Laplacian coefficient, matching polynomial and incidence energy of trees with described maximum degree

Abstract

Let $\mathcal {L}(T,\lambda )=\sum _{k=0}^n (-1)^{k}c_{k}(T)\lambda ^{n-k}$ be the characteristic polynomial of its Laplacian matrix of a tree T. This paper studied some properties of the generating function of the coefficients sequence $(c_0, \ldots , c_n)$ which are related with the matching polynomials of division tree of T. These results, in turn, are used to characterize all extremal trees having the minimum Laplacian coefficient generation function and the minimum incidence energy of trees with described maximum degree, respectively.

Appendix

In here, we present detail proof of Lemmas 3.5 and 3.6

Lemma 3.5 Let S(T) be an optimal tree in $\mathcal {S(T)}_{n, d+1}$. If degree of each vertex is 1 or $d+1$ in V(T), and there is a vertex v such that there are $1\le h\le d-1$ pendent vertices in N(v), then T is greedy tree $T_{d+1}^*$.

Proof

If the branches $T_1, \ldots , T_{d+1}$ of $T-u$ contains no $C_3$, then by Lemma 3.4, T is a greedy tree. Now assume that the branches $T_1, \ldots , T_{d+1}$ of $T-u$ contains $C_3$, say $T_{d+1}\supseteq C_3$, then by Lemma 3.4, $T_1=C_2$ and $T_i=C_1$ or $C_2$ for $i=2, \ldots , d$. Let P (see Fig. 1) be the longest path which goes through v and terminates at non-pendent vertices with $v_2=v$. Assume the length of P is 2k. Let $L_{1}^i,\ldots ,L_{d}^i$ be the branches of $T-v_i$ containing no $w_1$ and $L_{1}^i$ contains $v_2$ for $i=3, \ldots , k$. Let $R_{1}^i,\ldots ,R_{d}^i$ be the branches of $T-w_i$ containing no $v_1$, and $R_{1}^i$ contains $w_1$ for $2\le i\le k$. Clearly $R_{1}^2=R_{2}^2=\cdots =R_{d}^2=C_2$.

Claim

• (1) $\tau (S(C_t),x)<\tau (S(L_{1}^t),x)< \tau (S(C_{t-1}),x),~ t=3, \dots , k+1$.

• (2) $L_{i}^t= C_{t-1}$ or $~C_{t} ,~ t=3, \ldots , k+1,~ i=2, \ldots , d$.

• (3) $R_{1}^t=R_{2}^t=\cdots =R_{d}^t=C_t,~ t=3, \ldots , k$.

We prove Claim by the induction on t. For $t=3$, by Lemma 3.4,

$$\begin{aligned} \tau (S(C_t),x)= & {} \frac{1}{1+\sum _{j=1}^d \frac{x}{1+x\tau (S(C_2),x)}} \\< & {} \frac{1}{1+\sum _{j=1}^d \frac{x}{1+x\tau (S(L_j^2),x)}}=\tau (S(L_1^3),x)\\< & {} \frac{1}{1+\sum _{j=1}^d \frac{x}{1+x\tau (S(C_1),x)}} =\tau (S(C_2),x). \end{aligned}$$

So Claim (1) holds for $t=3$. By Lemma 3.4, Claim (2) holds for $t=3$. Moreover, by $\tau (S(L_1^3),x)>\tau (S(C_3),x)=\tau (S(R_1^3),x)$ and Corollary 2.6, $\tau (S(L_1^3),x)\ge \max \{\tau (S(R_1^3),x), \ldots , \tau (S(R_d^3),x)$. Combining with Lemma 3.4 and $\tau (S(C_2),x)>\tau (S(L_1^3),x)$, we have $R_i^3=C_3$ for $i=2, \ldots , d$. Therefore Claim (3) holds for $t=3$.

