Spanning Trees of Connected $K_{1,t}$-free Graphs Whose Stems Have a Few Leaves

Abstract

Let T be a tree, a vertex of degree one is called a leaf. The set of leaves of T is denoted by ${\mathrm{Leaf}}(T)$. The subtree $T-{\mathrm{Leaf}}(T)$ of T is called the stem of T and denoted by ${\mathrm{Stem}}(T)$. In this paper, we give a sharp sufficient condition to show that a $K_{1,t}$-free graph has a spanning tree whose stem has a few leaves. By applying the main result, we give improvements of previous related results.

1 Introduction

In this paper, we only consider finite simple graphs. Let G be a graph with vertex set V(G) and edge set E(G). For any vertex $v\in V(G)$, we use $N_G(v)$ and $\deg _G(v)$ to denote the set of neighbors of v and the degree of v in G, respectively. For any $X\subseteq V(G)$, we denote by |X| the cardinality of X. We define $G-uv$ to be the graph obtained from G by deleting the edge $uv\in E(G)$, and $G+uv$ to be the graph obtained from G by adding an edge uv between two non-adjacent vertices u and v of G. For two vertices u and v of G, the distance between u and v in G is denoted by $d_{G}(u, v)$.

For an integer $m\geqslant 2,$ let $\alpha ^{m}(G)$ denote the number defined by

$$\begin{aligned} \alpha ^{m}(G)=\max \{ |S|:S\subseteq V(G),d_{G}(x,y)\geqslant m\,\ \text {for all distinct vertices }\,x,y\in S\}. \end{aligned}$$

For an integer $p\geqslant 2$, we define

$$\begin{aligned} \sigma _p^{m}(G)= & {} \min \left\{ \sum _{a\in S}\deg _{G}(a):S\subseteq V(G),|S|=p, \right. \\&\left. d_{G}(x,y)\geqslant m\ \text {for all distinct vertices}\;x,y\in S\right\} . \end{aligned}$$

For convenience, we define $\sigma ^{m}_{p}(G)=+\infty$ if $\alpha ^{m}(G)<p$. We note that $\alpha ^{2}(G)$ is often written $\alpha (G)$, which is the independence number of G,  and $\sigma _p^{2}(G)$ is often written $\sigma _{p}(G)$, which is the minimum degree sum of p independent vertices.

Let T be a tree, a vertex of degree one is called a leaf. The set of leaves of T is denoted by ${\mathrm{Leaf}}(T)$. The subtree $T-{\mathrm{Leaf}}(T)$ of T is called the ${\mathrm{stem}}$ of T and is denoted by ${\mathrm{Stem}}(T)$. A tree having at most l leaves is called an l-ended tree and a stem having at most l leaves is called an l-ended stem. There are several well-known conditions (such as the independence number conditions and the degree sum conditions) ensuring that a graph G contains a spanning tree with a bounded number of leaves or branch vertices (see the survey paper [5] and the references cited therein for details). Win [7] obtained a sufficient condition related to the independence number for k-connected graphs, which confirms a conjecture of Las Vergnas [4]. Broersma and Tuinstra [1] gave a degree sum condition for a connected graph to contain a spanning tree with at most l leaves.

Theorem 1.1

(Win [7]) Let $k\ge 1$ and $l\ge 2$ be integers and let G be a k-connected graph. If $\alpha (G)\le k+l-1$, then G has a spanning l-ended tree.

Theorem 1.2

(Broerma and Tuinstra [1]) Let G be a connected graph and let $l\ge 2$ be an integer. If $\sigma _2(G)\ge |G|-l+1$, then G has a spanning l-ended tree.

Recently, many researches are studied on spanning trees in connected graphs whose stems have a bounded number of leaves or branch vertices (see [2, 3, 6, 8] for more details). We introduce here some results on spanning trees whose stems have a few leaves.

Theorem 1.3

(Tsugaki and Zhang [6]) Let G be a connected graph and let $l\geqslant 2$ be an integer. If $\sigma _{3}(G)\geqslant |G|-2l+1$, then G has a spanning tree with l-ended stem.

Theorem 1.4

(Kano and Yan [2]) Let G be a connected graph and let $l\geqslant 2$ be an integer. If $\sigma _{l+1}(G)\geqslant |G|-l-1$, then G has a spanning tree with l-ended stem.

Moreover, for a positive integer $t \ge 3$, a graph G is said to be $K_{1,t}$- free graph if it contains no $K_{1,t}$ as an induced subgraph. If $t=3,$ the $K_{1,3}$- free graph is also called the claw-free graph. Kano and Yan also gave the following result.

