原题链接在这里：https://leetcode.com/problems/tree-diameter/
题目：
Given an undirected tree, return its diameter: the number of edges in a longest path in that tree.
The tree is given as an array of edges where edges[i] = [u, v] is a bidirectional edge between nodes u and v. Each node has labels in the set {0, 1, ..., edges.length}.
Example 1:
Input: edges = [[0,1],[0,2]] Output: 2 Explanation: A longest path of the tree is the path 1 - 0 - 2.
Example 2:
Input: edges = [[0,1],[1,2],[2,3],[1,4],[4,5]] Output: 4 Explanation: A longest path of the tree is the path 3 - 2 - 1 - 4 - 5.
Constraints:
0 <= edges.length < 10^4edges[i][0] != edges[i][1]0 <= edges[i][j] <= edges.length题解：
If it is a tree, then its nodes n = edges.length + 1.
Build the tree graph first. Then start DFS from node 0.
DFS returns the deepest depth. DFS state needs current node, tree graph, and parent.
For current node, for all its neighbors, as long as it is not parent, get the depth from it, pick 2 largest.
Use these 2 largest to update the diameter.
And returns the largest depth + 1.
Time Complexity: O(n). n = edges.length.
Space: O(n).
AC Java:
1 class Solution { 2 int res = 0; 3 4 public int treeDiameter(int[][] edges) { 5 if(edges == null || edges.length == 0){ 6 return 0; 7 } 8 9 int n = edges.length + 1; 10 ArrayList<Integer>[] graph = new ArrayList[n]; 11 for(int i = 0; i < n; i++){ 12 graph[i] = new ArrayList<Integer>(); 13 } 14 15 for(int [] e : edges){ 16 graph[e[0]].add(e[1]); 17 graph[e[1]].add(e[0]); 18 } 19 20 dfs(0, graph, -1); 21 22 return res; 23 } 24 25 private int dfs(int root, ArrayList<Integer>[] graph, int parent){ 26 int max1 = 0; 27 int max2 = 0; 28 for(int next : graph[root]){ 29 if(next != parent){ 30 int depth = dfs(next, graph, root); 31 if(depth > max1){ 32 max2 = max1; 33 max1 = depth; 34 }else if(depth > max2){ 35 max2 = depth; 36 } 37 } 38 } 39 40 res = Math.max(res, max1 + max2); 41 return max1 + 1; 42 } 43 }
原文：https://www.cnblogs.com/Dylan-Java-NYC/p/12365031.html