Assume that Claim holds for less than t and we consider Claim for t. By the induction prothesis,

$$\begin{aligned} \tau (S(C_{t-1}),x)\le \tau (S(L_{1}^{t-1}),x)\le \tau (S(C_{t-2}),x) \end{aligned}$$

and $\tau (S(C_{t-1}),x)\le \tau (S(L_{j}^{t-1}),x)\le \tau (S(C_{t-2}),x)$ for $j=2, \ldots , d$. It follows from (11) and (21) that

$$\begin{aligned} \tau (S(C_{t}),x)= & {} \frac{1}{1+\sum _{i=1}^d\frac{x}{1+x\tau (S(C_{t-1}),x)}}\\< & {} \frac{1}{1+\sum _{i=1}^d\frac{x}{1+x\tau (S(L_{i}^{t-1}),x)}}\\= & {} \tau (S(L_{1}^t),x)\\< & {} \frac{1}{1+\sum _{i=1}^d\frac{x}{1+x\tau (S(C_{t-2}),x)}}.\\= & {} \tau (S(C_{t-1}),x) \end{aligned}$$

Hence (1) holds for t. In order to prove (2) holds for t, we first prove the following several Claims

Claim 3.1 $\tau (S(C_t), x)\le \tau (S(L_i^t), x)\le \tau (S(C_{t-1}, x),$ for $i=2, \ldots , d$.

In fact, there are d branches $L_1^t, \ldots , L_d^t$ containing no $w_t$ in $T_{d+1}^*-v_t$ and there are d branches $C_t, R_2^t,\ldots , R_d^t$ containing no $v_t$ in $T_{d+1}^*-w_t$. By Claim (1), we have $\tau (S(C_t),x)<\tau (S(L_{1}^t),x)$. Hence by Corollary 2.6, $\min \{\tau (S(L_1^t), x), \ldots , \tau (S(L_d^t), x)\}\ge \tau (S(C_t),x)$. On the other hand, there are d branches $L_1^t, \ldots , L_d^t$ containing no $w_{t-1}$ in $T_{d+1}^*-v_t$ and there are d branches $C_{t-1}, \ldots , C_{t-1}$ containing no $v_t$ in $T_{d+1}^*-w_{t-1}$. By Claim (1), $\tau (S(L_{1}^t),x)<\tau (S(C_{t-1}),x).$ Hence by Corollary 2.6, $\max \{S(L_{2}^t),x), \ldots , S(L_{d}^t),x)\}\le \tau (S(C_{t-1}),x).$ So Claim 3.1 holds.

Let the maximum distance between $v_t$ and any vertex in $L_1^t, \ldots , L_d^t$ is $l+1$. Denote by

$$\begin{aligned} V_j=\left\{ u\ |\ dist(u, v_t)=l-j+1, u\in \bigcup _{i=1}^d V(L_i^t)\right\} , j=0, \ldots , l. \end{aligned}$$

Claim 3.2 For any $u\in V_{l-j}$, there are d the branches $L_1^u, \ldots , L_d^u$ containing no $v_t$ in $T_{d+1}^*-u$ such that

$$\begin{aligned} \tau (S(C_{t-j-1}), x)\le \tau (S(L_i^u), x) \le \tau (S(C_{t-j-2}), x) , \ \ i=1, \ldots , d, \end{aligned}$$

(24)

where $j=0, \ldots , \min \{t, l\}-2$.

We prove Claim 3.2 by the induction on j. Let $L_1^u, \ldots , L_d^u$ be d the branches containing no $v_{t}$ in $T_{d+1}^*-u$ and $T^u$ be the subtree consisting of u and $L_1^u, \ldots , L_d^u$. For $j=0$, there exists a $1\le p\le d$ such that $T^u=L_p^t$. If there exists an $1\le i\le d$ such that $\tau (S(L_i^u), x) < \tau (S(C_{t-1}), x)$, let $R_1^{t-1}=C_{t-1}, \ldots , R_d^{t-1}$ be d the branches containing no $v_t$ in $T_{d+1}^*-w_{t-1}$. Hence by Corollary 2.6, $\max \{ \tau (S(L_1^u), x), \ldots , \tau (S(L_d^u), x)\}\le \tau (S(C_{t-1}), x)$. Then

$$\begin{aligned} \tau (S(L_p^{t}), x)= & {} \tau (S(T^u), x)=\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(L_q^u), x)}} <\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(C_{t-1}), x)}}\\= & {} \tau (S(C_{t}), x), \end{aligned}$$

which contradicts Claim 3.1. Therefore,

$$\begin{aligned} \tau (S(C_{t-1}), x)\le \tau (S(L_i^u), x), i=1, \ldots , d. \end{aligned}$$

