Prove by induction that a tree, $T$ has either $1$ centre or $2$ adjacent centres Let $T = ( V,E)$ be a tree. A vertex $v \in V$ is called a centre if the maximum distance (i.e. the length of the shortest path) to any of the other vertices in $V$ is as small as possible. 
Show by induction on the number of vertices that $T$ either has $1$ centre or $2$ adjacent centres.
I fully understand this question and logically I can see why there will only be 1 centre or 2 adjacent centres, but can't seem to use induction to prove it. It is trivial that when $n$ (the total number of vertices) is equal to $1 \: \text{or} \: 2 $ that the statement holds true, hence we assume it true for all $n$ and show for $n+1$.
 A: Hint: Consider the tree $T^\prime$ obtained from $T$ be deleting all leaves of $T$. Show that a leaf of $T$ can't be a center of $T$ and that the center(s) of $T^\prime$ are the also centers of $T$. The base cases here are the trees with a single vertex and two vertices. 
A: We know that the maximum distance $\text{max }\text{d}(v,w)$ from a given vertex $v$ to any other vertex $w$ occurs only when $w$ is of degree $1$.
Now, let $T$ be a tree with $n\geq 2$ vertices.
So, $T$ must have at least two vertices of unit degree. Delete all such vertices from $T$, and the resulting graph $T^\prime$ is still a tree with the same centers.
Repeat this process to get trees $T\to T^\prime \to T^{\prime \prime}\to \dots$ with the same centers.
Since there are only finite number of vertices in $T$, this process will eventually halt at either a point or a line. In the first case, we have one center, and in the second case, we have two adjacent centers.
This completes the proof.
It is very easy to write this same idea in the "proof by Induction format".