Theorem 1.5

(Kano and Yan [2]) Let G be a connected claw-free graph and let $l\geqslant 2$ be an integer. If $\sigma _{l+1}(G)\geqslant |G|-2l-1$, then G has a spanning tree with l-ended stem.

On the other hand, if the maximum degree of a graph G is denoted by $\Delta (G)$, then G is nothing but a $K_{1,t}$-free graph for all $t\ge \Delta (G)+1.$ Then, we may generalize the above theorems by studying the $K_{1,t}$-free graph. In this paper, we would like to study on the spanning tree of a $K_{1,t}$-free graph whose stem has a bounded number of leaves. Firstly, we want to prove the following result.

Theorem 1.6

For a positive integer $t \ge 3,$ let G be a connected $K_{1,t}$-free graph and let $l\ge 2\ (l\not = t-2)$ be an integer. If

$$\begin{aligned} \sigma ^{4}_{l+1}(G)\geqslant |G|-\left\lfloor \dfrac{l(t-1)}{t-2}\right\rfloor -1, \end{aligned}$$

(1.1)

then G has a spanning tree with l-ended stem. Here, the notation $\lfloor r\rfloor$ stands for the biggest integer not exceed the real number r.

We also note that the reason why we consider $\sigma ^{4}_{l+1}(G)$ is based on the following theorem of Kano and Yan. They proved that if a connected graph G satisfies that $|S|\le l$ for every $S\subseteq V(G)$ such that $d_{G}(x,y)\geqslant 4\ \text {for all distinct vertices}\;x,y\in S$, then G has a spanning tree with l-ended stem.

Theorem 1.7

(Kano and Yan [2]) Let G be a connected graph and let $l\geqslant 2$ be an integer. If $\alpha ^{4}(G)\le l$, then G has a spanning tree with l-ended stem.

By using Theorem 1.6 when $t=3,$ we have Theorem 1.5. Moreover, Theorem 1.6 is an improvement of Theorem 1.4 when we consider the positive integer t big enough.

We now construct two examples to show that the conditions of Theorem 1.6 are sharp.

Let t, k, m be integers such that $t\ge 3, k\ge 2, m \ge 1$ and let $l=k(t-2)$. Let D be a complete graph with $k+1$ vertices $u_1, u_2, \ldots , u_{k+1}$. Let $D_1, D_2, \ldots , D_{k(t-2)+1}$ be copies of the graph $K_m$. Let $v_1, v_2, \ldots , v_{k(t-2)+1}$ be vertices which are not in $V(D)\cup V(D_1)\cup V(D_2)\cup \cdots \cup V(D_{k(t-2)+1})$. For each $i\in \{1,2,\ldots ,k\},$ join $u_i$ to all vertices of the graphs $D_{(i-1)(t-2)+1}, D_{(i-1)(t-2)+2}, \ldots , D_{i(t-2)}$ and join $u_{k+1}$ to all vertices of the graph $D_{k(t-2)+1}$. Join $v_j$ to all vertices of $D_j$ for all $j\in \{1,2,\ldots , k(t-2)+1\}$. Then, the resulting graph G is a $K_{1,t}$-free graph (see Fig. 1).

Fig. 1

Graph G

Moreover, we have $|G|=k+1+(k(t-2)+1)(m+1)$ and

$$\begin{aligned} \sigma ^4_{l+1}(G)&=\sigma ^4_{k(t-2)+1}(G)=\sum \limits _{i=1}^{k(t-2)+1}\deg _G(v_i)\\&=(k(t-2)+1)m=|G|-k(t-1)-2=|G|-\left\lfloor \dfrac{l(t-1)}{t-2}\right\rfloor -2. \end{aligned}$$

But G has no spanning tree with l-ended stem. Hence, the condition (1.1) is sharp.

On the other hand, when $l=t-2,$ let $E_i(1\le i \le l+1)$ be connected graphs. For each $i \in \{1, 2, \ldots ,l+1\}$, denote by $K_i$ the set of vertex v in $V(E_i)$ such that $N_{E_i}(v)=V(E_i)-\{v\}$. Suppose that $K_i\not = \emptyset$ for all indices $1\le i \le l+1.$ Let u be a vertex which is not in $V(E_1)\cup V(E_2)\cup \cdots \cup V(E_{l+1}).$ For each $i\in \{1,2,\ldots ,l+1\},$ join u to all vertices in $V(E_i)-K_i$ and some vertices in $K_i$ excepting at least one vertex in $K_i$. The resulting graph is denoted by M. We only consider the case M is a $K_{1,t}$-free graph. By the definition of M, we may obtain that each subset $S\subseteq V(M)$ such that $|S|=l+1$ and $d_{M}(x,y)\geqslant 4$ must contain one and only one vertex in $K_i$ for all $1\le i \le l+1.$ Hence,

$$\begin{aligned} \sigma ^4_{l+1}(M)&=\sum \limits _{i=1}^{l+1}\deg _{M}(\text {one vertex}\ v_i \in K_i\ \text {such that}\ v_iu \not \in E(M))=\sum \limits _{i=1}^{l+1}(|E_i|-1)\\&=|M|-(l+2)=|M|-\left\lfloor \dfrac{l(t-1)}{t-2}\right\rfloor -1. \end{aligned}$$