On the other hand, if there exists $1\le i\le d$ such that $\tau (S(L_i^u), x) > \tau (S(C_{t-2}), x)$. let $R_1^{t-2}=C_{t-2}, \ldots , R_d^{t-2}$ be d the branches containing no $v_t$ in $T_{d+1}^*-w_{t-2}$. By Corollary 2.6,

$$\begin{aligned} \min \{ \tau (S(L_1^u), x), \ldots , \tau (S(L_d^u), x)\}\le \tau (S(C_{t-2}), x). \end{aligned}$$

Then

$$\begin{aligned} \tau (S(L_p^{t}), x)= & {} \tau (S(T^u), x)=\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(L_q^u), x)}} >\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(C_{t-1}), x)}}\\= & {} \tau (S(C_{t-1}), x), \end{aligned}$$

which contradicts Claim 3.1. Hence Claim 3.2 holds for $j=0$. Now assume that Claim 3.2 holds for j and consider the claim for $j+1$. For any $u\in V_{l-(j+1)}$, let $L_1^u, \ldots , L_d^u$ be d the branches containing no $v_t$ in $T_{d+1}^*-u$ and $T^u$ be the subtree consisting of u and $L_1^u, \ldots , L_d^u$. Clearly there exists a $u^{\prime }\in V_{l-j}$ such that there exists a branch $L_1^{u^{\prime }}$ in $T_{d+1}^*-u^{\prime }$ such that $T^u=L_1^{u^{\prime }}$.

If there exists an $1\le i\le d$ such that $\tau (S(L_i^u), x) < \tau (S(C_{t-j-2}), x)$, let $R_1^{t-j-2}=C_{t-j-2}, \ldots , R_d^{t-j-2}$ be d the branches containing no u in $T_{d+1}^*-w_{t-j-1}$. By Corollary 2.6,

$$\begin{aligned} \max \{ \tau (S(L_1^u), x), \ldots , \tau (S(L_d^u), x)\}\le \tau (S(C_{t-j-2}), x). \end{aligned}$$

Then

$$\begin{aligned} \tau (S(L_1^{u^{\prime }}), x)= & {} \tau (S(T^{u}), x)=\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(L_q^u), x)}} <\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(C_{t-j-2}), x)}}\\= & {} \tau (S(C_{t-j-1}), x), \end{aligned}$$

which contradicts the induction hypothesis. Therefore, for any $u\in V_{l-j-1}$,

$$\begin{aligned} \tau (S(C_{t-j-2}), x)\le \tau (S(L_i^u), x), i=1, \ldots , d. \end{aligned}$$

On the other hand, if there exists $1\le i\le d$ such that $\tau (S(L_i^u), x) > \tau (S(C_{t-j-3}), x)$. Let $R_1^{t-j-3}=C_{t-j-3}, \ldots , R_d^{t-j-3}$ be d the branches containing no u in $T_{d+1}^*-w_{t-j-2}$. By Corollary 2.6,

$$\begin{aligned} \min \{\tau (S(L_1^u), x), \ldots , \tau (S(L_d^u), x)\}\ge \tau (S(C_{t-j-3}), x). \end{aligned}$$

Then

$$\begin{aligned} \tau (S(L_1^{u^{\prime }}), x)= & {} \tau (S(T^u), x)=\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(L_q^u), x)}} >\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(C_{t-j-3}), x)}}\\= & {} \tau (S(C_{t-j-2}), x), \end{aligned}$$

which contradicts the induction hypothesis. Hence Claim 3.2 holds for $j+1$. Therefore Claim 3.2 holds.

Claim 3.3 $l=t-1$.

If $l>t-1$, by Claim 3.2, for any $u\in V_{l-t+2},$

$$\begin{aligned} \tau (S(C_{1}), x) \le \tau (S(L_i^u),x), \ i=1, \ldots , d. \end{aligned}$$

On the other hand, there exists a $u^{\prime }\in V_{l-t+2}$ such that the largest distance between $u^{\prime }$ and the pendent vertex is at least 2, then $C_2$ is a proper subgraph $L_1^{u^{\prime }}$, which implies $\tau (S(L_1^{u^{\prime }}), x)\le \tau (S(C_{2}), x)$. it is a contradiction. Hence $l\le t-1$. Since $l\ge t-1$, then $l=t-1$.

