How to prove that in a tree with $3$ leaves there's exactly one vertex of degree $3$?

Question

Given a tree of $x$ vertices which has exactly $3$ leaves, prove that the tree has exactly one vertex of degree $3$.

I thought to prove this by contradiction: suppose there're at least $2$ vertices $v$ such that $\deg(v) = 3$. Then there're two cases:

1. The vertices don't share an edge. Then there're at least $4$ leaves in the tree.
2. The vertices share an edge like in the illustration below:

But then we can rearrange the tree to have $4$ vertices.

I realize that my second case is not strong because I need to prove that all trees where at least $2$ vertices have a degree of $3$ are isomorphic to the tree in my illustration, that's why I'm not sure how to finish the proof.