining leaf edges at u from u to x d − . Note that bw T (cid:48)(cid:48) ( x i ) = (cid:106) | L ( T ) | +12 (cid:107) ≤ | L ( T (cid:48)(cid:48) ) | . Thus x i is also an L ( T (cid:48)(cid:48) )-centroid vertex by Lemma 2.2. So ls ( T (cid:48)(cid:48) ) = s T ( x i , L ( T )) > s T ( x i , L ( T )) = ls ( T ), a contradiction. This completes the proof Claim 2.By Claim 2, T is a caterpillar. Then | L ( T ) | = n − d + 1. Note that a = δ T ( x ) −
1. Let b = δ T ( x d − ) − Case 1. n − d is odd.We want to show i = 1 or d −
1. Suppose that this is not true. By the choice of T , each vertex from x , . . . , x d − different from x i has degree two in T . Note thatthe number of leaves at x i in T is n − d + 1 − a − b . By Lemma 2.2, a, b ≤ n − d +12 .Suppose that a < n − d +12 . We form a tree T (cid:48) by moving n − d +12 − a leaf edgesat x i from x i to x . Evidently, ls ( T (cid:48) ) = s T (cid:48) ( x i , L ( T (cid:48) )) > s T ( x i , L ( T )) = ls ( T ),a contradiction. So a = n − d +12 . Similarly, b = n − d +12 . Thus T ∼ = T n, n − d +12 and ls ( T ) = ( n − d +1) d . Case 2. n − d is even.If i = 1 or d −
1, then with a similar argument as in Case 1, we have T ∼ = T n ; n − d , n − d +22 and ls ( T ) = ( n − d ) d + 1. Suppose that i (cid:54) = 1 , d −
1. Then d ≥ T , each vertex from x , . . . , x d − different from x i has degreetwo in T . By Lemma 2.2, a, b ≤ n − d . Suppose that a < n − d . Then a tree T (cid:48) can be formed by moving n − d − a leaf edges at x i from x i to x . Thus ls ( T (cid:48) ) = s T (cid:48) ( x i , L ( T (cid:48) )) > s T ( x i , L ( T )) = ls ( T ), a contradiction. Thus a = n − d .Similarly, b = n − d . Thus T is isomorphic to a tree obtained from T n − , n − d byadding a leaf edge at a vertex of degree two, ls ( T ) = ( n − d ) d + 1.The result follows by combining the above two cases. Theorem 4.1.
Let T be a tree of order n . Then LS ( T ) ≥ n − with equality if and only if T ∼ = P n .Proof. If T ∼ = P n , then LS ( T ) = n − T is not a path. Then n ≥
4. Let x be a leaf peripherian vertexof T . By Lemma 2.3, x ∈ L ( T ). So LS ( T ) = s T ( x, L ( T )) = (cid:80) y ∈ L ( T ) \{ x } d T ( x, y ).Let u be a vertex of degree at least three such that d T ( x, u ) is as small as possible.Then the unique path from x to any other leaf of T contains the path from x to u . On the other hand, every edge of T lies on some path connecting x and someother leaf of T . So LS ( T ) = (cid:88) y ∈ L ( T ) \{ x } d T ( x, y )12 d T ( x, u )( | L ( T ) | −
1) + n − − d T ( x, u ) ≥ | L ( T ) | − n − > n − . So the result follows.For integers n , a with 1 ≤ a ≤ n −
2, let P n,a be the tree of order n obtainedby identifying the center of a star S a +1 and a terminal vertex of a path P n − a .Particularly, P n, = P n and P n,n − = S n . Theorem 4.2.
