Diameter of a tree $$T=(V,E) \text{ tree }$$
$$\text{diameter of a tree } = \max_{u,v \in V} \delta(u,v)$$
$$\delta(u,v)=\text{the length of the shortest path from the vertex u to the vertex v}$$
How can we calculate the diameter of a tree,when we are given the algorithm of the Breadth-first-search ?

Breadthfirstsearch(G,s)
for each u ∈ V \ {s}
     color[u]<-white
     d[u]<-oo
     p[u]<-Ø
color[s]<-gray
d[s]<-0
p[s]<-Ø
Q<-Ø 
Insert(Q,s)
while Q ≠ Ø
       u<-Del(Q)
       for each v  ∈ Adj(u)
          if color[v]=white then
             color[v]<-gray
             d[v]=d[u]+1
             p[v]<-u
             Insert(Q,v)
       color[u]<-black

According to my notes,we could do the following:


*

*We implement the Breadth-first search from a given initial node $s$.
  We calculate $d$ for all the vertices.
  Let $u$ the vertex,such that: $$\delta(s,u)=d[u]=\max_{v \in V} d[v]$$

*We implement the Breadth-first search with the vertex $u$,as the initial node.
Let $w$ the vertex,such that:
$$\delta(u,w)=d[w]=\max_{v \in V} d[v]$$
Then,the diameter of $T$ is equal to $\delta(u,w)$.
Could you explain me why we calculate the diameter in this way? 
 A: I tried to find this result online to no avail. If someone knows where this is originally proved, please let me know and I'll edit this post.
Let $s$ be some generic vertex of the tree $T$. Let $u$ be a vertex so that 
$$
\delta(s,u)=\max_v \delta(s,v)
$$
and $w$ be a vertex so that 
$$
\delta(u,w)=\max_v \delta(u,v).
$$
So we want to show that $\delta(u,w)$ is the diameter of $T$. One important fact that we'll be using is that between any two vertices in a tree, there is a $unique$ path between them. 
Suppose that $P:(u=v_0, v_1, \ldots, v_{\ell}=w)$ are the vertices, in order, of the path from $u$ to $w$. We partition the vertices based on which $v_i$ they are closest to. In particular, $V_i:=\{v \in V: \delta(v,v_i) = \min_j\delta(u,v_j)\}$. Note that these $V_i$ are disjoint (or else there would be cycles). I claim that for each $0\leq i \leq \ell$, $$d_i:=\max_{v \in V_i} \delta(v_i, v) \leq \min\{i, \ell-i\}.$$
First, let $k$ be so that $s \in V_k$. Note that $k \geq \ell/2$ or else $w$ would be further than $s$ than $u$. So for $k$, we only need to show that $d_k \leq \ell-k$. 
For $i \neq k$, if $d_i >i$, then there exists a vertex $v \in V_i$, so that the path from $s$ to $v$, which goes from $s$ to $v_k$ to $v_i$ to $v$ is longer than the path from $s$ to $u$, which is a contradiction. 
Now if there is a $d_i$ so that $d_i > \ell-i$, then there would be a path from $u$ to $v_i$ to this vertex which has length longer than the path from $u$ to $w$, which also is a contradiction.
To finish this proof off, we need to show why this condition on $d_i$ forces the diameter to be $\ell$. Suppose that $v\in V_i$ and $t \in V_j$ for some $0\leq i \leq j \leq \ell$. Then there is a path from $v$ to $v_i$ of length at most $i$, a path from $v_i$ to $v_j$ of length $j-i$, and a path from $v_j$ to $t$ of length at most $\ell-j$. Adding these paths together, gives us a walk of length at most $\ell$.
