Tree diameter LeetCode 5098.

Address  https://leetcode-cn.com/contest/biweekly-contest-12/problems/tree-diameter/

You tree "No to the tree," you measure and return to its "diameter": the number of sides this tree longest simple path.

We use an array of all "sides" of the composition  edges expressed an undirected tree, which  edges[i] = [u, v] represents a node  u and  v two-way between the edges.

The tree nodes have been used  {0, 1, ..., edges.length} to make the tag number, marked on each node is unique.

Example 1

Input: Edges = [[ 0 , . 1 ], [ 0 , 2 ]]
Output: 2
Explanation:
This tree is the longest path . 1 - 0 - 2 , the number of sides is 2 .

Example 2

Input: Edges = [[ 0 , . 1 ], [ . 1 , 2 ], [ 2 , . 3 ], [ . 1 , . 4 ], [ . 4 , . 5 ]]
Output: 4
Explanation: 
This tree is the longest path . 3 - 2 - . 1 - 4 - . 5 , the number of sides of 4 .

Algorithm 1
seeking maximum distance no tree is divided into two steps to
a point BFS or DFS optionally acquire the point furthest away from the point
2 to the point from the first step or starting again BFS DFS acquired farthest point. Both distance is the maximum distance

C ++ code

 1 class Solution {
 2 public:
 3     pair<int, int> bfs(vector<vector<int>>& e, int start) {
 4         vector<int> d(e.size(), -1);
 5 
 6         queue<int> Q;
 7         Q.push(start);
 8         d[start] = 0;
 9 
10         pair<int, int> ret;
11 
12         while (!Q.empty()) {
13             int x = Q.front();
14             Q.pop();
15             ret.first = x;
16             ret.second = d[x];
17             for (auto& y : e[x]) {
18                 if (d[y] == -1) {
19                     d[y] = d[x] + 1;
20                     Q.push(y);
21                 }
22             }
23         }
24         return ret;
25     }
26 
27     int treeDiameter(vector<vector<int>>& edges) {
28         int n = edges.size() + 1;
29         vector<vector<int>> e(n, vector<int>());
30         for (auto& edge: edges) {
31             e[edge[0]].push_back(edge[1]);
32             e[edge[1]].push_back(edge[0]);
33         }
34 
35         pair<int, int> p;
36         p = bfs(e, 0);
37         p = bfs(e, p.first);
38 
39         return p.second;
40     }
41 };

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