4. Minimum status and number of pendant vertices

https://doi.org/10.1051/ro/2020015 www.rairo-ro.org

MINIMUM STATUS OF TREES WITH GIVEN PARAMETERS

Zhene Peng and Bo Zhou

Abstract.The status of a vertex vin a connected graph Gis defined as the sum of the distances from vto all other vertices in G. The minimum status of Gis the minimum of status of all vertices ofG. We give the smallest and largest values for the minimum status of a tree with fixed parameters such as the diameter, the number of pendant vertices, the number of odd vertices, and the number of vertices of degree two, and characterize the unique extremal trees.

Mathematics Subject Classification. 05C12, 05C35, 05C90.

Received September 29, 2019. Accepted February 12, 2020.

1. Introduction

We consider simple and undirected graphs. LetGbe a connected graph of ordernwith vertex setV(G). The distance between verticesuandv inG, denoted byd_G(u, v), is the length of a shortest path connectinguand v inG. The status of a vertexuin Gis defined as [9,15]

s_G(u) = X

v∈V(G)

d_G(u, v).

The minimum status ofG, denoted bys(G), is defined as [9,24,27]

s(G) = min{sG(u) : u∈V(G)}.

In the literature, the status of a vertex is also known as the distance [12,26], the transmission [21,25,29,30], or the total distance [6] of a vertex, and if n≥2, the normalized status _n−1¹ sG(u) of vertex uinG is called the average distance from vertexu to all other vertices ofG, see [2]. The status or normalized status of a vertex may be used to measure its closeness centrality in the network [13,14]. The minimum status of a graphGis also studied through the proximity ofG, defined as _n−1¹ s(G), see,e.g., [1–5]. Recall that the mean vertex derivation ofG, defined as ¹_ns(G), was also studied in [31].

Aouchiche and Hansen [2] gave sharp lower and upper bounds for the minimum status of a graph as a function of its order, and characterized the extremal graphs, and they also gave a sharp lower bound for the minimum status of a graph with fixed order and diameter. A trivial lower bound for the minimum status of a connected graph of ordernisn−1, which is achieved if and only if there is a vertex of degreen−1. LetGbe a connected

Keywords.Minimum status, closeness centrality, diameter, pendant vertices, odd vertices, series-reduced trees.

School of Mathematical Sciences, South China Normal University, Guangzhou 510631, P.R. China.

∗Corresponding author:zhoubo@scnu.edu.cn

Article published by EDP Sciences c EDP Sciences, ROADEF, SMAI 2021

graph of order n. They showed that s(G)≤ bⁿ₄²cwith equality if and only if Gis either the cycle Cn or the pathPn. If Gis a tree, then this result has also been given in [31]. Aouchiche and Hansen [1] gave Nordhaus- Gaddum-type inequalities for the minimum status of a graph,i.e., lower and upper bounds fors(G) +s(G) and s(G)s(G) in terms of the order of G with both G and Gbeing connected, where G is the complement of G.

Lin et al. [24] established sharp lower and upper bounds for the minimum status of a graph with maximum degree, and they characterized extremal graphs for the lower bound and gave a necessary condition for graphs attaining the upper bound. Rissner and Burkard [27] gave alternate and simple proofs for the results in [24], and showed that the minimum status and the radius achieve their minimum (maximum, respectively) values at the same type of trees when order and maximum degree are given. Liang et al. [22] gave sharp lower and upper bounds for the minimum status of a tree with fixed matching number (domination, respectively), and characterize the extremal trees. Related work may be found in [2,3,10,28].

A vertex in a graph is a pendant vertex if its degree is one, and a vertex in a graph is an odd (even, respectively) vertex if its degree is odd (even, respectively). By Handshaking Theorem, a graph possesses an even number of odd vertices. A vertex is an internal vertex if its degree is at least two. A tree whose internal vertices all have degree at least three is called a series-reduced tree or a homeomorphically irreducible tree [7,16,17]. That is, a series-reduced tree has no vertices of degree two.

Following the above work, in this paper, we give sharp lower and upper bounds for the minimum status of trees using parameters, including the diameter, the number of pendant vertices, the number of odd vertices, and the number of vertices of degree two, and we also characterize all extremal cases. In some cases, the bounds may be extended to connected graphs.

2. Preliminaries

For a proper subset U of vertices of a graph G, G−U denotes the subgraph of G obtained by deleting the vertices from U (and the incident edges), and in particular, we write G−v forG− {v} if U ={v}. For a subset E₁ of edges ofG,G−E₁ denotes the subgraph obtained from Gby deleting all the edges inE₁, and in particular, we writeG−uvforG− {uv}ifE₁={uv}. For a subsetE₂ of unordered vertex pairs of distinct vertices of G, if each element of E2 is not an edge of G, then G+E2 denotes the graph obtained from Gby adding all elements ofE2 as edges, and in particular, we writeG+uvforG+{uv}ifE2={uv}.

For a vertexv in a graphG, NG(v) denotes the set of vertices adjacent tov (i.e., neighbors ofv) in G, and the degree of v, denoted by δ_G(v), is equal to |N_G(v)|. For k ≥1, we say a path P =v₀. . . v_k in a graph G is a pendant path (of length k) at v0 ifδG(v0)>2, δG(vk) = 1, and ifk ≥2,δG(v1) =· · · =δG(vk−1) = 2.

A pendant path of length one is also known as a pendant edge.

IfPis a pendant path of length`at vertexvin a graphG, thenGis said to be obtained fromG−(V(P)\{v}) by attaching a pendant path of length ` at v. Particularly, if `= 1, say P =vw, then we also say thatG is obtained fromG−wby attaching a pendant edge atv(or a pendant vertex to v).

A caterpillar is a tree with a path such that each vertex of the tree either lies on this path or is adjacent to a vertex in this path. Note thatSn andPn are caterpillars.

A quasi-pendant vertex is a vertex that is adjacent to a pendant vertex.

The median of a connected graphGis the set of vertices of Gwith minimum status. The median of a tree consists of either one vertex or two adjacent vertices [12,18,31].

Lemma 2.1([19,20,31]). Let T be a nontrivial tree of order n. Then a vertexxis in the median of T if and only if|V(C)| ≤ⁿ₂ for every componentC of T−x.

It follows from Lemma2.1that any pendant vertex in a tree with at least 3 vertices can not be in the median of this tree.

For a graphGwith an cut edgeuv, letG_uv=G− {vw:w∈N_G(v)\ {u}}+{uw:w∈N_G(v)\ {u}}.

Lemma 2.2([22]). Let Gbe a connected graph and uv be a cut edge of G. If uv is not a pendant edge of G, thens(G)> s(G_uv).

Lemma 2.3([22]). Let T be a tree with u∈V(T)andN_T(u) ={u1, . . . , u_k}, wherek≥3. For1≤i≤k, let Ti be the component ofT−ucontainingui. Forw∈V(T2), letT⁰ =T− {uui: 3≤i≤t}+{wui: 3≤i≤t}, where3≤t≤k. If |V(T1)| ≥ |V(T2)|, thens(T⁰)> s(T).

