首页 > 其他 > 详细

559. Maximum Depth of N-ary Tree

时间:2018-08-14 21:40:57      阅读:118      评论:0      收藏:0      [点我收藏+]

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:

 

技术分享图片

 

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 };

 

559. Maximum Depth of N-ary Tree

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

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!