Claim 3.4 For any $u\in V_{t-j-1}, ~j=0, \ldots , t-3$. Let $L_1^u, \ldots , \ldots , L_d^u$ be the d branches containing no $v_t$ in $T_{d+1}^*-u$ and $T^u$ consist of u and d branches $L_1^u, \ldots , \ldots , L_d^u$. Then $L_1^u=\cdots =L_d^u=C_{t-j-1}$ or $L_1^u=\cdots =L_d^u=C_{t-j-2}$, i.e., $T^u=C_{t-j}$ or $T^u=C_{t-j-1}$.

We use induction for $t-j-1$. In fact, for $j=t-3$ and $u\in V_{2},$ by Claim 3.2, $\tau (S(C_{2}), x)\le \tau (S(L_i^u), x)\le \tau (S(C_{1}), x)$ for $i=1, \ldots , d$. Hence $L_i^u=C_2$ or $L_i^u=C_1$ for $i=1, \ldots , d$. If, say $L_1^u=C_2$ and $L_2^u=C_1$, then by $\tau (S(L_1^2), x)>\tau (S(L_1^u),x)$ and Corollary 2.6, $\tau (S(L_1^2), x)\ge \max \{\tau (S(L_1^u), x), \ldots , \tau (S(L_d^u), x)\}\ge \tau (S(C_1), x)$, which is a contradiction. Hence $L_1^u=\cdots =L_d^u=C_{2}$ or $L_1^u=\cdots =L_d^u=C_{1}$, i.e., $T^u=C_{3}$ or $T^u=C_{2}$ for $u\in V_2$. Assume that Claim 3.4 hold for any vertex in $V_{t-j-2}$. Now for $u\in V_{t-j-1}$. Let $z_1, \ldots , z_d\in V_{t-j-1}$ be the roots of $L_1^u, \ldots , L_d^u$, respectively. By the induction hypothesis, $L_1^u, \ldots , L_d^u\in \{C_{t-j-1}, C_{t-j-2}\}$. Further $L_1^u=\cdots = L_d^u= C_{t-j-1}$ or $L_1^u=\cdots = L_d^u= C_{t-j-2}$. In fact, if, say $L_1^u=C_{t-j-1}$ and $L_2^u=C_{t-j-2}$, By $\tau (S(L_1^{t-j-1}),x)>\tau (S(C_{t-j-1}),x)$ and Corollary 2.6,

$$\begin{aligned} \tau (S(L_1^{t-j-1}),x)\ge \max \{\tau (S(C_{t-j-1}),x), \tau (S(C_{t-j-2}),x)\ge \tau (S(C_{t-j-2}),x), \end{aligned}$$

which contradiction to Claim 3.1. Hence $L_1^u=\cdots = L_d^u= C_{t-j-1}$ or $L_1^u=\cdots = L_d^u= C_{t-j-2}$, i.e., $T^u=C_{t-j}$ or $T^u=C_{t-j-1}$. So Claim 3.4 holds. Hence $L_i^t=C_t$ or $L_i=C_{t-1}$ for $i=2, \ldots , d$. In other words, Claim (2) holds.

Similarly, we can prove Claim (3) and omit the detail. It is easy from Claim that $T_{d+1}^*$ is greedy tree. If the length of P is odd, using similar way to prove this assertion. So we finish our proof.$\square$

Lemma 3.6 Let S(T) be an optimal tree in $\mathcal {S(T)}_{n, d+1}$. If degree of each vertex is 1 or $d+1$ in V(T), and there are d pendente vertices or no pendent vertices in N(u) for any $u\in V(T)$. Then T is greedy tree $T_{d+1}^*$.

Proof

Let $U=\{ u\ |\ deg (u)=1, u\in V(T)\}$ and $dist(v, U)=\min \{dist (v, u), u\in U\}$.

$$\begin{aligned} U_i=\{v\ |\ dist(v, U)=i\},~i=1,2,\ldots . \end{aligned}$$

Clearly, for any $v\in U_{1}, T-v$ has d branches $C_1, \ldots , C_1$. If for any vertex $v\in U_{i}, T-v$ has d branches $C_i, \ldots , C_i$ for $i=1, 2, \ldots ,$ then T is the greedy tree $T_{d+1}^*$, and we complete the proof. Next assume that there exists a vertex $v\in U_{i}$ such that $T-v$ has at least two branches different from $C_{i}$. Let $t_0$ be the smallest integer such that for any vertex $v\in U_{i}, T-v$ has d branches $C_i, \ldots , C_i$ for $i=1, \ldots , t_0-1$ and for some vertex $v\in U_{t_0}, T-v$ has at least two branches different from $C_{t_0}$. Let P be the longest path through $v=v_{s}\in U_{t_0}$ (see Fig 2). Then by $v\in V_{t_0}$, we have $s\ge t_0+1$. Assume that the length of P is even.