But M has no spanning tree with l-ended stem.

A natural question is whether we can find all graphs so that the claim of Theorem 1.6 is not correct in the case $l=t-2$. We will give an answer for this question. In particular, we state the following theorem which is an improvement of Theorem 1.6.

Theorem 1.8

For a positive integer $t \ge 3,$ let G be a connected $K_{1,t}$-free graph and let $l\ge 2$ be an integer. If

$$\begin{aligned} \sigma ^{4}_{l+1}(G)\geqslant |G|-\left\lfloor \dfrac{l(t-1)}{t-2}\right\rfloor -1, \end{aligned}$$

then G has a spanning tree with l-ended stem except for the case $l=t-2$ and G is isomorphic to a graph M.

Remark 1.9

Since $\sigma ^{4}_{l+1}(M)=|M|-\lfloor \dfrac{l(t-1)}{t-2}\rfloor -1$, it follows from Theorem 1.8 that if G is a connected $K_{1,t}$-free graph such that $\sigma ^{4}_{l+1}(G)\geqslant |G|-\lfloor \dfrac{l(t-1)}{t-2}\rfloor$, then G has a spanning tree with l-ended stem.

2 Proof of Theorem 1.8

We prove the theorem by contradiction. Suppose to the contrary that G contains no spanning tree with l-ended stem. Let T be a tree such that $|{\mathrm{Leaf}}({\mathrm{Stem}}(T))| \le l.$ Choose a tree T so that

• (T1) |T| is as large as possible, and

• (T2) $|{\mathrm{Leaf}}(T)|$ is as large as possible, subject to (T1).

By the maximality of T, we have the following three claims. We note that their proofs are written in [2, 8], but for the convenience of readers, we reintroduce them here.

Claim 2.1

For every $v\in V(G)-V(T), N_{G}(v)\subseteq {\mathrm{Leaf}}(T)\cup (V(G)-V(T))$.

Because G is connected and T is not a spanning tree of G and by Claim 2.1, there exist two vertices $v_1\in V(G)-V(T)$ and $v_2\in {\mathrm{Leaf}}(T)$ such that $v_1v_2\in E(G)$. We may obtain that ${\mathrm{Stem}}(T)$ has exactly l leaves. Indeed, otherwise we consider the tree $T'=T+v_1v_2.$ Then, $T'$ has l-ended stem and $|T'|>|T|$, this implies a contradiction with the maximality of T. Let $\{x_1, x_2, \ldots , x_l\}$ be the leaf set of ${\mathrm{Stem}}(T)$.

Claim 2.2

For every $x_{i}(1 \leqslant i \leqslant l)$, there exists a vertex $y_{i}\in {\mathrm{Leaf}}(T)$ such that $y_{i}$ is adjacent to $x_{i}$ and $N_{G}(y_{i})\subseteq {\mathrm{Leaf}}(T)\cup \left\{ x_{i}\right\}$.

Proof

By the definition of ${\mathrm{Stem}}(T),$ it is easy to see that for each leaf $x\in {\mathrm{Leaf}}({\mathrm{Stem}}(T)),$ there exists at least a vertex y in ${\mathrm{Leaf}}(T)$ such that y is adjacent to x. Suppose that for some $1\leqslant i\leqslant l$, each leave $y_{i_j}$ of T adjacent to $x_{i},$ is also adjacent to a vertex $z_{i_j}\in (V({\mathrm{Stem}}(T))-\left\{ x_{i}\right\} )$. Then, we consider $T'$ to be the tree obtained from T by removing the edge $y_{i_j}x_{i}$ and adding the edge $y_{i_{j}}z_{i_{j}}$. Hence, $T'$ is a tree with l-end stem such that $|T'|=|T|$ and ${\mathrm{Leaf}}(T')={\mathrm{Leaf}}(T)\cup \{x_i\}$, which contradicts the condition (T2). Therefore, for each $x_i$, there exists a leaf $y_i\in N_G(x_i)$ such that $N_G(y_i)\cap (V({\mathrm{Stem}}(T))-\{x_i\})=\emptyset$. By the maximality of T we also see that $N_{G}(y_i)\cap (V(G)-V(T))=\emptyset .$ The claim holds. $\square$

Claim 2.3

For any two distinct vertices $y,z\in \left\{ v_1,y_{1},y_{2},\dots ,y_{l}\right\} ,d_{G}\left( y,z\right) \geqslant 4$.

