Let $T$ be a tree with even number of vertices. Prove that $T$ has "exactly" one spanning subgraph in which all vertices have odd degree.
I could only think of this result: In a tree with even number of vertices, the number of vertices with an odd number of children is odd.
But I don't know how to proceed. I'm self-studying discrete math and I found this question in a discrete math book.
EDIT:
The post linked in the comments proves existence of such tree. This question also requires the proof of uniqueness.
Here's a quick proof of uniqueness; the linked post already covers existence in multiple ways.
Let $G_1, G_2$ be two odd subgraphs of $T$. Then their symmetric difference $G_1 \mathbin{\Delta} G_2$ (which has every edge found in exactly one of $G_1$ or $G_2$) is an even subgraph of $T$: all vertices have even degree, possibly $0$.
Suppose that the even subgraph has any vertices of positive degree. Then start at one such vertex and take a walk, without repeating an edge, until you've returned to a vertex you've seen before. (You'll never get stuck, because no vertex of $G_1 \mathbin{\Delta} G_2$ has degree $1$. If you enter a vertex, you can leave it.)
But now we've found a cycle in $T$! This is impossible: trees don't have cycles. So $G_1 \mathbin{\Delta} G_2$ cannot have any vertices of positive degree - in other words, it's the empty graph. Therefore $G_1 = G_2$.