The sum of distances from a vertex in a Tree to all other vertex have the same parity. Let $G$ be a tree on $n$ vertices when $n$ is even. Then for each vertex, the sum of distances from it to all other vertex is computed. It is interesting to note that all of them have the same parity.
I saw some examples of trees and the above fact holds true. But I am unable to prove that.
I was thinking if we can use the fact that there is always a unique path from one vertex to another in a tree.
 A: For any connected bipartite graph $G=(A,B;E)$, $V(G)=A\cup B$, the following statement is true.

For any vertex $v\in A$ the sum of distances from $v$ to other
vertices of the graph and $|B|$ have the same parity.

Therefore if we take a tree with odd number of vertices, not all such sums will have the same parity.
Edit.
If the number of vertices of the tree $G$ is even, then $G$ is a bipartite graph for which at any partition $V(G)=A\cup B$ the numbers $|A|$ and $|B|$ have the same parity. Therefore in this case the above sums will have the same parity. This immediately follows from the above statement.
A: Hint
For any vertex $v\in G$, let $S(v)$ be the sum of the distances from $v$ to all other vertices. It suffices to prove that whenever $v$ and $w$ are adjacent, that $S(v)$ and $S(w)$ have the same parity.
When you remove the edge $v-w$, the rest of $G$ has two connected components, one with $v$, and one with $w$.

*

*For each vertex $x$ in the $v$ component, how does the distance of $x$ to $v$ compare to that of $x$ to $w$?


*After answering the same question for each $y$ in the $w$ component, what does this mean for the computation of $S(v)$ and $S(w)$?
You should be to give a simple description for the quantity $S(v)-S(w)$, in terms of these components. Then, the fact that $n$ is even will allow you to prove this difference is also even.