Proof

First, we show that $d_{G}\left( v_1,y_{i}\right) \geqslant 4$ for every $1\leqslant i\leqslant l$. Let $P_i$ be the shortest path connecting $v_1$ and $y_{i}$ in G. If all the vertices of $P_i$ between $v_1$ and $y_{i}$ are contained in ${\mathrm{Leaf}}(T)\cup (V(G)-V(T))\cup \left\{ x_{i}\right\} ,$ add $P_{i}$ to T (if $P_{i}$ passes through $x_{i}$, we just add the segment of $P_{i}$ between $v_1$ and $x_{i})$ and remove the edges of T joining $V(P_{i})\cap {\mathrm{Leaf}}(T)$ to $V({\mathrm{Stem}}(T))$ except the edge $y_{i}x_{i}.$ The resulting tree is denoted by $T'.$ Then, $T'$ is a tree in G with l-ended stem and $|T'|>|T|$, which contradicts to the maximality of T. So we conclude that for each $1\le j \le l,$ if P is a shortest path connecting $v_{1}$ and $y_{j}$ in G, then $V(P)\cap (V({\mathrm{Stem}}(T))-\{x_j\})\not = \emptyset$.

Hence, we may choose the vertex s in $V({\mathrm{Stem}}(T))\cap V(P_i)$ such that it is nearest to $v_1$ in $P_i$. If $s=x_j$ for some $1\le j\le l$, then we add the segment of $P_{i}$ between $v_1$ and $x_j$ (which is denoted by Q) to T and remove the edges of T joining $V(Q)\cap {\mathrm{Leaf}}(T)$ to $V({\mathrm{Stem}}(T))$ except $x_jy_j$. Hence, the resulting tree has l-ended stem and its order is greater then |T|, contradicting the maximality of T. Thus, $s\in V({\mathrm{Stem}}(T))-\{x_1,\ldots ,x_l\}$. By Claims 2.1 and 2.2, we have $d_G(v_1,s)\ge 2$, $d_G(s,y_i)\ge 2.$ Therefore, we conclude that $d_G(v_1,y_i)=|P_i|-1 \ge d_G(v_1,s)+d_G(s,y_i)\ge 4$.

Next, we show that $d_{G}(y_{i},y_{j})\geqslant 4$ for all $1\leqslant i<j\leqslant l$. Let $P_{{\textit{ij}}}$ be the shortest path connecting $y_{i}$ and $y_{j}$ in G. We note that if $P_{{\textit{ij}}}$ passes through $x_{i}$ (or $x_{j}$), then $y_{i}x_{i}\in E(P_{{\textit{ij}}})$ (or $y_{j}x_{j}\in E(P_{{\textit{ij}}})$), respectively. We consider the following two cases.

Case 1. All vertices of $P_{{\textit{ij}}}$ between $y_{i}$ and $y_{j}$ are contained in ${\mathrm{Leaf}}(T)\cup (V(G)-V(T))\cup \left\{ x_{i},x_{j}\right\}$. Then, we add $P_{{\textit{ij}}}$ to T and remove the edges of T joining $V(P_{ij })\cap {\mathrm{Leaf}}(T)$ to $V({\mathrm{Stem}}(T))$ except the edges $y_{i}x_{i}$ and $y_{j}x_{j}$. Hence, the resulting graph has exactly a cycle, which contains an edge e of ${\mathrm{Stem}}(T)$ incident with a branch vertex in ${\mathrm{Stem}}(T)$. By removing the edge e and by adding an edge $v_1v_2$, we have a resulting tree $T'$ with l-ended stem of order greater than |T|, which contradicts the maximality of T. So we conclude that for every $1\le i < j \le l,$ if P is a shortest path connecting $y_{i}$ and $y_{j}$ in G, then $V(P)\cap (V({\mathrm{Stem}}(T))-\{x_i, x_j\})\not = \emptyset .$

Case 2. There exists a vertex $s \in V(P_{{\textit{ij}}})\cap (V({\mathrm{Stem}}(T))-\{x_i,x_j\}).$ Then, $d_G(y_i,s)\ge 2, d_G(s,y_j)\ge 2$ by Claim 2.2. This concludes that $d_G(y_i,y_j)=|P_{{\textit{ij}}}| -1\ge d_G(y_i,s)+d_G(s,y_j)\ge 4$.