For integers n, rand t with r ≥t≥0 and r+t+ 2≤n, we denote by D_n(r, t) the caterpillar formed by attaching rpendant edges to one terminal vertex and t pendant edges to the other terminal vertex of a path P_n−r−t. Ifr+t=n−1, thenDn(r, t) is just the starSn. Note thatDn(1,1)∼=Pn ∼=Dn(1,0).

Lemma 2.4([22]). If r+t+ 2≤n andr≥t≥2, then s(D_n(r, t))> s(D_n(r+ 1, t−1)).

Lemma 2.5. If 2≤a≤n−1, thens(Dn(d^a₂e,b^a₂c)) =b^n2−a₄^2+2ac.

Proof. It is evident fora= 2, n−1. Suppose that 3≤a≤n−2.

Note that the diameter of Dn(d^a₂e,b^a₂c) is n−a+ 1. By Lemma 2.1, the vertex of distance b^n−a+1₂ c from pendant vertex adjacent to a vertex of degreed^a₂e+ 1 is in the median ofDn(d^a₂e,b^a₂c). Thus, ifn−ais odd, then

s Dn

la 2 m

,ja 2

k

= 2

n−a+1 2 −1

X

j=1

j+la 2

m·n−a+ 1 2 +ja

2

k·n−a+ 1 2

= n²−a²+ 2a−1

4 ,

and ifn−ais even, then s

D_nla 2 m

,ja 2

k

=

n−a 2 −1

X

j=1

j+

n−a 2

X

j=1

j+la 2

m·n−a 2 +ja

2 k·

n−a 2 + 1

=

n−a 2

²

+a(n−a) 2 +la

2 m·

Thus

s D_nla

2 m,ja

2 k=











n²−a²+2a−1

4 ifn−ais odd

n²−a²+2a

4 ifn−ais even andais even

n²−a²+2a−2

4 ifn−ais even andais odd

=

n²−a²+ 2a 4

,

as desired.

For a nontrivial treeT withu∈V(T) and integersp, q≥1, letTu;p,q be the tree obtained by attaching two pendant paths, one of lengthpand one of lengthq, atu.

Lemma 2.6. Let T be a nontrivial tree withu∈V(T). Ifp≥q+ 2 andq≥1, thens(Tu;p,q)> s(Tu;p−1,q+1).

Proof. LetH =Tu;p,q. Let P =v0. . . vp and Q=w0. . . wq be the two pendant paths at u(=v0=w0) inH. LetH⁰ =H−v_p−1v_p+w_qv_p. Letxbe a vertex in the median ofH. Asp > q, it follows from Lemma2.1that x6∈V(Q)\ {u}.

Case 1.x∈V(T).

By Lemma 2.1, x is also in the median of H⁰. As we go from H to H⁰, the distance between xand vp is decreased byp−(q+1) (which is at least 1) and the distance betweenxand any other vertex remains unchanged.

Therefore

s(H⁰)−s(H) =sH⁰(x)−sH(x)≤ −1<0, implying thats(H)> s(H⁰).

Figure 1.The treeT(r, t).

Case 2.x∈V(P)\ {u}.

Assume that x=vi, and if two vertices in V(P)\ {u} are in the median of H, we choose vi with smaller indexi. Asδ_H(x) = 2, there are exactly two components inH−x. Letnbe the order ofH. By Lemma2.1, we have d_H(x, v_p) =bⁿ₂c,i.e.,p−i=bⁿ₂c. By Lemma2.1,v_i−1 is in the median ofH⁰. Note thatd_H0(v_i−1, v_p)− dH(vi, vp) =i+q−_n

2

,dH⁰(vi−1, z)−dH(vi, z) = 1 forz∈ {vi+1, . . . , vp−1}, anddH⁰(vi−1, z)−dH(vi, z) =−1 forz∈V(H)\ {v_i−1, . . . , vp}. Thus, we have

s(H⁰)−s(H) =s_H⁰(v_i−1)−s_H(v_i)

= X

z∈V(H⁰)

d_H⁰(v_i−1, z)− X

z∈V(H)

d_H(v_i, z)

=

i+q−jn 2

k

+ 1·(p−1−i)−1·(n−p−2 +i)

=p+q−n+ 1<0,

implying thats(H)> s(H⁰).

Let r and t be positive integers. For a nontrivial tree T and a caterpillar C of degrees one or three with a diametral path u₁. . . u_r+t+1, where v_i is the unique neighbor of u_i outside path u₁. . . u_r+t+1 for each 2 ≤ i ≤ r+t, let T(r, t) be the tree of order 2(r+t)−1 +|V(T)| consisting of T and C such that they share a unique common vertexvr+1, see Figure1. Note thatT(r, t)−(V(T)\ {vr+1}) is a caterpillar of degrees one or three with diametral pathu1. . . ur+t+1, wherevi is the unique neighbor ofui outside pathu1. . . ur+t+1

for each 2≤i≤r+t.

Lemma 2.7. Let T be a nontrivial tree. Ifr≥t≥2, then

s(T(r+ 1, t−1))> s(T(r, t)).

Proof. LetH =T(r, t). LetT₁ andT₂ be the components ofG−u_r+1 containingu₁ andu_r+t+1, respectively.

Let

H⁰=H− {ur+1vr+1, ur+2vr+2}+{ur+1vr+2, ur+2vr+1}.

ThenH⁰∼=T(r+ 1, t−1). As|V(T2)|= 2t−1≤2r−1 =|V(T1)|, we havex /∈V(T2) by Lemma2.1.

Case 1.x∈V(T).

By Lemma2.1, x is also in the median of H⁰. Note that dH⁰(x, z)−dH(x, z) = 1 for z ∈ V(T1)∪ {ur+1}, dH⁰(x, z)−dH(x, z) = −1 for z∈V(T2)\ {vr+2}, and the distance betweenxand a vertex of V(T)∪ {vr+2} remains unchanged. Thus

s(H⁰)−s(H) =sH⁰(x)−sH(x)

= 1·(|V(T₁)|+ 1)−1·(|V(T₂)| −1)

=|V(T1)| − |V(T2)|+ 2>0, implying thats(H⁰)> s(H).

Case 2.x∈V(T₁), orx=u_r+1 and|V(T₁)|=bⁿ₂c,bⁿ₂c −1.

By Lemma2.1,xis also in the median ofH⁰. As we go fromH toH⁰, the distance betweenxand a vertex ofV(T) is increased by 1, the distance betweenxandvr+2 is decreased by 1, and the distance betweenxand any other vertex remains unchanged. Thuss(H⁰)−s(H) =|V(T)| −1>0,i.e.,s(H⁰)> s(H).

Case 3.x=ur+1 and|V(T1)| ≤ bⁿ₂c −2.

