Prove: Let $T$ be a tree, show that each Hamiltonian path is also Eulerian

Question

I want to be sure if my proof is correct.

Let's assume that in $T$ (tree graph) there is an Hamiltonian path. Show that the path is also Eulerian.

Let's look on the subgraph of the Hamiltonian path. It's a simple path: it has $2$ vertices with degree $1$ and $n-2$ vertices with degree $2$. The degree sum of this subgraph is $2n-2$.

Now, If it's not Eulerian path then there is at least one more edge in $T$ graph. If so, the degree sum of $T$ is at least $n$.

This is a contradiction because $T$ is a tree (degree sum of every tree is $2n-2$).

Therefore, this path must be also Eulerian path.

Does it make any sense?

Thanks.

Answer 1

Yes, your proof is correct.

For completeness, here's a slightly simpler proof:

Assume that a tree $T$ has a Hamiltonian path $P$. All vertices of $T$ must be on $P$. Then any edge of $T$ not in $P$ must be between two vertices of $P$, but this would create a cycle in $T$, a contradiction with $T$ being a tree.