So the assertion of the claim holds. $\square$

Denote $Y=\left\{ y_{1},y_2,\ldots ,y_{l}\right\}$. By Claims 2.1–2.3, we have

$$\begin{aligned} N_{G}(v_1)&\subseteq (V(G)-V(T)-\left\{ v_1\right\} )\cup (N_{G}(v_1)\cap ({\mathrm{Leaf}}(T)-Y)),\\ \displaystyle \bigcup _{i=1}^{l}N_{G}(y_{i})&\subseteq ({\mathrm{Leaf}}(T)-Y-N_{G}(v_1))\cup \left\{ x_{1},\dots ,x_{l}\right\} . \end{aligned}$$

Hence by setting $q=|N_{G}(v_1)\cap ({\mathrm{Leaf}}(T)-Y)|$, we obtain

$$\begin{aligned} \begin{aligned} \deg _{G}(v_1)+\displaystyle \sum \limits _{i=1}^{l}\deg _{G}(y_{i})&\leqslant (|G|-|T|-1+q)+(|{\mathrm{Leaf}}(T)|-l-q)+l\\&=|G|-|{\mathrm{Stem}}(T)|-1. \end{aligned} \end{aligned}$$

On the other hand, by the assumption of Theorem 1.8, and by Claim 2.3, we have

$$\begin{aligned} |G|-\left\lfloor \dfrac{l(t-1)}{t-2}\right\rfloor -1\le \sigma ^4_{l+1}(G)\le \deg _{G}(v_1)+\displaystyle \sum \limits _{i=1}^{l}\deg _{G}(y_{i}). \end{aligned}$$

Therefore, we obtain $|{\mathrm{Stem}}(T)|\le \lfloor \dfrac{l(t-1)}{t-2}\rfloor$. By combining with $|{\mathrm{Leaf}}({\mathrm{Stem}}(T))|=l$, we conclude that

$$\begin{aligned} |{\mathrm{Stem}}({\mathrm{Stem}}(T))|\le \left\lfloor \dfrac{l}{t-2}\right\rfloor . \end{aligned}$$

(2.1)

Claim 2.4

$N_G(v_2)\cap \{x_1,x_2,\ldots ,x_l\}=\emptyset$.

Proof

Suppose the assertion of the claim is false. Then, there exists some $i \in \{1,\ldots ,l\}$ such that $v_2x_i \in E(G).$ Combining with the fact that $v_2v_1 \in E(G)$ and $x_iy_i \in E(G)$ (by Claim 2.2), we obtain that $d_G(v_1, y_i) \le 3.$ This contradicts Claim 2.3. Claim 2.4 is proved. $\square$

Now, we complete the Proof of Theorem 1.8 by considering the following two steps.

Step 1. $|{\mathrm{Stem}}({\mathrm{Stem}}(T))|=1$.

We assume that ${\mathrm{Stem}}({\mathrm{Stem}}(T))=\{u\}$. By $t\ge 3$ and $|{\mathrm{Stem}}({\mathrm{Stem}}(T))|\le \lfloor \dfrac{l}{t-2}\rfloor$, we obtain $l\ge t-2.$ We consider the following two cases.

Case 1. $l \ge t- 1.$

By combining with Claims 2.3 and 2.4, G induced a $K_{1,t}$ subgraph with the vertex set $\{u, x_1, x_2,\ldots ,x_{t-1},v_2\}$, this gives a contradiction.

Case 2.$l= t-2$. In this case, we will show that G is isomorphic to a graph M.

For each $X\subseteq V(G),$ we denoted by G[X] the subgraph of G induced by X. For each $j \in \{1, 2,\ldots , l\}$, we set $E_j = G[(N_G(x_j) - \{u\})\cup N_G(y_j)]$ and $E_{l+1}=G[(V(G)-\cup _{i=1}^lV(E_{i}))- \{u\}]$. For each $1\le j \le l,$ by the maximality of T, we obtain that $N_G(x_j) \subseteq V(T)$. Moreover, since $d_G(y_i, y_j) \ge 4$ for all $1\le i \not =j \le l$, we obtain $x_ix_j\not \in E(G).$ Hence, $N_G(x_j) \subseteq V(T)-\{x_1,x_2,\ldots ,x_l\}.$ Then, combining with Claim 2.2 and the definition of $E_j$, we conclude $V(E_j)\subseteq {\mathrm{Leaf}}(T)\cup \{x_j\}$ for each $j\in \{1,2,\ldots ,l\}.$ Hence, $V(G)-V(T) \subseteq V(E_{l+1}).$ On the other hand, since $d_G(v_1, y_j) \ge 4$, it implies that $v_2\not \in V(E_j)$ for all $1\le j \le l.$ Hence $v_2 \in V(E_{l+1})$ (see Fig. 2).