By Lemma 2.1, u_r+2 is a vertex in the median of H⁰. Note that d_H⁰(u_r+2, z)−d_H(u_r+1, z) = 1 for z ∈ V(T₁), d_H⁰(u_r+2, z)−d_H(u_r+1, z) = −1 for z ∈ V(T₂)\ {ur+2, v_r+2}, and d_H⁰(u_r+2, z) = d_H(u_r+1, z) for z∈V(T)∪ {vr+2}. Thus

s(H⁰)−s(H) =sH⁰(ur+2)−sH(ur+1) =|V(T1)| −(|V(T2)| −2)>0,

implying thats(H⁰)> s(H).

Lemma 2.8. Let T be a caterpillar with the largest minimum status in the class of caterpillars of ordernwith t ≥ 1 vertices of degree two and maximum degree three, or in the class of caterpillars of order n with t ≥1 vertices of degree two and one or two vertices of maximum degree four, where each vertex of degree four has three pendant neighbors. LetU be the set of vertices of degree two inT. ThenU induces a path inT.

Proof. Suppose thatU does not induce a path inT. LetP =u0. . . ud be a diametral path ofT. Then there are three vertices u_i, u_j, u<sub>`</sub> with 1 ≤i < j < ` ≤d−1 withδ_T(u_j) = 3 and u_i, u_{`</sub> ∈ U. Let T<sub>1</sub> and T<sub>2</sub> be the nontrivial components ofT −u<sub>j</sub> containingu<sub>i</sub> andu<sub>`}, respectively. Letv_j be the pendant neighbor ofu_j. Assume that|V(T1)| ≥ |V(T2)|. LetT⁰ =T−ujvj+u`vj. By Lemma2.3,s(T⁰)> s(T), a contradiction.

Fora, b≥0 with 2(a+b+ 1)≤n, letC_n(a, b) be the tree obtained from the pathP_n−a−b with consecutive vertices u₁, . . . , u_n−a−b by attaching a pendant vertex v_i to u_i for each i with 2≤i ≤a+ 1 and each i with n−a−2b≤i≤n−a−b−1. In particular Cn(0,0) is just the path Pn.

Lemma 2.9. If b≥0,a≥b+ 2, and2(a+b) + 2< n, then

s(C_n(a−1, b+ 1))> s(C_n(a, b)).

Proof. LetT =Cn(a, b). LetT⁰ =T−ua+1va+1+un−a−2b−1va+1. Then T⁰ ∼=Cn(a−1, b+ 1).

Case 1.2a≥_n

2

.

Note thatT−udⁿ⁻¹4 e⁺¹contains exactly two nontrivial components of orders 2dⁿ⁻¹₄ e −1 andn−1−2dⁿ⁻¹₄ e, respectively. By Lemma2.1,udⁿ⁻¹4 e⁺¹is in the median ofT andT⁰. As we go fromT toT⁰, the distance between udⁿ⁻¹4 e⁺¹ and v_a+1 is increased by n−a−2b−1− _n−1

4

+ 1

− a+ 1−(_n−1

4

+ 1)

(which is equal to n−2(a+b)−2, larger than 0), and the distance betweenudⁿ⁺¹₄ e⁺¹ and any other vertex remains unchanged.

Thuss(T⁰)> s(T), as desired.

Case 2.2a <_n

2

.

Letx=udⁿ2e^−a and x⁰ =udⁿ2e^−a+1. By Lemma2.1,xis in the median ofT andx⁰ is in the median ofT⁰. LetT1be the component ofT−xcontainingu1, andT2the component ofT−x⁰ containingu_n−a−b. Note that

dT⁰(x⁰, z)−dT(x, z) =

(1 ifz∈V(T₁)\ {v_a+1}

−1 ifz∈V(T₂) and

dT⁰(x⁰, va+1)−dT(x, va+1) =n−a−2b−1−ln 2

m−a+ 1

−ln 2

m−a−(a+ 1)

=n−2ln 2 m

+ 2(a−b)−1≥2.

If 2a+ 1 =_n

2

, thenx=u_a+1,v_a+16∈V(T₁), and thus s(T⁰)−s(T) =sT⁰(x⁰)−sT(x)

=|V(T₁)| − |V(T₂)|+d_T⁰(x⁰, v_a+1)−d_T(x, v_a+1)

≥ln 2

m−2

−jn 2

k−1

+ 2>0, and otherwise, we have

s(T⁰)−s(T) =sT⁰(x⁰)−sT(x)

=|V(T1)| −1− |V(T2)|+dT⁰(x⁰, va+1)−dT(x, va+1)

≥ln 2

m−1

−1−jn 2

k−1

+ 2>0.

It follows thats(T⁰)> s(T), as desired.

3. Minimum status and diameter

In this section, we give sharp lower and upper bounds for the minimum status of a tree with fixed diameter, and characterize the unique trees achieving these bounds.

LetFn,dbe the caterpillar obtained by attachingn−d−1 pendant vertices tov_dd

2eof the pathPd+1=v0. . . vd, where 2≤d≤n−1. DefineF_n,n−1=Pn. Particularly,Fn,2=Sn andFn,3=Dn(n−3,1). The lower bound in the following theorem has been obtained in [2] for a connected graph of ordern and diameterd. However, we include a proof here for completeness.

Theorem 3.1. Suppose that T is a tree of order n≥4 with diameterd, where2≤d≤n−1. Then s(T)≥n−1−d+

(d+ 1)² 4

with equality if and only if T ∼=F_n,d.

Proof. Letxbe a vertex in the median ofT andP a diametral path ofT. Then s(T) = X

v∈V(P)

d_T(x, v) + X

v∈V(T)\V(P)

d_T(x, v)

≥s(P) + X

v∈V(T)\V(P)

1

=n−1−d+

(d+ 1)² 4

with equality if and only if x is in the median of P and all vertices outside P are adjacent to x, i.e.,

T ∼=Fn,d.

Corollary 3.2. Suppose thatT is a tree of order n, and T Sn. Then s(T)≥n with equality if and only if T∼=Dn(n−3,1). Moreover, ifTDn(n−3,1), thens(T)≥n+1with equality if and only ifT∼=Fn,4, Dn(n−4,2).

Proof. Letd be the diameter of T. Letf(d) =n−1−d+j_(d+1)2

4

k

for 2 ≤d≤n−1. It is easy to see that f(d+ 1) > f(d) for 2 ≤ d ≤ n−2. As T Sn, we have d ≥ 3. If d ≥ 4, then by Theorem 3.1, we have s(T)≥f(d)≥f(4) =n+ 1 with equalities if and only ifT ∼=Fn,4. Ifd= 3, thenT ∼=Dn(r, n−2−r) for some rwith ⁿ⁻²₂ ≤r≤n−3, and in this case, s(T) = 2n−3−r, which isnifr=n−3,n+ 1 ifr=n−4 and at

leastn+ 2 ifr≤n−5.

Theorem 3.3. Suppose that T is a tree of order n≥4 with diameterd, where2≤d≤n−1. Then s(T)≤

d(2n−d) + 1 4

with equality if and only if T ∼=Dn(d^n−d+1₂ e,b^n−d+1₂ c).

