标签：int tco vector ble des color alt find height
题目描述：
Given a n-ary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
For example, given a 3-ary tree:
We should return its max depth, which is 3.
Note:
1000.5000.
解题思路：
使用DFS方法。
代码：
1 /* 2 // Definition for a Node. 3 class Node { 4 public: 5 int val; 6 vector<Node*> children; 7 8 Node() {} 9 10 Node(int _val, vector<Node*> _children) { 11 val = _val; 12 children = _children; 13 } 14 }; 15 */ 16 class Solution { 17 public: 18 int maxDepth(Node* root) { 19 int depth = 0; 20 dfs(root, depth); 21 return max; 22 } 23 void dfs(Node* root, int& depth) { 24 if (root == NULL) 25 return; 26 depth += 1; 27 for (auto child : root->children) { 28 dfs(child, depth); 29 depth -= 1; 30 } 31 if (root->children.size() == 0) { 32 if (depth > max) { 33 max = depth; 34 } 35 } 36 } 37 int max = 0; 38 };
559. Maximum Depth of N-ary Tree
标签：int tco vector ble des color alt find height
原文地址：https://www.cnblogs.com/gsz-/p/9477886.html
