Number of leaves in a tree with maximum degree $k$

Question

Let $G$ be a tree with a maximum degree of the vertices equal to $k$. At least how many vertices with a degree of $1$ can be in $G$ and why?

I think the answer must be $k$ but I don't know how to prove it.

Answer 1

HINT. Let $G$ be a smallest tree that meets the requirements. Could adding nodes to $G$ decrease the answer?

Answer 2

For $d\in\{1,\dots,k\}$, let $n_d$ be the number of nodes of degree $d$. By the handshake lemma, we have $$\sum_{d=1}^k d n_d = 2\left(\sum_{d=1}^k n_d - 1\right),$$ which implies that $$n_1 = 2 + \sum_{d=3}^k (d-2) n_d \ge 2 + (k-2) n_k \ge 2 + (k-2) 1 = k.$$ This lower bound is attained by a star with $k+1$ nodes.

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Alternatively, perform a depth-first search rooted at a node with degree $k$. Each branch yields at least one leaf.

Answer 3

Hint Any tree on $\geq 2$ vertices has at least two leafs. Consider a tree with maximal degree $k$ and delete a vertex $v$ with degree $k$. You are left with a forest consisting of $k$ components, coming with a distinct neighbor $v_i$ of $v$ in $T$. Finish by making a case distinction on whether the components have 1 or more vertices and using the first sentence.