Proof. By Lemma2.5, we haves(D_n(d^n−d+1₂ e,b^n−d+1₂ c)) =b^d(2n−d)+1₄ c.

LetT be a tree of ordernwith diameterdsuch that its minimum status is as large as possible. By the value ofs(Dn(d^n−d+1₂ e,b^n−d+1₂ c)), we only need to show thatT ∼=Dn(d^n−d+1₂ e,b^n−d+1₂ c).

It is trivial ifd= 2, n−1.

Suppose that 3≤d≤n−2. Letαbe the number of quasi-pendant vertices inT. Then there are at least two pendant vertices with different neighbors,i.e., α≥2.

We claim thatα= 2. Otherwise,α≥3. LetP =v0. . . vdbe a diametral path ofT. Asv0 andvdare pendant vertices,v1andv_d−1are quasi-pendant vertices inT. Sinceα≥3,δT(vi)≥3 for someiwith 2≤i≤d−2. For z∈N_T(v_i), letT_zbe the component ofT−v_icontainingz. Letwbe a neighbor ofv_ioutsideP such that the order ofT_wis maximum among the components ofT−viexceptT_vi−1 andT_vi+1. Letρ= max{dT(v_i, x) :x∈V(T_w)}.

Asd=dT(v0, vd) =dT(vi, v0) +dT(vi, vd), we haveρ≤dT(vi, v0) =iandρ≤dT(vi, vd) =d−i. Assume that

|V(Tvi−1)| ≥ |V(Tv_i+1)|. Suppose thatρ < d−i. LetT⁰ =T−viw+vi+1w. ThenP is still a diametral path of T⁰,i.e.,T⁰ is a tree of ordernwith diameter d. By Lemma2.3,s(T⁰)> s(T), a contradiction. Thusρ=d−i.

Letxbe a vertex in the median ofT.

Suppose that|V(T_vi−1)| ≥ ⁿ₂. LetT⁰be the tree obtained fromT−viw by deleting all edges inT_wand adding all edges in {vd−1z : z ∈V(Tw)}. By Lemma 2.1, xmay be chosen in V(Tvi−1), and xis also in the median ofT⁰. As we go fromT toT⁰, the distance betweenxand any vertex different from wis increased or remains unchanged. Thuss(T⁰)−s(T) =sT⁰(x)−sT(x)≥dT⁰(x, w)−dT(x, w) =d−i−1>0, implying thats(T⁰)> s(T), a contradiction. It follows that |V(T_vi−1)|< ⁿ₂. Similarly, if |V(T_w)| ≥ ⁿ₂, then, as above, we may form a tree T⁰ from T−vivi+1 by deleting all edges inTv_i+1 and adding all edges in {v1z:z∈V(Tv_i+1)}, andx∈V(Tw) is in the median ofT and T⁰, such thats(T⁰)−s(T)≥dT⁰(x, vi+1)−dT(x, vi+1) =i−1>0, a contradiction.

So|V(Tw)|< ⁿ₂. By Lemma2.1,x=vi.

Leta=|V(T_vi−1)|andc=|V(T_w)|. LetV₀=V(T_w) ifa+c≤ bⁿ₂cand letV₀be a subset ofV(T_w) consisting of bⁿ₂c −avertices otherwise. Let T⁰⁰ be the tree obtained from T −v_iw by deleting all the edges in T_w and adding all edges in{v1z:z∈V0} ∪ {vd−1z:z∈V(Tw)\V0}. Thenviis also in the median ofT⁰⁰by Lemma2.1.

This is true fora+c >bⁿ₂cbecause froma+c+|V(Tv_i+1)| ≤n−1 we have|V(Tv_i+1)|+a+c− bⁿ₂c ≤ ⁿ₂. As we go from T to T⁰⁰, the distance between vi and any vertex different from w is increased or remains unchanged. Thuss(T⁰⁰)−s(T) =s_T⁰⁰(v_i)−s_T(v_i)≥d_T⁰⁰(v_i, w)−d_T(v_i, w) = min{i, d−i} −1 =d−i−1>0, implying thats(T⁰⁰)> s(T), a contradiction. Now we have proved thatα= 2. That is,T =Dn(p, q) for some p,qwithp+q=n−d+ 1. By Lemma 2.4, we haveT ∼=Dn(d^n−d+1₂ e,b^n−d+1₂ c).

For 3 ≤ d ≤ n−1, let F = D_n(d^n−d+1₂ e,b^n−d+1₂ c) and h(n, d) = s(F). Let xy be the edge of F with δF(x) ≥ 2 and δF(y) = 1 +b^n−d+1₂ c. Then Fxy ∼= Dn(d^{n−(d−1)+1}₂ e,b^{n−(d−1)+1}₂ c). By Lemma 2.2, we have h(n, d)> h(n, d−1).

Let P_n,i be the tree obtained from the path P_n−1 =v₁. . . v_n−1 by attaching a pendant edge at vertex v_i, where 2≤i≤ bⁿ₂c. Note thatPn,2∼=Dn(2,1).

Corollary 3.4. Suppose thatT is a tree of ordern, andTP_n. Thens(T)≤ bⁿ²₄⁻³cwith equality if and only ifT ∼=Dn(2,1). Moreover, ifT Dn(2,1), thens(T)≤ bⁿ²₄⁻⁸cwith equality if and only if T ∼=Dn(2,2),Pn,3. Proof. Let d be the diameter of T. Thend ≤ n−2. If d ≤ n−3, then by Theorem 3.3, s(T) ≤ h(n, d) ≤ h(n, n−3) =bⁿ²₄⁻⁸cwith equalities if and only ifT ∼=Dn(2,2). If d=n−2, thenT ∼=Pn,i with 2≤i≤ bⁿ₂c, ands(T) is equal tobⁿ²₄⁻³cifi= 2,bⁿ²₄⁻⁸cifi= 3, and is at most bⁿ²₄⁻⁸c −1 ifi≥4.

Note that, ifuandv are two nonadjacent vertices in a connected graphG, thens(G)≥s(G+uv).

Suppose thatGis a unicyclic graph of ordern≥4 with diameterd, where 2≤d≤n−2. By Theorem3.3, s(G)≤

d(2n−d) + 1 4

with equality if and only if Gis isomorphic to a graph with diameterd obtained fromD_n(d^n−d+1₂ e,b^n−d+1₂ c) by adding an edge between two pendant vertices.

The eccentricityeG(u) of vertexuin a connected graphGis its distance to a farthest vertex. Ifuis a vertex such thatdG(u, v) =eG(v), then uis called an eccentric vertex ofv. Recall that the radius ofGis defined to be the minimum eccentricities of all vertices of G. Let diam(G) and r(G) be the diameter and radius of G, respectively. A vertex v is central if e_G(v) = r(G). Buckley and Lewinter [8] characterized graphs that have diameter-preserving spanning trees. They showed that a connected graphGhas a diameter-preserving spanning tree if either

(1) diam(G) = 2r(G), or

(2) diam(G) = 2r(G) andGcontains a pair of adjacent central verticesxandythat have no common eccentric vertex.

