On Trees

A tree $T$ is $(V,E)$ such that $|E|=|V|-1$ and $T$ is connected.
A tree is a connected acyclic graph. Meaning that between any two vertices, there is a unique path.

The diameter of a tree is the length of the longest path between any two leaves of the tree. It is the longest path in the tree itself.
Theorem: $\mathrm{\exists }a,b\in V$ where $a,b$ are unique and are leaves such that $a\to b$ is the diametric path.
proof: if $a\to b^{\prime }$ is a diametric path, a is not a leaf, and $b\to a\to b^{\prime }$ is a longer leaf to leaf path. if $a^{\prime }\to b^{\prime }$ is a diametric path then $a\to b\to a^{\prime }\to b^{\prime }$ is a longer path, both cases are contradictions.

def: Centre of a tree. The centre of a tree is any vertex $v\ast$ such that the maximum of all path lengths from $v\ast$ to any other node is minimum. denote path length as $L(u\to v)$.

Theorem, Any Centre of a tree lies on the diametric path.

The picture above shows the centre $C$ not on the diametric path, but is connected to some $P_i$ on the diametric path. let $Z$ denote the larger of the two distances $P_i\to P_1$ or $P_i\to P_{D+1}$.
and $max(y)=Y$ the maximum of all path lengths from $C$ to vertices not on the diametric path. since $Z\ge \lceil \frac{D}{2}\rceil$, it has to be that $x+Y\le Z$. Therefore, the longest path from $C$ is $C\to P_i\to P_{D+1}$ of length $(x+Z)$ but given $x+Y\le Z$, the longest path length from $P_i$ to any vertex is $Z$.
Hence $C$ is not a centre, giving the desired contradiction.
Moreover, the centre $C$ lies in the middle of the diametric path.
That is if $D$ is odd, there are two centres, both have a depth (max path length to any other node) of $\lceil \frac{D}{2}\rceil$.
if $D$ is even, then there is centre at $D/2$.