559. Maximum Depth of N-ary Tree

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:

1. The depth of the tree is at most 1000.
2. The total number of nodes is at most 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 };

原文地址：https://www.cnblogs.com/gsz-/p/9477886.html