Let $L_{1}^i,\ldots ,L_{d}^i$ be the branches of $T-v_i$ containing no $w_1$ and $L_{1}^i$ contains $v_1$ for $i=2, \ldots , k$. Let $R_{1}^i,\ldots ,R_{d}^i$ be the branches of $T-w_i$ containing no $v_1$, and $R_{1}^i$ contains $w_1$ for $2\le i\le k$. By the definition of $v=v_{s}, C_{t_0}$ is a branch of $T-v_{s}$ and $C_{t_0+1}$ is a subtree of $L_j^{s}$ if $L_i^{s}\ne C_{t_0},~i=1,2,\ldots ,d$. Then $\tau (S(L_j^{s}),x)\le \tau (S(C_{t_0+1}),x)$ for $L_i^{s}\ne C_{t_0}, i=1, \ldots . d$. Similarly, $C_{t_0}$ is a branch of $T-w_{t_0}$ and $C_{t_0+1}$ is a subtree of $R_j^{t_0}$ if $R_i^{t_0}\ne C_{t_0},~i=1,2,\ldots ,d$. Then $\tau (S(R_j^{t_0}),x)\le \tau (S(C_{t_0+1}),x)$ for $R_i^{t_0}\ne C_{t_0}, i=1, \ldots . d$. If there is a branch of $R_1^{t_0},R_2^{t_0},\ldots ,R_d^{t_0}$ is not $C_{t_0}$, say $R_1^{t_0}\ne C_{t_0}$. By Corollary 2.6, we have

$$\begin{aligned} \tau (S(R_1^{t_0}), x)\ge & {} \min \{ \tau (S(R_1^{t_0}), x), \ldots , \tau (S(R_d^{t_0}), x)\} \\\ge & {} \max \{ \tau (S(L_1^s), x), \ldots , \tau (S(L_d^s), x)\}\ge \tau (S(C_{t_0}),x). \end{aligned}$$

It contradicts $\tau (S(R_1^{t_0}),x)\le \tau (S(C_{t_0+1}),x)$. Thus $R_1^{t_0}=\cdots =R_d^{t_0}=C_{t_0}$. Which will imply that $R_1^{t_0+1}=C_{t_0+1}$. Since $T-v_i$ has branches $L_{1}^s,\ldots ,L_{d}^s$ containing no $w_1$ and $T-w_{t_0+1}$ has branches $R_{1}^{t_0+1},\ldots ,R_{d}^{t_0+1}$ containing no $v_1$, by $\tau (S(C_{t_0+1}),x)<\tau (S(C_{t_0}),x)$ and Corollary 2.6, we have

$$\begin{aligned} \tau (S(C_{t_0+1}),x)\le & {} \max \{ \tau (S(R_1^{t_0+1}), x), \ldots , \tau (S(R_d^{t_0+1}), x)\} \\\le & {} \min \{ \tau (S(L_1^s), x), \ldots , \tau (S(L_d^s), x)\}\le \tau (S(C_{t_0+1}),x). \end{aligned}$$

Since $C_{t_0+1}$ is a subtree of $L_i^{s}$ if $L_i^{s}\ne C_{t_0},~i=1,2,\ldots ,d$, then $L_i^{s}=C_{t_0+1}$ if $L_i^{s}\ne C_{t_0},~i=1,2,\ldots ,d$. This implies $s=t_0+1$. Without loss of generality, we can assume $L_1^{t_0+1}=C_{t_0+1}$ and $L_2^{t_0+1}=C_{t_0}$.

Claim

• (1) $\tau (S(C_t),x)<\tau (S(L_{1}^t),x)< \tau (S(C_{t-1}),x),~ t=t_0+2, \ldots , k+1, \tau (S(L_{i}^{t_0+1}),x)=\tau (S(C_{t_0+1}),x) ~\text{ or }~\tau (S(C_{t_0}),x),~i=1, \ldots , d$.