LetGbe a connected graph with diameter-preserving spanning trees, and letdbe the diameter ofG, where 2 ≤d≤n−1. Then s(G) ≤s(T) for a spanning tree T of diameter d ofG. Thus, by Theorem 3.3, we have s(G)≤s(T)≤ b^d(2n−d)+1₄ cwith equality ifG∼=Dn(d^n−d+1₂ e,b^n−d+1₂ c).

4. Minimum status and number of pendant vertices

In this section, we give sharp lower and upper bounds for the minimum status of a tree with fixed number of pendant vertices, and characterize the unique trees achieving these bounds.

For 2≤p≤n−1, letP(n, p) be the set of trees of ordernwithppendant vertices. Particularly,P(n,2) ={Pn} andP(n, n−1) ={Sn}.

For 3≤p≤n−1, let Sn,p be the tree withppendant paths of almost equal lengths (i.e.,n−1−pbⁿ⁻¹_p c pendant paths of length bⁿ⁻¹_p c+ 1 and p+pbⁿ⁻¹_p c −(n−1) pendant paths of length bⁿ⁻¹_p c) at a common vertex. Particularly,S_n,n−1=S_n. LetS_n,2=P_n.

Theorem 4.1. Suppose that T ∈P(n, p), where2≤p≤n−1. Then

s(T)≥n−1−pbⁿ⁻¹_p c²−(2n−p−2)bⁿ⁻¹_p c 2

with equality if and only if T ∼=Sn,p.

Proof. It is trivial forp= 2, n−1. Suppose that 3≤p≤n−2.

Let xbe the vertex in S_n,p of degree p. By Lemma 2.1, x is in the median of S_n,p. By direct calculation, we have

s(Sn,p) =p

bⁿ⁻¹_p c

X

j=1

j+

n−1−p n−1

p

n−1 p

+ 1

=p·(1 +bⁿ⁻¹_p c)bⁿ⁻¹_p c

2 + (n−1−p) n−1

p

−p n−1

p ²

+n−1

=n−1−pbⁿ⁻¹_p c²−(2n−p−2)bⁿ⁻¹_p c

2 ·

Let T be a tree inP(n, p) such that its minimum status is as small as possible. By the value of s(S_n,p), it suffices to show thatT ∼=Sn,p.

Suppose that there are at least two vertices with degree at least three inT. Then we may choose two vertices, sayw1 and w2, inT with degree at least three, such that dT(w1, w2) is as small as possible. Let P =v0. . . v`

be the unique path connectingw₁andw₂, wherev₀=w₁ andv<sub>`</sub>=w<sub>2</sub>. If the length ofP is at least two, then any internal vertex of P has degree two inT. LetT<sub>w1</sub> be the component of T−v<sub>1</sub> containingw<sub>1</sub> andT<sub>w2</sub> the component ofT −v`−1 containingw2.

LetNT(w1)\ {v1}={u1, . . . , us}. Fori= 1, . . . , s, letTu_i be the component ofT−w1containingui. Assume that|V(Tu₁)| ≤ · · · ≤ |V(Tu_s)|. Denote a=Ps

i=2|V(Tu_i)|.

Assume that |V(T_w1)| ≤ |V(T_w2)|. Let T⁰ = T − {w₁u_i : i = 2, . . . , s}+{w₂u_i : i = 2, . . . , s}. Note that the pendant vertices ofT⁰ are just pendant vertices of T, i.e., T⁰ ∈P(n, p). Letxbe a vertex in the median of T. Then x∈V(Tw₂)∪n

v_b`

2c, . . . , v_`−1o

, which follows from Lemma2.1and the fact that |V(Tw₂)|+d<sup>`</sup><sub>2</sub>e ≥ <sup>n</sup><sub>2</sub> if`≥2, and follows from Lemma2.1if`= 1.

Suppose thatx∈V(Tw₂). By Lemma 2.1,xis also in the median ofT⁰. As we go fromT toT⁰, the distance betweenxand a vertex of ∪^s_i=2V(Tu_i) is decreased bydT(w1, w2), and the distance betweenxand any other vertex remains unchanged. Thus

s(T⁰)−s(T) =s_T⁰(x)−s_T(x) =−d_T(w₁, w₂)·a <0,

implying thats(T⁰)< s(T), a contradiction. It follows that x=v_i for some iwithb₂<sup>`</sup>c ≤i≤`−1.

Note thats≥2 and|V(Tw1)|= 1 +Ps

j=1|V(Tuj)| ≥ |V(Tu1)|+s. Thus, we have |V(Tu1)| ≤ |V(Tw2)| −2, implying that|V(Tu₁)| − |V(Tw₂)|+ 1<0.

Suppose that a ≤ `−i−1, i.e., i+a ≤`−1. Then ` ≥ 2. By Lemma 2.1, vi+a is in the median of T⁰. Note that

dT⁰(vi+a, z)−dT(vi, z) =







a ifz∈ {v0, . . . , v_i−1} ∪V(Tu₁)

−a ifz∈V(Tw₂)∪ {vi+a+1, . . . , v_`−1}

`−2i−a ifz∈V(T_u2)∪. . .∪V(T_us) and

i+a−1

X

j=i+1

dT⁰(vi+a, vj) =

i+a−1

X

j=i+1

dT(vi, vj).

Thus

s(T⁰)−s(T) =sT⁰(vi+a)−sT(vi)

=a(i+|V(T_u1)|)−a(|V(T_w2)|+`−i−a−1) + (`−2i−a)a

=a(i+|V(Tu₁)| − |V(Tw₂)| −`+i+a+ 1 +`−2i−a)

=a(|V(Tu₁)| − |V(Tw₂)|+ 1)<0,

implying thats(T⁰)< s(T), a contradiction. It follows thata≥`−i, where`≥1. Thenw2is in the median of T⁰ by Lemma2.1. Note that

d_T⁰(w₂, z)−d_T(v_i, z) =







`−i ifz∈ {v0, . . . , vi−1} ∪V(Tu₁)

−(`−i) ifz∈V(Tw₂)\ {w2}

−i ifz∈V(Tu₂)∪. . .∪V(Tu_s) and

`−1

X

j=i+1

dT⁰(w2, vj) =

`−1

X

j=i+1

dT(vi, vj).

Thus

s(T⁰)−s(T) =s_T⁰(w₂)−s_T(v_i)

= (`−i)(i+|V(Tu<sub>1</sub>)|)−(`−i)(|V(Tw₂)| −1)−ai

≤(`−i)(i+|V(Tu₁)| − |V(Tw₂)|+ 1−i)

= (`−i)(|V(T_u1)| − |V(T_w2)|+ 1)<0,

implying that s(T⁰) < s(T), also a contradiction. Therefore, there is exactly one vertex with degree at least three in T. AsT ∈P(n, p), T consists ofp pendant paths at a common vertex. Now by Lemma 2.6, we have

T ∼=Sn,p.