Let T be a tree of order n ≥ . Then LS ( T ) ≤ (cid:22) n (cid:23) with equality if and only if T ∼ = P n, n for even n , and T ∼ = P n, n − , P n, n +12 for odd n .Proof. If n = 4, then T ∼ = P or S (= P , ) and LS ( P ) = 3 < LS ( P , ).If n = 5, then T ∼ = P , P , or S (= P , ) and LS ( P ) = 4 < LS ( P , ) = LS ( P , ). So the result holds if n = 4 , n ≥
6. Let T be a tree of order n that maximizes the maximumleaf status. Note that the maximum leaf status of P n, n for even n , and P n, n − or P n, n +12 for odd n is (cid:106) n (cid:107) . If T ∼ = P n , then LS ( T ) = n − < (cid:106) n (cid:107) . If T ∼ = S n ,then LS ( T ) = 2( n − < (cid:106) n (cid:107) . So T is neither a path nor a star.Let x be a leaf peripherian vertex of T . By Lemma 2.3, x ∈ L ( T ). Let P := x . . . x r be a longest path of T starting from x = x . Then 3 ≤ r ≤ n − δ T ( x i ) ≥ i with 1 ≤ i ≤ r −
1. By the choice of T , δ T ( x i ) = 2for i = 1 , . . . , r −
2, as, otherwise, by moving an edge outside P from x i to x r − weget a tree T (cid:48) , for which we have LS ( T (cid:48) ) ≥ s T (cid:48) ( x, L ( T (cid:48) )) > s T ( x, L ( T )) = LS ( T ),which is a contradiction. As P is a longest path from x , all neighbors of x r − except x r − are leaves. Let a = δ T ( x r − ) −
1. Then T ∼ = P n,a , and LS ( T ) = s T ( x, L ( T )) = a ( n − a )which is maximized to (cid:98) n (cid:99) if and only if a = n if n is even, and a = n − , n +12 if n is odd. Theorem 4.3.
Let T be a tree of order n with diameter d , where ≤ d ≤ n − .Let t = (cid:108) n − − d ) d (cid:109) for even d and t = (cid:108) n − − d ) d − (cid:109) for odd d . Then LS ( T ) ≥ n − (cid:24) dt (cid:25) with equality if and only if T is a tree with a diametric path between two leaves x and y and exactly t hanging paths at vertices of the diametric path such that LS ( T ) = s T ( x, L ( T )) = s T ( y, L ( T )) for even dt and LS ( T ) = max { s T ( x, L ( T )) , s T ( y, L ( T )) } and | s T ( x, L ( T )) − s T ( y, L ( T )) | = 1 for odd dt . roof. Let P := v v . . . v d be a diametric path in T . Let L ∗ ( T ) = L ( T ) \ { v , v d } .Note that (cid:88) w ∈ L ∗ ( T ) d T ( w, P ) ≥ | E ( T ) | − d = n − − d with equality if and only if each vertex outside P has degree one or two in T .For convenience, denote by t = (cid:100) n − − d ) d (cid:101) for even d and t = (cid:100) n − − d ) d − (cid:101) for odd d . By a result of Qiao and Zhan [14], | L ( T ) | ≥ t + 2. So | L ∗ ( T ) | ≥ t . Then LS ( T ) ≥ max { s T ( v , L ( T )) , s T ( v d , L ( T )) }≥ s T ( v , L ( T )) + s T ( v d , L ( T ))2= d + (cid:88) w ∈ L ∗ ( T ) d T ( v , w ) + d T ( v d , w )2= d + | L ∗ ( T ) | d (cid:88) w ∈ L ∗ ( T ) d T ( w, P ) (4.1) ≥ d + dt n − − d = n − dt . So LS ( T ) ≥ n − (cid:24) dt (cid:25) . Suppose that LS ( T ) = n − (cid:6) dt (cid:7) . By the proof of (4.1), each vertexoutside P has degree one or two in T . Note also that | L ∗ ( T ) | = t . Otherwise, | L ∗ ( T ) | ≥ t + 1. So, by (4.1), n − (cid:24) dt (cid:25) = LS ( T ) ≥ d + | L ∗ ( T ) | d n − − d ≥ n − d ( t + 1)2 , a contradiction. So T is a tree with a diametric path P and exactly t hangingpaths at vertices of P .If dt is even, then the three inequalities in (4.1) must be equalities, so LS ( T ) = s T ( v , L ( T )) = s T ( v d , L ( T ))Suppose next that dt is odd. From (4.1), we havemax { s T ( v , L ( T )) , s T ( v d , L ( T )) } − s T ( v , L ( T )) + s T ( v d , L ( T ))2 ≤ , i.e., | s T ( v , L ( T )) − s T ( v d , L ( T )) | = 0 ,
1. As | L ∗ ( T ) | = t , we have | s T ( v , L ( T )) − s T ( v d , L ( T )) | = 1 by (4.1).From (4.1), we also have LS ( T ) = max { s T ( v , L ( T )) , s T ( v d , L ( T )) } , as other-wise, n − (cid:24) dt (cid:25) = LS ( T )14 max { s T ( v , L ( T )) , s T ( v d , L ( T )) } = s T ( v , L ( T )) + s T ( v d , L ( T )) + 12= n − (cid:24) dt (cid:25) , a contradiction.Conversely, if T is a tree with a diametric path between two leaves x and y and exactly t hanging paths at vertices of the diametric path such that LS ( T ) = s T ( x, L ( T )) = s T ( y, L ( T )) for even dt and | s T ( x, L ( T )) − s T ( y, L ( T )) | = 1 and LS ( T ) = max { s T ( x, L ( T )) , s T ( y, L ( T )) } for odd dt , then LS ( T ) = max { s T ( x, L ( T )) , s T ( y, L ( T )) } = (cid:24) s T ( x, L ( T )) + s T ( y, L ( T ))2 (cid:25) = n − (cid:24) dt (cid:25) , as desired.We give an example on trees of order 15 with diameter 8. The three trees inFig. 1 are the ones that minimize the maximum leaf status.Fig. 1: Three trees of order 15 with diameter 8. Theorem 4.4.