Fig. 2

$E_j$ ($1\le j \le l+1$)

By using the same arguments in the proofs of Claim 2.3, we conclude again the following fact.

Fact 1

For each $1\le j \le l,$ if P is a shortest path connecting $v_{1}$ and $y_{j}$ in G, then $V(P)\cap (V({\mathrm{Stem}}(T))-\{x_j\})\not = \emptyset ,$ and for every $1\le i < j \le l,$ if P is a shortest path connecting $y_{i}$ and $y_{j}$ in G, then $V(P)\cap (V({\mathrm{Stem}}(T))-\{x_i, x_j\})\not = \emptyset .$

We now give the following facts:

Fact 2

For every $1\le i < j \le l+1,$ then $V(E_i) \cap V(E_j) =\emptyset$.

Proof

By the definition of $E_{l+1}$ we obtain that $V(E_i) \cap V(E_{l+1}) =\emptyset$ for all $1\le i \le l$.

Now, assume that there exists a vertex $x\in V(E_i)\cap V(E_j)$ for some $1\le i < j \le l.$ If $x\in N_G(x_i)\cap N_G(x_j)$, then $x\in {\mathrm{Leaf}}(T).$ Consider the path P in G with its vertex set $\{y_i, x_i, x, x_j, y_j\}.$ By combining with $d_G(y_i, y_j) \ge 4$, we obtain that P is a shortest path connecting $y_i$ and $y_j$ in G. But $V(P)\cap (V({\mathrm{Stem}}(T))-\{x_i, x_j\})= \emptyset ,$ which contradicts Fact 1. Otherwise, without loss of generality, we may assume that $x\in N_G(y_i)\cap N_G(x_j)$ (or $x\in N_G(y_i)\cap N_G(y_j)$). Then, $d_G(y_i,y_j)\le 3$ (or $d_G(y_i,y_j)\le 2$ respectively), this contradicts Claim 2.3. Fact 2 is proved. $\square$

Fact 3

For each $1\le i < j \le l+1,$ if $x\in V(E_i), y\in V(E_j)$, then $xy\not \in E(G)$.

Proof

Suppose to the contrary that there exist two vertices $x\in V(E_i), y\in V(E_j) (1\le i < j \le l+1)$ such that $xy\in E(G).$

Subcase 1. $1\le i < j \le l.$

If $x \in N_G(y_i)$ and $y\in N_G(y_j)$, then $d_G(y_i, y_j) \le 3.$ This contradicts Claim 2.3.

If $x \in N_G(y_i)$ and $y\in N_G(x_j),$ we consider the path P in G with its vertex set $\{y_i, x, y, x_j, y_j\}.$ By combining with $d_G(y_i, y_j) \ge 4$, we obtain $y\not =y_j$, and then P is a shortest path connecting $y_i$ and $y_j$ in G. But $V(P)\cap (V({\mathrm{Stem}}(T))-\{x_i, x_j\})= \emptyset ,$ which contradicts Fact 1. By the same arguments, we also give a contradiction if $x \in N_G(x_i)$ and $y\in N_G(y_j)$.

If $x \in N_G(x_i)$ and $y\in N_G(x_j),$ remove the edges connecting x and y to $V({\mathrm{Stem}}(T))$ in T. After that add the edges $x_ix, xy, yx_j$ and $v_1v_2$ and remove the edge $x_ju.$ Then, the resulting tree $T'$ has l-ended stem and $|T'|>|T|$, this contradicts the maximality of T.

The subcase 1 is proved.

Subcase 2.$1\le i < j=l+1$.

Firstly, we show that $N_G(v_1)\cap V(E_a)=\emptyset$ for all $1\le a \le l.$ Indeed, suppose to the contrary that there exists a vertex $z\in N_G(v_1)\cap V(E_a).$ If $z\in N_G(x_a)$, then we consider the path P in G with its vertex set $\{v_1, z, x_a, y_a\}.$ This is a contradiction with $d_G(v_1, y_a) \ge 4.$ Otherwise, $z\in N_G(y_a)$, then $d_G(v_1, y_a) \le 2.$ This also gives a contradiction with $d_G(v_1, y_a)\ge 4.$ Therefore, we conclude that $N_G(v_1)\cap V(E_a)=\emptyset$ for all $1\le a \le l$. In particular, we obtain $N_G(v_1)\subseteq V(E_{l+1})-\{v_1\}$.

