Longest path length in an undirected tree, can we prove this algorithm is correct (which it is)?
I think this might help. Copied from https://stackoverflow.com/questions/20010472/proof-of-correctness-algorithm-for-diameter-of-a-tree-in-graph-theory.

Let s, t be a maximally distant pair. Let u be the arbitrary vertex. We have a schematic like,

    u
    |
    |
    |
    x
   / \
  /   \
 /     \
s       t 


where x is the junction of s, t, u (i.e. the unique vertex that lies on each of the three paths between these vertices).

Suppose that v is a vertex maximally distant from u. If the schematic now looks like

    u
    |
    |
    |
    x   v
   / \ /
  /   *
 /     \
s       t ,


then

d(s, t) = d(s, x) + d(x, t) &lt;= d(s, x) + d(x, v) = d(s, v),
where the inequality holds because d(u, t) = d(u, x) + d(x, t) and d(u, v) = d(u, x) + d(x, v). 
There is a symmetric case where v attaches between s and x instead of between x and t.

The other case looks like

    u
    |
    *---v
    |
    x
   / \
  /   \
 /     \
s       t .


Now,

d(u, s) &lt;= d(u, v) &lt;= d(u, x) + d(x, v)
d(u, t) &lt;= d(u, v) &lt;= d(u, x) + d(x, v)

d(s, t)  = d(s, x) + d(x, t)
         = d(u, s) + d(u, t) - 2 d(u, x)
        &lt;= 2 d(x, v)

2 d(s, t) &lt;= d(s, t) + 2 d(x, v)
           = d(s, x) + d(x, v) + d(v, x) + d(x, t)
           = d(v, s) + d(v, t),
so max(d(v, s), d(v, t)) >= d(s, t) by an averaging argument, and v belongs to a maximally distant pair.

PS: I saw many answers after searching the question. I do not have enough repuation to comment otherwise I would have done so.