Let T be a tree of order n with diameter d , where ≤ d ≤ n − .Then LS ( T ) ≤ d ( n − d ) with equality if and only if T ∼ = P n,n − d .Proof. The result is trivial if d = 2 , n −
1. Suppose that 3 ≤ d ≤ n −
2. Let T bea tree of order n with diameter d that maximizes the maximum leaf status.Let x be a leaf peripherian vertex of T . By Lemma 2.3, x ∈ L ( T ). We wantto show that x lies on some diametric path of T . Suppose that this is not true.That is, x lies outside any diametric path. Let P := x x . . . x d be an arbitrarydiametric path of T . Then d T ( x, P ) = d T ( x, x i ) for some i with 2 ≤ i ≤ d − ≤ d T ( x, x i ) ≤ min { i − , d − i − } . By the choice of T , any vertex in the path Q connecting x and x i except x and x i (if any exists) has degree two in T , as,otherwise, we move an edge outside Q from this vertex to x i to form a tree T (cid:48) , andfor T (cid:48) , its diameter is still d , but LS ( T (cid:48) ) ≥ s T (cid:48) ( x, L ( T (cid:48) )) > s T ( x, L ( T )) = LS ( T ),15hich is a contradiction. Let r = δ T ( x i ). Denote by z , . . . , z r the neighbors of x i in T , where z lies on the path Q , z = x i − and z = x i +1 . For i = 1 , . . . , r , let B i be the branch of T at x i containing z i , and let a i = | V ( B i ) ∩ L ( T ) | . Supposethat r ≥
4. Let T (cid:48) = T − x i z + x i − z . and T (cid:48)(cid:48) = T − x i z + x i +1 z . Note thediameters of T (cid:48) and T (cid:48)(cid:48) are both d . By direct calculation, s T (cid:48) ( x, L ( T (cid:48) )) − s T ( x, L ( T )) = r (cid:88) j =3 (cid:88) w ∈ V ( B j ) ∩ L ( T ) − (cid:88) w ∈ V ( B ) ∩ L ( T ) r (cid:88) j =3 a j − a = r (cid:88) j =2 a j − a , and similarly, s T (cid:48)(cid:48) ( x, L ( T (cid:48)(cid:48) )) − s T ( x, L ( T )) = r (cid:88) j =2 a j − a . So s T (cid:48) ( x, L ( T (cid:48) )) − s T ( x, L ( T )) + s T (cid:48)(cid:48) ( x, L ( T (cid:48)(cid:48) )) − s T ( x, L ( T )) = 2 (cid:80) rj =4 a j >
0, im-plying that either LS ( T (cid:48) ) ≥ s T (cid:48) ( x, L ( T (cid:48) )) > LS ( T ) or LS ( T (cid:48)(cid:48) ) ≥ s T (cid:48)(cid:48) ( x, L ( T (cid:48)(cid:48) )) >LS ( T ), a contradiction. It follows that r = 3. Suppose that there is a leaf of T different from x such that it is not adjacent to x or x d − . Choose such a leaf y of T such that d T ( y, P ) is as large as possible. Assume that d T ( y, P ) = d T ( y, x j ),where 2 ≤ j ≤ d − j (cid:54) = i . Assume that j < i . Let y (cid:48) be the uniqueneighbor of y in T . By moving the leaf edges at y (cid:48) from y (cid:48) to x we get a tree T (cid:48) ,for which we have LS ( T (cid:48) ) − LS ( T ) ≥ s T (cid:48) ( x, L ( T (cid:48) )) − s T ( x, L ( T )) ≥ d T ( x, y (cid:48) ) > LS ( T (cid:48) ) > LS ( T ), a contradiction. This shows that δ T ( x k ) = 2 for any2 ≤ k ≤ d − k (cid:54) = i .Let (cid:96) = d T ( x, x i ). Assume that i ≤ (cid:98) d (cid:99) . By the choice of T , δ T ( x ) = 2 and δ T ( x d − ) = n − d − (cid:96) . So LS ( T ) = s T ( x, L ( T )) = (cid:96) + i + ( (cid:96) + d − i )( n − d − (cid:96) − s T ( x , L ( T )) = ( n − d − (cid:96) − d + (cid:96) + i . But s T ( x , L ( T )) − LS ( T ) =( n − d − (cid:96) − i − (cid:96) ) >
0, a contradiction. Thus x lies on some diametric pathof T .Let P (cid:48) := xv . . . v d be a diametric path starting from x . By the choice of T and the argument as above, all leaves of T different from x are adjacent to v d − ,i.e., T ∼ = P n,n − d . The result follows by noting that LS ( T ) = d ( n − d ). Proposition 5.1.