Secondly, we prove that $\deg _G(y_a)=|E_{a}|-1$ for all $1\le a \le l$ and $\deg _G(v_1)=|E_{l+1}|-1.$ Indeed, for each $1\le a \le l,$ by Claim 2.2 and the definition of $E_a$, we obtain $N_G(y_a)\subseteq V(E_a)-\{y_a\}.$ By combining the assumptions of Theorem 1.8, Claim 2.3, Fact 2 and $N_G(v_{1})\subseteq V(E_{l+1})-\{v_1\}$, we have

$$\begin{aligned} |G|-\left\lfloor \dfrac{l(t-1)}{t-2}\right\rfloor -1&\le \sigma ^4_{l+1}(G) \le \sum \limits _{a=1}^{l}\deg _{G}(y_a)+\deg _G(v_1)\\&\le \sum \limits _{a=1}^{l+1}(|E_a|-1)=|G|-l-2 =|G|-\left\lfloor \dfrac{l(t-1)}{t-2}\right\rfloor -1. \end{aligned}$$

Therefore, the equalities happen. Hence, $\deg _{G}(y_a)=|E_a|-1$ for every $1\le a \le l$ and $\deg _G(v_1)=|E_{l+1}|-1,$ in particular we obtain $N_G(v_1)=V(E_{l+1})-\{v_1\}$.

Finally, since $x\in V(E_i)$ and $y\in V(E_{l+1})$ such that $xy \in E(G),$ then $y\not =v_1$ (by $N_G(v_1)\cap V(E_i)=\emptyset$) and $yv_1 \in E(G)$ (by $N_G(v_1)=V(E_{l+1})-\{v_1\}$). If $x\in N_G(x_i)$, then we consider the path P in G with its vertex set $\{v_1, y, x, x_i, y_i\}.$ Since $d_G(v_1, y_i) \ge 4$, this implies that P is a shortest path connecting $v_1$ and $y_i$ in G. This is a contradiction with Fact 1. Otherwise, $x\in N_G(y_i)$, then $d_G(v_1, y_i) \le 3$. This also gives a contradiction with $d_G(v_1, y_i)\ge 4$. Therefore, we obtain that $xy\not \in E(G)$ for all $x\in V(E_i), y\in V(E_{l+1})$. This completes the proof of the subcase 2.

Therefore, Fact 3 holds. $\square$

Fact 4

For each $j\in \{1, 2, \ldots , l+1\},$$E_j$ is connected. Moreover, for every $w\in V(E_j)$ such that $uw \not \in E(G),$ then $N_G(w)= V(E_j) - \{w\}$ and $\deg _G(w)=|E_j|-1$.

Proof

Set $y_{l+1}= v_1$. In the proof of subclaim 2 of Fact 3, we conclude that $\deg _{G}(y_j)=|E_j|-1$. This implies that $N_G(y_j)=V(E_j)-\{y_j\}$. Hence, $E_j$ is connected.

Now, by Fact 2, Fact 3, the definition of $E_a$ and $E_a$ is connected for all $a\in \{1, 2, \ldots ,l+1\}$, the graph $G[V(G)-\{u\}]$ is disconnected and has $l+1$ components $E_1,\ldots , E_{l+1}$. Then, for every $1\le a < b\le l+1,$ if P is a path connecting two vertices $x\in E_a$ and $y\in E_{b}$, then P must pass through u. So for every $x\in E_a, y\in E_b$ such that $xu\not \in E(G)$ and $yu\not \in E(G)$, then $d_G(x,y) \ge 4.$ In particular, $d_G(w,y_a) \ge 4$ for all $1\le a\le l+1, a \not = j.$ Moreover, by Fact 3 and $wu\not \in E(G)$, we have $N_G(w)\subseteq V(E_j)-\{w\}.$ Hence, by Facts 2 and 3 and the assumptions of Theorem 1.8, we obtain

$$\begin{aligned} |G|-\left\lfloor \dfrac{l(t-1)}{t-2}\right\rfloor -1&\le \sigma ^4_{l+1}(G) \le \deg _{G}(w)+\sum \limits _{a=1,a\not =j}^{l+1}\deg _{G}(y_a) \\&\le \sum \limits _{a=1}^{l+1}(|E_a|-1)=|G|-l-2 =|G|-\left\lfloor \dfrac{l(t-1)}{t-2}\right\rfloor -1. \end{aligned}$$

Therefore, the equalities happen. So $\deg _G(w)=|E_j|-1$, and we thus also obtain $N_G(w)= V(E_j) - \{w\}$. These complete the proof of Fact 4. $\square$