• (2) $L_{i}^t= C_{t-1}$ or $~C_{t} ,~ t=t_0+1, \ldots , k+1,~ i=2, \ldots , d$.

• (3) $R_{1}^t=R_{2}^t=\cdots =R_{d}^t=C_t,~ t=t_0+1, \ldots , k$.

We prove Claim by the induction on t. For $t=t_0+1$, by the above argument, we can find Claim holds. Assume that Claim holds for the number less than $t>t_0+1$ and we consider Claim for t. By the induction prothesis,

$$\begin{aligned} \tau (S(C_{t-1}),x)\le \tau (S(L_{1}^{t-1}),x)\le \tau (S(C_{t-2}),x) \end{aligned}$$

and $\tau (S(C_{t-1}),x)\le \tau (S(L_{j}^{t-1}),x)\le \tau (S(C_{t-2}),x)$ for $j=2, \ldots , d$. It follows from (11) and (21) that

$$\begin{aligned} \tau (S(C_{t}),x)= & {} \frac{1}{1+\sum _{i=1}^d\frac{x}{1+x\tau (S(C_{t-1}),x)}}\\< & {} \frac{1}{1+\sum _{i=1}^d\frac{x}{1+x\tau (S(L_{i}^{t-1}),x)}}\\= & {} \tau (S(L_{1}^t),x)\\< & {} \frac{1}{1+\sum _{i=1}^d\frac{x}{1+x\tau (S(C_{t-2}),x)}}.\\= & {} \tau (S(C_{t-1}),x) \end{aligned}$$

Hence (1) holds for t. In order to prove (2) holds for t, we first prove the following several Claims.

Claim 3.1 $\tau (S(C_t), x)\le \tau (S(L_i^t), x)\le \tau (S(C_{t-1}, x)$ for $i=2, \ldots , d$.

In fact, there are d branches $L_1^t, \ldots , L_d^t$ containing no $w_t$ in $T_{d+1}^*-v_t$ and there are d branches $R_1^t=C_t, R_2^t,\ldots , R_d^t$ containing no $v_t$ in $T_{d+1}^*-w_t$. By (1) of the Claim, we have $\tau (S(C_t),x)<\tau (S(L_{1}^t),x)$. Hence by Corollary 2.6, $\min \{\tau (S(L_1^t), x), \ldots , \tau (S(L_d^t), x)\}\ge \tau (S(C_t),x)$. On the other hand, there are d branches $L_1^t, \ldots , L_d^t$ containing no $w_{t-1}$ in $T_{d+1}^*-v_t$ and there are d branches $C_{t-1}, \ldots , C_{t-1}$ containing no $v_t$ in $T_{d+1}^*-w_{t-1}$. By (1) of the Claim, $\tau (S(L_{1}^t),x)<\tau (S(C_{t-1}),x).$ Hence by Corollary 2.6, $\max \{S(L_{2}^t),x), \ldots , S(L_{d}^t),x)\}\le \tau (S(C_{t-1}),x).$ So Claim 3.1 holds.

Let the maximum distance between $v_t$ and any vertex in $L_1^t, \ldots , L_d^t$ is $l+1$. By the definition of P, we can find that $l\ge t-1$. Denote by

$$\begin{aligned} V_j=\left\{ u\ |\ dist(u, v_t)=l-j+1, u\in \bigcup _{i=1}^d V(L_i^t)\right\} , j=0, \ldots , l. \end{aligned}$$

Claim 3.2 For any $u\in V_{l-j}$, there are d branches $L_1^u, \ldots , L_d^u$ containing no $v_t$ in $T_{d+1}^*-u$ such that

$$\begin{aligned} \tau (S(C_{t-j-1}), x)\le \tau (S(L_i^u), x) \le \tau (S(C_{t-j-2}), x) , \ \ i=1, \ldots , d, \end{aligned}$$

(25)

where $j=0, \ldots , \min \{t, l\}-2$.