Corollary 4.2. If2≤p≤n−2, thens(S_n,p)> s(S_n,p+1).

Proof. Letxbe the vertex of degree pin Sn,p and xy be an edge of a longest pendant path in Sn,p, where if p= 2,xis any vertex of degree two andxyis in the longer sub-path with a terminal vertexx. By Lemma2.2, s(S_n,p)> s

(S_n,p)_xy

. Note that (S_n,p)_xy ∈P(n, p+ 1). By Theorem4.1,s

(S_n,p)_xy

≥s(S_n,p+1). It follows

thats(Sn,p)> s(Sn,p+1).

Theorem 4.3. Suppose that T ∈P(n, p), where2≤p≤n−1. Then s(T)≤

n²−p²+ 2p 4

with equality if and only if T ∼=Dn(d^p₂e,b^p₂c).

Proof. By Lemma2.5, we haves(Dn(d^p₂e,b^p₂c)) =b^n2−p₄^2+2pc.

LetT be a tree inP(n, p) such that its minimum status is as large as possible. From the value of the status of D_n(d^p₂e,b^p₂c), it suffices to show thatT ∼=D_n(d^p₂e,b^p₂c). It is trivial forp= 2, n−1. Suppose that 3≤p≤n−2.

Letαbe the number of quasi-pendant vertices inT. Since the diameter ofT is at least three, we haveα≥2.

Suppose thatα≥3. Then there are at least three components, sayTv, Tw andTz, inT−ufor some vertexu, and at least two of them are not nontrivial, wherev, w, z ∈NT(u). Assume that|V(Tv)| ≥ |V(Tw)| ≥ |V(Tz)|.

Then T_v andT_w are nontrivial. Let u₁ be a quasi-pendant vertex of T_w. ThenT⁰ =T−uz+u₁z is a tree in P(n, p). By Lemma2.3,s(T⁰)> s(T), a contradiction. Therefore, we haveα= 2. That is,T ∼=D_n(`<sub>1</sub>, `₂), where

`1≥`2≥1 and`1+`2=p. By Lemma2.4, we haveT ∼=Dn(d^p₂e,b^p₂c).

We note that Theorem 4.3 follows also from Theorem 3.3. Suppose that T ∈ P(n, p), where 2 ≤ p ≤ n−1. Let d be the diameter ofT. As a diametral path contains exactly two pendant vertices, we have d≤ n−p+ 1. By Theorem 3.3, s(T) ≤ h(n, d) ≤ h(n, n−p+ 1) = b^n2−p₄^2+2pc with equalities if and only if T ∼=Dn(d^n−d+1₂ e,b^n−d+1₂ c) withd=n−p+ 1,i.e., T ∼=Dn(d^p₂e,b^p₂c).

5. Minimum status and number of odd vertices

In this section, we give sharp lower and upper bounds for the minimum status of a tree with fixed number of odd vertices, and characterize the unique trees achieving these bounds.

For integersnandkwith 1≤k≤ bⁿ₂c, letO(n, k) be the set of trees of ordernwith 2kodd vertices [23].

Theorem 5.1. Suppose that T ∈O(n, k), where1≤k≤ bⁿ₂c. Then s(T)≥n−1−k

n−1 2k

2

+ (n−k−1) n−1

2k

with equality if and only if T ∼=S_n,2k, where ifk=ⁿ₂, thenS_n,n=S_n.

Proof. The casek= 1 is trivial as in this case,T ∼=P_n∼=S_n,2. The casek=bⁿ₂cis also trivial asS_n∼=S_n,2bⁿ

2c

is the unique tree of ordernwith smallest minimum statusn−1.

Suppose that 2≤k <bⁿ₂c,i.e., 4≤2k≤n−2. Letpbe the number of pendant vertices ofT. Thenp≤2k.

By Theorem4.1and Corollary4.2, we have

s(T)≥s(S_n,p)≥s(S_n,2k) =n−1−k n−1

2k ²

+ (n−k−1) n−1

2k

with equalities if and only ifT ∼=Sn,p andp= 2k,i.e.,T ∼=Sn,2k. Theorem 5.2. Suppose that T ∈O(n, k), where1≤k≤ bⁿ₂c. Then

s(T)≤







j_(n+1−k)2

4

k +j

2nk−3k²+6k−2n 4

k

ifkis odd j_(n+1−k)2

4

k +j

2nk−3k²+6k−2n 4

k−1 ifkis even with equality if and only if T ∼=Cn(d^k−1₂ e,b^k−1₂ c).

Proof. Let a = d^k−1₂ e, b = b^k−1₂ c and c = d^n+1−k₂ e. Then a + b = k −1 and a = b, b + 1. Let H =C_n(a, b), whose vertices are labelled as before. Let x=v_c. Thenxis in the median of H by Lemma 2.1.

LetU ={ui:i= 1, . . . , n+ 1−k}andW ={vi:i= 2, . . . , a+ 1, n+ 1−k−b, . . . , n−k}. By direct calculation, we have

X

u∈U

d(x, u) =s(Pn+1−k) =

(n+ 1−k)² 4

and

X

u∈W

d(x, u) =

a

X

i=1

(c−1−a+i) +

b

X

i=1

(n+ 1−k−b−c+i)

=a(c−1−a) +a(a+ 1)

2 + (n+ 1−k−b−c)b+b(b+ 1) 2

=

(a(n−k−2a) +a(a+ 1) ifa=b (a−1)(n+ 1−k−2a) +c−1−a−a² ifa=b+ 1

=

((k−1)(2n−3k+3)

4 ifkis odd,

(k−2)(2n+2−3k)

4 +d^n+1−k₂ e −1 ifkis even.

Thus

s(H) =X

u∈U

d(x, u) + X

u∈W

d(x, u)

=







j_(n+1−k)2

4

k +j

2nk−3k²+6k−2n 4

k

ifkis odd, j(n+1−k)²

4

k +j

2nk−3k²+6k−2n 4

k−1 ifkis even.

Let T be a tree in O(n, k) such that its minimum status is as large as possible. By the value of s(Cn(d^k−1₂ e,b^k−1₂ c)), it suffices to show thatT ∼=Cn(d^k−1₂ e,b^k−1₂ c).

Ifk= 1, then it is obvious thatT ∼=Pn∼=Cn(0,0).

Suppose that k ≥ 2. We claim that the maximum odd degree of T is 3. Otherwise, δ_T(u) = 2t+ 1 from some u ∈ V(T) and t ≥ 2. Let N_T(u) = {u1, . . . , u_2t+1}. Let T_i be the component of T −u containing u_i,

where 1 ≤ i ≤ 2t+ 1. Assume that |V(T₁)| ≥ |V(T₂)|. Let z be a pendant vertex of T in V(T₂), and let T⁰ = T − {uui : 3 ≤ i ≤ 2t}+{zui : 3 ≤ i ≤ 2t}. Note that the degrees of u and z are still odd in T⁰. ThenT⁰∈O(n, k). By Lemma2.3,s(T⁰)> s(T), a contradiction. Thus, the maximum odd degree ofT is 3, as claimed.

