Show that if a tree is balanced than each of its partite sets contains a leaf. I am asked to show that the following holds. Show that if a tree is balanced than each of its partite sets contains a leaf. Note balanced means that $|U| = |W| $ where $U$ and $W$ are $G$'s partite sets. Now what i originally tried to do was use the fact that in a tree basically every vertex is a cut vertex but apparently i'm not allowed to do that because in this particular course we haven't learned what a cut vertex is yet. i should be allowed to use the concept of removing a vertex from the graph just not it being a cut vertex specifically. i tried considering the longest path and showing that there was a leaf on each end of it but then i couldn't show that one end was in $U$ and the other was in $W$. Any advice on how to complete this?

Notice that the sum of the degrees in each part is the same, since every edge crosses from one part to the other. We know that a tree with $n$ nodes has $n{-}1$ edges so the sum of the degrees in each part is also $n{-}1$, since every edge crosses from one part to the other. Then in each part we have $n/2$ nodes with degree values that sum to $n{-}1$ and by a version of the pigeonhole principle, there must be at least one vertex of degree one (a leaf) in each part.