Every tree a path exists which connects two leaves and removing that path the remaining graph is still a tree?

Question

I wanted to prove that that in any tree $G$ with $2k$ leaves, there are $k$ edge-disjoint paths with endpoints being leaves of $G$.

To solve this I tried to use induction. In the induction step there are $2k + 2$ leaves. Now I want to find a path $P$ which connects two leaves and removing it leaves behind a tree with $2k$ or $2k +1$ leaves.

Such a path has to contain at most one vertex with degree $\ge 3$. Otherwise the remaining graph could be a forest.

So I have tried to solve it algorithmically. Let $P$ be a path which connects two leaves $a$ and $b$. If $P$ contains multiple vertices with degree $\ge 3$, remove the edge which connects the one vertex $s$ with degree $\ge 3$ with next one. So we have now a path from $a$ to $s$. The $s$' degree is at least $2$. So there is path to another leaf $c$. If there is again a vertex $t$ we remove the edge leading to $s$. And now we have a path from $c$ to $t$. If we repeat it multiple times we will get a path with at most one vertex with degree $\ge 3$.

Removing the path means that we remove everything except the vertex with degree $\ge 3$.

Answer 1

What does your theorem state about this tree?

Answer 2

This is false (even for undirected trees, which is presumably the intended setting). To see this, take a star with at least five vertices. Any path between leaves has three vertices, and after removing the path the remaining graph isn't even connected.