We prove Claim 3.2 by the induction on j. Let $L_1^u, \ldots , L_d^u$ be d the branches containing no $v_{t}$ in $T_{d+1}^*-u$ and $T^u$ be the subtree consisting of u and $L_1^u, \ldots , L_d^u$. For $j=0$, there exists a $1\le p\le d$ such that $T^u=L_p^t$. If there exists an $1\le i\le d$ such that $\tau (S(L_i^u), x) < \tau (S(C_{t-1}), x)$, let $R_1^{t-1}=C_{t-1}, \ldots , R_d^{t-1}$ be d the branches containing no $v_t$ in $T_{d+1}^*-w_{t-1}$. Hence by Corollary 2.6, $\max \{ \tau (S(L_1^u), x), \ldots , \tau (S(L_d^u), x)\}\le \tau (S(C_{t-1}), x)$. Then

$$\begin{aligned} \tau (S(L_p^{t}), x)= & {} \tau (S(T^u), x)=\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(L_q^u), x)}} <\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(C_{t-1}), x)}}\\= & {} \tau (S(C_{t}), x), \end{aligned}$$

which contradicts Claim 3.1. Therefore,

$$\begin{aligned} \tau (S(C_{t-1}), x)\le \tau (S(L_i^u), x), i=1, \ldots , d. \end{aligned}$$

On the other hand, if there exists $1\le i\le d$ such that $\tau (S(L_i^u), x) > \tau (S(C_{t-2}), x)$. let $R_1^{t-2}=C_{t-2}, \ldots , R_d^{t-2}$ be d the branches containing no $v_t$ in $T_{d+1}^*-w_{t-2}$. By Corollary 2.6,

$$\begin{aligned} \min \{ \tau (S(L_1^u), x), \ldots , \tau (S(L_d^u), x)\}\le \tau (S(C_{t-2}), x). \end{aligned}$$

Then

$$\begin{aligned} \tau (S(L_p^{t}), x)= & {} \tau (S(T^u), x)=\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(L_q^u), x)}} >\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(C_{t-1}), x)}}\\= & {} \tau (S(C_{t-1}), x), \end{aligned}$$

which contradicts Claim 3.1. Hence Claim 3.2 holds for $j=0$. Now assume that Claim 3.2 holds for j and consider the claim for $j+1$. For any $u\in V_{l-(j+1)}$, let $L_1^u, \ldots , L_d^u$ be d the branches containing no $v_t$ in $T_{d+1}^*-u$ and $T^u$ be the subtree consisting of u and $L_1^u, \ldots , L_d^u$. Clearly there exists a $u^{\prime }\in V_{l-j}$ such that there exists a branch $L_1^{u^{\prime }}$ in $T_{d+1}^*-u^{\prime }$ such that $T^u=L_1^{u^{\prime }}$.

If there exists an $1\le i\le d$ such that $\tau (S(L_i^u), x) < \tau (S(C_{t-j-2}), x)$, let $R_1^{t-j-2}=C_{t-j-2}, \ldots , R_d^{t-j-2}$ be d the branches containing no u in $T_{d+1}^*-w_{t-j-1}$. By Corollary 2.6,

$$\begin{aligned} \max \{ \tau (S(L_1^u), x), \ldots , \tau (S(L_d^u), x)\}\le \tau (S(C_{t-j-2}), x). \end{aligned}$$

Then

$$\begin{aligned} \tau (S(L_1^{u^{\prime }}), x)= & {} \tau (S(T^{u}), x)=\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(L_q^u), x)}} <\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(C_{t-j-2}), x)}}\\= & {} \tau (S(C_{t-j-1}), x), \end{aligned}$$

which contradicts the induction hypothesis. Therefore, for any $u\in V_{l-j-1}$,

$$\begin{aligned} \tau (S(C_{t-j-2}), x)\le \tau (S(L_i^u), x), i=1, \ldots , d. \end{aligned}$$

On the other hand, if there exists $1\le i\le d$ such that $\tau (S(L_i^u), x) > \tau (S(C_{t-j-3}), x)$. Let $R_1^{t-j-3}=C_{t-j-3}, \ldots , R_d^{t-j-3}$ be d the branches containing no u in $T_{d+1}^*-w_{t-j-2}$. By Corollary 2.6,

$$\begin{aligned} \min \{ \tau (S(L_1^u), x), \ldots , \tau (S(L_d^u), x)\}\ge \tau (S(C_{t-j-3}), x). \end{aligned}$$

Then

$$\begin{aligned} \tau (S(L_1^{u^{\prime }}), x)= & {} \tau (S(T^u), x)=\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(L_q^u), x)}} >\frac{1}{1+\sum _{q=1}^d\frac{x}{1+x\tau (S(C_{t-j-3}), x)}}\\= & {} \tau (S(C_{t-j-2}), x), \end{aligned}$$

which contradicts the induction hypothesis. Hence Claim 3.2 holds for $j+1$. Therefore Claim 3.2 holds.

