Prove that in a tree graph $T$, a path is a hamilton path iff it is an euler path.
so I said this:
==>: Let $<x_1,x_2,...,x_n>$ be a hamilton path and let us suppose by contradiction that it is not an euler path. Then there is an edge that starts in a vertix $x_i$ that we visit twice or more, as in $<x_1,x_2,...,x_i,...,x_k,x_i,...,x_n>$, and then we have a circle $<x_i,...,x_k,x_n>$, in contradiction that $T$ is a tree.
<==: Let $<x_1,x_2,...,x_k>$ be an euler path in the tree. An euler path passes through all the edges. A tree is connected, and the degree of any vertix is $\geq 1$, so this euler path has to visit every vertix $x_i$ in the graph in order to get into it and exit out of it, thus it is a hamilton path.
Is this proof good enough?