I am trying to prove that the number of edges in a tree is $n-1$ where $n$ is the number of vertices. I do not wish to use induction.
I already have established that a tree is a planar graph. Now my plan is as follows. (It is motivated by the proof here). Put a $+1$ charge on each vertex and a $-1$ charge on each edge. Put a $+1$ charge on the outside region. Now show that except for the two end vertices of the longest path in the tree, the net charge due to the vertices and the edges is $-1$. This cancels with the $+1$ charge on the outside region. So the total charge is $0$. Bear in mind that we have not considered the two positive charges on account of the two end vertices of the longest path.
Hence total positive charge = $(n-2)+1$ = Total negative charge = $e$ and so $e=n-1$.
In this proof how do I formally prove that except for the two end vertices of the longest path in the tree, the net charge due to the vertices and the edges is $-1$?