Claim 3.3 $l=t-1$.

If $l>t-1$, by Claim 3.2, for any $u\in V_{l-t+2},$

$$\begin{aligned} \tau (S(C_{1}), x) \le \tau (S(L_i^u),x), \ i=1, \ldots , d. \end{aligned}$$

On the other hand, there exists a $u^{\prime }\in V_{l-t+2}$ such that the largest distance between $u^{\prime }$ and the pendent vertex is at least 2, then $C_2$ is a proper subgraph $L_1^{u^{\prime }}$, which implies $\tau (S(L_1^{u^{\prime }}), x)\le \tau (S(C_{2}), x)$. it is a contradiction. Hence $l\le t-1$. Since $l\ge t-1$, then $l=t-1$.

Claim 3.4 For any $u\in V_{t-j-1}, ~j=0, \ldots , t-3$. Let $L_1^u, \ldots , \ldots , L_d^u$ be the d branches containing no $v_t$ in $T_{d+1}^*-u$ and $T^u$ consist of u and d branches $L_1^u, \ldots , \ldots , L_d^u$. Then $L_1^u=\cdots =L_d^u=C_{t-j-1}$ or $L_1^u=\cdots =L_d^u=C_{t-j-2}$, i.e., $T^u=C_{t-j}$ or $T^u=C_{t-j-1}$.

We use induction for $t-j-1$. In fact, for $j=t-3$ and $u\in V_{2},$ by Claim 3.2, $\tau (S(C_{2}), x)\le \tau (S(L_i^u), x)\le \tau (S(C_{1}), x)$ for $i=1, \ldots , d$. Hence $L_i^u=C_2$ or $L_i^u=C_1$ for $i=1, \ldots , d$. If, say $L_1^u=C_2$ and $L_2^u=C_1$, then by $\tau (S(L_1^2), x)>\tau (S(L_1^u),x)$ and Corollary 2.6, $\tau (S(L_1^2), x)\ge \max \{\tau (S(L_1^u), x), \ldots , \tau (S(L_d^u), x)\}\ge \tau (S(C_1), x)$, which is a contradiction. Hence $L_1^u=\cdots =L_d^u=C_{2}$ or $L_1^u=\cdots =L_d^u=C_{1}$, i.e., $T^u=C_{3}$ or $T^u=C_{2}$ for $u\in V_2$. Assume that Claim 3.4 hold for any vertex in $V_{t-j-2}$. Now for $u\in V_{t-j-1}$. Let $z_1, \ldots , z_d\in V_{t-j-1}$ be the roots of $L_1^u, \ldots , L_d^u$, respectively. By the induction hypothesis, $L_1^u, \ldots , L_d^u\in \{C_{t-j-1}, C_{t-j-2}\}$. Further $L_1^u=\cdots = L_d^u= C_{t-j-1}$ or $L_1^u=\cdots = L_d^u= C_{t-j-2}$. In fact, if, say $L_1^u=C_{t-j-1}$ and $L_2^u=C_{t-j-2}$, By $\tau (S(L_1^{t-j-1}),x)>\tau (S(C_{t-j-1}),x)$ and Corollary 2.6,

$$\begin{aligned} \tau (S(L_1^{t-j-1}),x)\ge \max \{\tau (S(C_{t-j-1}),x), \tau (S(C_{t-j-2}),x)\ge \tau (S(C_{t-j-2}),x), \end{aligned}$$

which contradiction to Claim 3.1. Hence $L_1^u=\cdots = L_d^u= C_{t-j-1}$ or $L_1^u=\cdots = L_d^u= C_{t-j-2}$, i.e., $T^u=C_{t-j}$ or $T^u=C_{t-j-1}$. So Claim 3.4 holds.

Hence $L_i^t=C_t$ or $L_i^t=C_{t-1}$ for $i=2, \ldots , d$. In other words, (2) of Claim holds.

Similarly, we can prove (3) of Claim, here we omit the detail. It is easy from Claim that $T_{d+1}^*$ is greedy tree. If the length of P is odd, using similar way to prove this assertion. So we finish our proof.$\square$