Ifk= ⁿ₂, then by Lemma2.7,T ∼=C_n d^k−1₂ e,b^k−1₂ c .

Suppose thatk < ⁿ₂. Then there is at least one even vertex inT. We claim that all even vertices have degree two. Otherwise, δT(w) = 2t for some w ∈ V(T), where t ≥ 2. Let NT(w) = {w1, . . . , w2t}. Let Ti be the component ofT−wcontainingwi, where 1≤i≤2t. Assume that|V(T1)| ≥ |V(T2)|. Letzbe a pendant vertex ofT inV(T2), and letT⁰⁰=T−ww3+zw3. Note that, inT⁰⁰, the degree ofwis odd and the degree ofzis even.

ThenT⁰⁰∈O(n, k). By Lemma2.3,s(T⁰⁰)> s(T), a contradiction. Thus, all even vertices ofT have degree two.

Now, we claim T is a caterpillar. Otherwise, as the maximum degree of T is three, there is a vertex u of degree three inT such thatustill has degree three in the tree obtained fromT by deleting all pendant vertices.

Let NT(u) = {u1, u2, u3}. Then δT(ui) ≥2 for i = 1,2,3. Let Ti be the component of T −ucontaining ui, where i= 1,2,3. LetU be the set of vertices of degree two inT. Suppose that U 6⊆V(T_i) for anyi= 1,2,3, say v₁ ∈ V(T₁) and v₂ ∈ V(T₂) for v₁, v₂ ∈ U. Assume that|V(T₁)| ≥ |V(T₂)|. Let T^∗ = T −uu₃+v₂u₃. AsδT^∗(u) = 2 andδT^∗(v2) = 3, we haveT^∗∈O(n, k). By Lemma2.3,s(T^∗)> s(T), a contradiction. Therefore, U ⊆ V(Ti) for some i = 1,2,3, say U ⊆ V(T1). Suppose that V(T2) (V(T3), respectively) contains r (t, respectively) vertices of degree three of T. Then T ∼=T1(r+ 1, t+ 1)∈ O(n, k), where r, t ≥1. Assume that r ≥ t. Note that T₁(r+ 2, t) ∈ O(n, k). By Lemma 2.7, s(T₁(r+ 2, t)) > s(T), a contradiction. Thus, T is a caterpillar, as claimed. By Lemma2.8, the set of all n−2k vertices of degree two induces a path, and thus T ∼=Cn(a, b) for someaandb witha+b=k−1. By Lemma2.9, we haveT ∼=Cn(d^k−1₂ e,b^k−1₂ c).

A vertex in a graph is called a branching vertex if its degree is at least three. LetT be a tree of ordernwith kbranching vertices. Letpbe the number of pendant vertices inT. Thenp+ 2(n−p−k) + 3k≤2(n−1),i.e., p≥k+ 2. This implies that 2k+ 2≤k+p≤n, and thus,k≤ⁿ₂ −1.

Corollary 5.3. Suppose that T is a tree of order n with k branching vertices, and 0 ≤ k ≤ ⁿ₂ −1. Then s(T)≤s(Cn(d^k₂e,b^k₂c))with equality if and only if T ∼=Cn(d^k₂e,b^k₂c).

Proof. Ifk= 0, then the result follows from the known fact thatPnis the unique tree of ordernwhose minimum status is maximum [2].

Suppose thatk≥1. LetT be a tree of ordernwithkbranching vertices such that its minimum status is as large as possible.

Let ∆ be the maximum degree ofT. Suppose that ∆≥4. Letu∈V(T) andNT(u) ={u1, . . . , u∆}. LetTi

be the component ofT−ucontainingui, where 1≤i≤∆. Assume that|V(T1)| ≥ |V(T2)|. Letwbe a pendant vertex ofT in V(T2), and letT⁰ =T−uu3+wu3. Then T⁰ is a tree of ordernwith kbranching vertices. By Lemma 2.3, s(T⁰)> s(T), a contradiction. Hence ∆ = 3. Let pbe the number of pendant vertices in T. As p+ 2(n−k−p) + 3k= 2(n−1), we havep=k+ 2, and thusT is a tree of ordernwith 2k+ 2 odd vertices.

By Theorem5.2, we haveT ∼=Cn(d^k₂e,b^k₂c).

6. Minimum status and number of vertices of degree two

For integers nand t with 0≤t ≤n−2, let H(n, t) be the set of trees of ordern with t vertices of degree two. Note thatH(n, n−2) ={Pn},H(n, n−3) =∅, andH(n,0) is the class of series-reduced trees of ordern.

So we only consider trees inH(n, t) with 0≤t≤n−4.

Theorem 6.1. Suppose that T ∈H(n, t), where0≤t≤n−4. Then

s(T)≥n−1−(n−t−1)b_n−t−1ⁿ⁻¹ c²−(n+t−1)b_n−t−1ⁿ⁻¹ c 2

with equality if and only if T ∼=S_n,n−t−1.

Proof. The caset= 0 is trivial as in this caseS_n(∼=S_n,n−1) is the unique tree of ordernwith smallest minimum statusn−1.

Suppose thatt≥1. Letpbe the number of pendant vertices ofT. As there is a vertex with degree at least 3, we havep≤n−t−1. By Theorem4.1and Corollary4.2, we have

s(T)≥s(Sn,p)≥s(S_n,n−t−1) =n−1−(n−t−1)b_n−t−1ⁿ⁻¹ c²−(n+t−1)b_n−t−1ⁿ⁻¹ c 2

with equalities if and only ifT ∼=S_n,p andp=n−t−1,i.e.,T ∼=S_n,n−t−1. Lemma 6.2. Let T be a tree in H(n, t) such thats(T)is maximum, where0≤t≤n−4. Then the maximum degree ofT is at most 4,T is caterpillar, and there are at most two vertices of degree four in T.

Proof. Let ∆ be the maximum degree of T. Suppose that ∆ ≥5. Let u∈ V(T) withδT(u) = ∆(T) and let NT(u) ={u1, . . . , u∆}. For 1≤i≤∆, letTi be the component of T−ucontainingui. Suppose without loss of generality|V(T₁)| ≥ |V(T₂)|. LetT⁰ =T − {uu_i : 4≤i≤∆}+{wu_i : 4≤i≤∆}, wherew is a pendant vertex ofT in V(T₂). Asδ_T⁰(u) = 3, δ_T⁰(w) = ∆−2≥3 and δ_T⁰(v) =δ_T(v) for v∈ V(T)\ {u, w}, we have T⁰∈H(n, t). By Lemma2.3,s(T⁰)> s(T), a contradiction. Therefore ∆≤4.

Suppose that T is not a caterpillar. Then for some vertex uof T, T −u has three nontrivial components.