Let T be a tree of order n ≥ .(i) is ( T ) ≥ with equality if and only if T ∼ = S n .(ii) If T (cid:29) S n , then is ( T ) ≥ with equality if and only if T is a double star. iii) If T (cid:29) S n , and T is not a double star, then is ( T ) ≥ with equality if andonly if T is a caterpillar of diameter .Proof. Item (i) follows from the fact that S n is the only tree with exactly oneinternal vertex. Item (ii) follows as the double stars are the only trees with exactlytwo (adjacent) internal vertices. Item (iii) follows as for any tree T of diameterat least 4 contains three internal vertices inducing a path P in T , and if T hasmore than 3 internal vertices, then the subtree induced by internal vertices in T contains S or P so that is ( T ) ≥ Proposition 5.2.
Suppose that T is a tree of order n with diameter d , where ≤ d ≤ n − . Then is ( T ) ≥ (cid:22) ( d − (cid:23) with equality if and only if T is caterpillar.Proof. By Lemma 2.2, an I ( T )-centroid vertex is an internal vertex of T .Let P := v . . . v d be a diametric path in T . Obviously, { v , . . . , v d − } inducesin T a path P d − . If T is a caterpillar, then is ( T ) = s ( P d − ). Suppose that T is not a caterpillar, i.e., I ( T ) \ V ( P ) (cid:54) = 0. For any i = 1 , . . . , d −
1, wehave s T ( v i , I ( T )) ≥ s ( P d − ) + (cid:80) w ∈ I ( T ) \ V ( P ) d T ( w, v i ) > s ( P d − ). Suppose that w ∈ I ( T ) \ V ( P ). Assume that d T ( w, v j ) = d T ( w, P ) for some 2 ≤ j ≤ d − s T ( w, I ( T )) ≥ s T ( w, I ( T ) ∩ V ( P )) > s T ( v j , I ( T ) ∩ V ( P )) ≥ s ( P d − ) . Therefore, is ( T ) ≥ s ( P d − ) = (cid:106) ( d − (cid:107) with equality if and only if T is caterpillar. Theorem 5.1.
Let T be a tree of order n ≥ . Then is ( T ) ≤ (cid:22) ( n − (cid:23) with equality if and only if T ∼ = P n .Proof. By Lemma 2.2, an I ( T )-centroid vertex is an internal vertex of T , so is ( T ) equals to the minimum status of T (cid:48) , i.e., is ( T ) = s ( T (cid:48) ), where T (cid:48) is thetree obtained from T by deleting all leaves. Let k = | L ( T ) | . By Proposition 2.1in [1], we have is ( T ) = s ( T (cid:48) ) ≤ (cid:22) ( n − k ) (cid:23) ≤ (cid:22) ( n − (cid:23) with equalities if and only if k = 2 and T ∼ = P n .17 Maximum internal status
Similarly to Proposition 5.1, we have
Proposition 6.1.
Let T be a tree of order n ≥ . The following statements aretrue.(i) IS ( T ) ≥ with equality if and only if T ∼ = S n .(ii) If T (cid:29) S n , then IS ( T ) ≥ with equality if and only if T is a double star.(iii) If T (cid:29) S n , and T is not a double star, then IS ( T ) ≥ with equality if andonly if T is a caterpillar of diameter . Furthermore, we have
Proposition 6.2.
