You can solve this problem using DP on trees, check this tutorial

solition in O(n log n)

lets find the answer for node V. V has O(logN) nodes in path from V to root. Let U — node in this path. then also let's find for each V max_d[V] — max len from V to some node in subtree of V.

let's try to update the answer for V. there's a exactly one son of U, who is in the path from V to root, let it be Z. then there's the algorithm to find the ans for V, if U is known.

for (int to : edges[U]) {
    if (to != Z)
        ans[V] = max(ans[V], len(V, U) + max_d[to] + 1);
}

i think it's pretty obvious why it works. so then whole alg is this:

ans[v] = max_d[v];
for (int U : path[V]) {
    for (int to : edges[U]) {
        if (to != Z)
            ans[V] = max(ans[V], len(V, U) + max_d[to] + 1);
    }
}