Note thatδ_T(u) = 3,4 as ∆≤4.

Case 1.δT(u) = 3.

Let NT(u) = {u1, u2, u3}. Let Ti be the component of T −u containing ui for i = 1,2,3. Assume that

|V(T1)| ≥ |V(T2)| ≥ |V(T3)|. Suppose first thatT has a vertexwof degree two inV(T2). LetT⁰=T−uu3+wu3. As δT⁰(u) = 2,δT⁰(w) = 3 and δT⁰(z) = δT(z) forz ∈V(T)\ {u, w}, we have T⁰ ∈ H(n, t). By Lemma 2.3, s(T⁰)> s(T), a contradiction. Therefore,T has no vertex of degree two inV(T₂). Letw₁be a pendant vertex of T inV(T₂) such thatd_T(u, w₁) = max{dT(u, s) :s∈V(T₂)}. LetT⁰ =T− {uu3, w₁w₂}+{uw1, w₂u₃}, where w2is the neighbor of w1. Note thatT⁰∈H(n, t).

Letxbe a vertex in the median ofT. As|V(T1)| ≥ |V(T2)| ≥ |V(T3)|, we havex∈V(T1)∪{u}by Lemma2.1.

Case 1.1.x∈V(T1).

By Lemma 2.1,xis also in the median of T⁰. As we go from T to T⁰, the distance betweenxand a vertex ofV(T₃) is increased byd_T(u, w₂), the distance betweenxandw₁is decreased by d_T(u, w₂), and the distance betweenxand any other vertex remains unchanged. Thus

s(T⁰)−s(T) =s_T⁰(x)−s_T(x)

=dT(u, w2)|V(T3)| −dT(u, w2)

=dT(u, w2)(|V(T3)| −1)>0, implying thats(T⁰)> s(T), a contradiction.

Case 1.2.x=u

Letx⁰ be a vertex in the median ofT⁰. By Lemma2.1, x⁰ lies on the path connectinguandw₂ inT⁰. Note that dT⁰(x⁰, s)−dT(x, s) = dT(u, x⁰) fors ∈ V(T1), dT⁰(x⁰, s)−dT(x, s)≥ −dT(u, x⁰) for s ∈ V(T2)\ {w1}, dT⁰(x⁰, s)−dT(x, s) =dT(w2, x⁰) fors∈V(T3), anddT⁰(x⁰, w1)−dT(x, w1) =−dT(w2, x⁰). Thus

s(T⁰)−s(T) =s_T⁰(x⁰)−s_T(x)

≥dT(u, x⁰)(|V(T1)| − |V(T2)|+ 1) +dT(x⁰, w2)(|V(T3)| −1)

≥dT(u, x⁰) +dT(x⁰, w2)>0, a contradiction.

Case 2.δ_T(u) = 4.

LetNT(u) ={u1, u2, u3, u4}. Let Ti be the component ofT −ucontainingui fori= 1, . . . ,4. Assume that T1,T2andT3are nontrivial, and|V(T1)| ≥ |V(T2)|. Suppose thatT has a vertex, sayw, of degree two inV(T2).

LetT⁰=T− {uu3, uu4}+{wu3, wu4}. AsδT⁰(u) = 2,δT⁰(w) = 4 andδT⁰(v) =δT(v) for anyv∈V(T)\ {u, w}, we have T⁰ ∈H(n, t). By Lemma 2.3, s(T⁰)> s(T), a contradiction. Thus, T has no vertex of degree two in V(T₂). Then there is a vertexz∈V(T₂) such thatδ_T(z) = 3,4. LetT⁰=T−uu₃+zu₃. ThenT⁰∈H(n, t). By Lemma2.3, s(T⁰)> s(T), a contradiction.

By combining Cases 1 and 2, we conclude thatT is a caterpillar.

Suppose that there are three vertices, sayv1, v2, andv3 inT with degree four. AsT is a caterpillar,v1, v2

and v₃ lie on a diametral path of T, and so one of them, say v₃, lies on the path P connecting the other two verticesv1andv2. LetTi be the component ofT−v3containingvi fori= 1,2. Assume that|V(T1)| ≥ |V(T2)|.

Letzbe a neighbor ofv3 outside the pathP inT. LetT⁰⁰=T−v3z+v2z. ThenT⁰⁰∈H(n, t). By Lemma2.3, s(T⁰⁰)> s(T), a contradiction. Therefore, there are at most two vertices of degree four inT.

Lemma 6.3. LetT ∈H(n, t)such thats(T)is maximum, where0≤t≤n−4. Letki be the number of vertices of degreei inT, where1≤i≤4. Then







k₁=^n−t+3₂ k3=^n−t−5₂ k4= 1

ifn−t is odd,

and







k₁=^n−t+2₂ k3=^n−t−2₂ k4= 0

or







k₁=^n−t+4₂ k3=^n−t−8₂ k4= 2

ifn−tis even.

Proof. Trivially, k2 = t. By Lemma 6.1, the maximum degree of T is at most 4 and k4 ≤ 2. Now the result follows from the facts thatk1+k2+k3+k4=nandk1+ 2k2+ 3k3+ 4k4= 2(n−1).

Lemma 6.4. Let T ∈H(n, t) such thats(T)is maximum, where 0 ≤t≤n−4. Ifu∈V(T) withδT(u) = 4, thenT−uhas at most one nontrivial component.

Proof. From Lemma6.1,T is a caterpillar with at most two vertices of degree four. LetNT(u) ={u1, u2, u3, u4}.

Suppose that T−uhas two nontrivial components, sayT1 andT2, withui ∈V(Ti) for i= 1,2. Assume that

|V(T1)| ≥ |V(T2)|. IfT has a vertexwof degree two inV(T2), then by settingT⁰=T−{uu3, uu4}+{wu3, wu4}, we haveT⁰∈H(n, t), and by Lemma2.3,s(T⁰)> s(T), which is a contradiction. ThusT has no vertex of degree two inV(T₂). Thenδ_T(w) = 3,4 for somew∈V(T₂). LetT⁰=T−uu3+wu₃. ThenT⁰∈H(n, t). By Lemma2.3,

s(T⁰)> s(T), a contradiction.

For nonnegative integersa, band positive integernwith 2(a+b) + 5≤n, letRn(a, b) be the tree of ordern obtained from a path u1. . . u_n−a−b by attaching pendant vertex v1 to u2, and then attaching pendant vertex v_i to u_i for each 2≤i≤a+ 2 and eachi withn−a−2b−2≤i≤n−a−b−3. The structure ofR_n(a, b) is shown in Figure2.

Lemma 6.5. Fora≥max{b,1} and2(a+b) + 5< n, we have

s(R_n(a−1, b+ 1))> s(R_n(a, b)).

Proof. LetR_n=R_n(a, b). LetR⁰_n =R_n−u_a+2v_a+2+u_{n−a−2b−3}v_a+2. ThenR_n⁰ ∼=R_n(a−1, b+ 1).