Suppose that T is a tree of order n with diameter d , where ≤ d ≤ n − . Then IS ( T ) ≥ d ( d − with equality if and only if T is caterpillar.Proof. Let P := v v . . . v d be a diametric path in T . Let I ∗ ( T ) = I ( T ) \{ v , . . . , v d − } . Then IS ( T ) ≥ max { s T ( v , I ( T )) , s T ( v d , I ( T )) }≥ s T ( v , I ( T )) + s T ( v d , I ( T ))2= 2 (cid:80) d − i =1 i + (cid:80) w ∈ I ∗ ( T ) ( d T ( v , w ) + d T ( v d , w ))2 ≥ d ( d − I ∗ ( T ) = ∅ , i.e., T is a caterpillar.Let T be a tree with u ∈ V ( T ). For positive integer p , we denote by T u ; p thetree consisting of T and a path P := uu . . . u p such that u is the only commonvertex of T and the path P . In this case, we also say that P is ‘hanging’ a pathof length p at u in T u ; p , though it is really a hanging path of length p at u in T u ; p only when δ T ( u ) ≥ G u ;0 = G . For nonnegative integer p and q , let G u ; p,q = ( G u ; p ) u ; q . Lemma 6.1.
Let T be a nontrivial tree with u ∈ V ( T ) Let p and q be positiveintegers with p ≥ q . Then IS ( T u ; p +1 ,q − ) = IS ( T u ; p,q ) if q ≥ and T is a starwith center u , otherwise IS ( T u ; p +1 ,q − ) > IS ( T u ; p,q ) .Proof. Let H = T u ; p,q and H (cid:48) = T u ; p +1 ,q − . Let u u . . . u p and v v . . . v q bethe two ‘hanging’ paths at u in H , where u = v = u . Let x be a internalperipherian vertex of H . Let L ∗ ( T ) = L ( T ) if u (cid:54)∈ L ( T ) and L ∗ ( T ) = L ( T ) \ { u } otherwise. Then x ∈ L ( H ) = L ∗ ( T ) ∪ { u p , v q } by Lemma 2.4. As p ≥ q , we have s H ( u p , I ( H )) ≥ s H ( v q , I ( H )). So we may assume that x ∈ L ∗ ( T ) ∪ { u p } . Let I ∗ ( T ) = I ( T ) if u ∈ I ( T ) and I ∗ ( T ) = I ( T ) ∪ { u } otherwise.18 ase 1. x ∈ L ∗ ( T ).If q ≥
2, then s H ( x, I ( H )) = (cid:88) w ∈ I ∗ ( T ) d H ( x, w ) + p − (cid:88) i =1 ( d H ( x, u ) + i ) + q − (cid:88) i =1 ( d H ( x, u ) + i )= (cid:88) w ∈ I ∗ ( T ) d H ( x, w ) + d H ( x, u )( p + q −
2) + p − (cid:88) i =1 i + q − (cid:88) i =1 i, and s H (cid:48) ( x, I ( H (cid:48) )) = (cid:88) w ∈ I ∗ ( T ) d H (cid:48) ( x, w ) + p (cid:88) i =1 ( d H (cid:48) ( x, u ) + i ) + q − (cid:88) i =1 ( d H (cid:48) ( x, u ) + i )= (cid:88) w ∈ I ∗ ( T ) d H (cid:48) ( x, w ) + d H (cid:48) ( x, u )( p + q −
2) + p (cid:88) i =1 i + q − (cid:88) i =1 i. Then s H (cid:48) ( x, I ( H (cid:48) )) − s H ( x, I ( H )) = p − q +1 > IS ( H (cid:48) ) ≥ s H (cid:48) ( x, I ( H (cid:48) )) >s H ( x, I ( H )) = IS ( H ).If q = 1, then s H ( x, I ( H )) = (cid:88) w ∈ I ∗ ( T ) d H ( x, w ) + p − (cid:88) i =1 ( d H ( x, u ) + i )= (cid:88) w ∈ I ∗ ( T ) d H ( x, w ) + d H ( x, u )( p −
1) + p − (cid:88) i =1 i, and s H (cid:48) ( x, I ( H (cid:48) )) = (cid:88) w ∈ I ∗ ( T ) d H (cid:48) ( x, w ) + p (cid:88) i =1 ( d H (cid:48) ( x, u ) + i )= (cid:88) w ∈ I ∗ ( T ) d H (cid:48) ( x, w ) + d H (cid:48) ( x, u ) p + p (cid:88) i =1 i. Then s H (cid:48) ( x, I ( H (cid:48) )) − s H ( x, I ( H )) = d H ( x, u )+ p > IS ( H (cid:48) ) ≥ s H (cid:48) ( x, I ( H (cid:48) )) >s H ( x, I ( H )) = IS ( H ). Case 2. x = u p .Let q ≥