For each $1\le i \le l+1,$ denote by $K_i$ the set of vertex w in $V(E_i)$ such that $N_{E_i}(w)=V(E_i)-\{w\}$. Then, for every $1\le i \le l+1,$$y_i \in K_i$ and in particular $K_i\not = \emptyset .$ On the other hand, by Facts 3 and 4, we can see that if $w\in V(E_i)$ such that $uw \not \in E(G),$ then $N_{E_i}(w)=N_G(w)=V(E_i)-\{w\}.$ Hence, $w \in K_i$ and u joins to all vertices in $V(E_i)-K_i$ for all $1\le i \le l+1.$ Therefore, using the definitions of $E_j\ (1\le j\le l+1)$ and Facts 1–4, we obtain that G is isomorphic to a graph M.

Hence, we conclude that if $l=t-2$, then G is isomorphic to a graph M.

Step 2. $|{\mathrm{Stem}}({\mathrm{Stem}}(T))|\ge 2$.

By Claim 2.4, there exists a vertex $v_3\in N_G(v_2)\cap V({\mathrm{Stem}}({\mathrm{Stem}}(T)))$.

Now, we conclude that $|N_T(v_3)\cap \{x_1,x_2,\ldots ,x_l\}|<t-2$. Indeed, otherwise, without loss of generality, we may assume $x_1,x_2,\ldots ,x_{t-2}\in N_T(v_3)$. Since $|{\mathrm{Stem}}({\mathrm{Stem}}(T))| \ge 2,$ there exists $s\in V({\mathrm{Stem}}({\mathrm{Stem}}(T)))\cap N_T(v_3)$. We consider the subgraph with the vertex set $\{v_3, v_2, s, x_1, x_2,\ldots ,x_{t-2}\}$ in G. By combining with Claim 2.4, the fact that G is $K_{1,t}$-free and since $\{x_1,\ldots ,x_{t-2}\}$ is an independent set by Claim 2.3, we have the following two cases.

Case 1.$sv_2\in E(G)$. This implies that the tree $T'=T+sv_2+v_2v_1-sv_3$ has l-ended stem and $|T'|>|T|$, this contradicts to the maximality of T.

Case 2.$x_js \in E(G)$ for some $j \in \{1, \ldots , t-2\}.$ Then, we consider the tree $T'=T+x_js+v_2v_1-sv_3.$ Hence, $T'$ has l-ended stem and $|T'|>|T|$, this also contradicts to the maximality of T.

Therefore, $|N_T(v_3)\cap \{x_1,x_2,\ldots ,x_l\}|<t-2$.

Now, if $|N_T(u)\cap \{x_1,x_2,\ldots ,x_l\}|\le t-2$ for all $u\in V({\mathrm{Stem}}({\mathrm{Stem}}(T)))-\{v_3\},$ then combining with $|N_T(v_3)\cap \{x_1,x_2,\ldots ,x_l\}|<t-2$, we have

$$\begin{aligned} l&=|{\mathrm{Leaf}}({\mathrm{Stem}}(T))|< (t-2)|{\mathrm{Stem}}({\mathrm{Stem}}(T))-\{v_3\}|+ t-2\\&=(t-2)|{\mathrm{Stem}}({\mathrm{Stem}}(T))|\\&\le (t-2)\left\lfloor \dfrac{l}{t-2}\right\rfloor \le l\ (\text {by }(2.1)). \end{aligned}$$

This is a contradiction. Hence, there exists a vertex $u\in V({\mathrm{Stem}}({\mathrm{Stem}}(T)))$ such that $|N_T(u)\cap \{x_1,x_2,\ldots ,x_l\}|\ge t-1$. Without loss of generality, we may assume $x_1,x_2,\ldots ,x_{t-1}\in N_T(u)$. Set $s\in V({\mathrm{Stem}}({\mathrm{Stem}}(T)))\cap N_T(u)$. Now, if $x_js \in E(G)$ for some $j \in \{1, \ldots , t-1\}$, then we consider the tree $T'=T+x_js+v_2v_1-su.$ Hence, $T'$ has l-ended stem and $|T'|>|T|$, this also contradicts to the maximality of T. Hence, we obtain $x_js \not \in E(G)$ for all $j \in \{1, \ldots , t-1\}.$ Then, G induces a $K_{1,t}$ subgraph with vertex set $\{u, s, x_1,x_2,\ldots ,x_{t-1}\}$. This gives a contradiction with the assumption of Theorem 1.8.

Therefore, we complete the Proof of Theorem 1.8.