2. Then s H ( x, I ( H )) = (cid:88) w ∈ I ∗ ( T ) ( p + d H ( u, w )) + p − (cid:88) i =1 i + q − (cid:88) i =1 ( p + i )= (cid:88) w ∈ I ∗ ( T ) ( p + d H ( u, w )) + p ( q −
1) + p − (cid:88) i =1 i + q − (cid:88) i =1 i, s H (cid:48) ( x, I ( H (cid:48) )) = (cid:88) w ∈ I ∗ ( T ) ( p + 1 + d H (cid:48) ( u, w )) + p (cid:88) i =1 i + q − (cid:88) i =1 ( p + 1 + i )= (cid:88) w ∈ I ∗ ( T ) ( p + 1 + d H (cid:48) ( u, w )) + ( p + 1)( q −
2) + p (cid:88) i =1 i + q − (cid:88) i =1 i. Then s H (cid:48) ( x, I ( H (cid:48) )) − s H ( x, I ( H )) = | I ∗ ( T ) | −
1. If | I ∗ ( T ) | ≥
2, then IS ( H (cid:48) ) ≥ s H (cid:48) ( x, I ( H (cid:48) )) > s H ( x, I ( H )) = IS ( H ). If | I ∗ ( T ) | = 1, then IS ( H (cid:48) ) = s H (cid:48) ( x, I ( H (cid:48) )) = s H ( x, I ( H )) = IS ( H ).Let q = 1. Then s H ( x, I ( H )) = (cid:88) w ∈ I ∗ ( T ) ( p + d H ( u, w )) + p − (cid:88) i =1 i, and s H (cid:48) ( x, I ( H (cid:48) )) = (cid:88) w ∈ I ∗ ( T ) ( p + 1 + d H (cid:48) ( u, w )) + p (cid:88) i =1 i. Then s H (cid:48) ( x, I ( H (cid:48) )) − s H ( x, I ( H )) = | I ∗ ( T ) | + p >
0, and so IS ( H (cid:48) ) ≥ s H (cid:48) ( x, I ( H (cid:48) )) >s H ( x, I ( H )) = IS ( H ).The result follows by combing the above two cases and noting that | I ∗ ( T ) | = 1if and only if T is a star with center u . Theorem 6.1.
Let T be a tree of order n ≥ . Then IS ( T ) ≤ n − n + 22 with equality if and only if T ∼ = P n .Proof. Let T be a tree of order n that maximizes the maximum internal status.Suppose that T is not a path. Let x ∈ L ( T ). Then we choose a vertex ofdegree at least three, say u , such that d T ( x, u ) is as large as possible. Thenthere are two hanging paths P and Q at u in T . By Lemma 6.1, we can obtaina tree T (cid:48) so that IS ( T (cid:48) ) > IS ( T ), a contradiction. Thus T ∼ = P n . Evidently, IS ( P n ) = (cid:80) n − i =1 i = n − n +22 . Theorem 6.2.
Let T be a tree of order n with maximum degree ∆ , where ≤ ∆ ≤ n − . Then IS ( T ) ≤ ( n − ∆)( n − ∆ + 1) with equality if and only if T isa starlike tree with at least ∆ − hanging paths being of length one.Proof. It is trivial if ∆ = 2. Suppose that ∆ ≥
3. Let T be a tree of order n with maximum degree ∆ that maximizes the maximum internal status. Let u be a vertex of degree ∆. If there is a vertex different from u with degree at20east three, then we may choose such a vertex v by requiring that d T ( u, v ) is aslarge as possible. This implies that there are two hanging paths at v in T . ByLemma 6.1, there is a tree of order n with maximum degree ∆ having largermaximum internal status, which is a contradiction. That is, u is the only vertexof degree at least three. In other words, T is a starlike tree. By Lemma 6.1again, ∆ − T is a starlike tree with atleast ∆ − IS ( T ) = IS ( P n, ∆ − ) = (cid:80) n − ∆ i =1 i = ( n − ∆)( n − ∆ + 1). Acknowledgement.
This work was supported by National Natural Science Foun-dation of China (No. 11671156).