【树】429. N叉树的层序遍历

题目：

解法：

广度优先遍历就好。

 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) {
11         val = _val;
12     }
13 
14     Node(int _val, vector<Node*> _children) {
15         val = _val;
16         children = _children;
17     }
18 };
19 */
20 
21 class Solution {
22 public:
23     vector<vector<int>> levelOrder(Node* root) 
24     {
25         if(NULL == root)
26         {
27             return {};
28         } 
29         
30         vector<vector<int>> ans;
31         
32         queue<Node*> que;
33         que.push(root);
34 
35         while(!que.empty())
36         {
37             vector<int> level;
38             for(int i=que.size(); i ; i--)
39             {
40                 //压入当前层
41                 Node* curr=que.front();
42                 que.pop();
43                 level.push_back(curr->val);
44 
45                 for(Node* it:curr->children)
46                 {
47                     que.push(it);
48                 }
49             }
50             ans.push_back(level);
51         }
52         return ans;
53     }
54 };

标签：Node,遍历,val,int,层序,429,vector,que,children
来源： https://www.cnblogs.com/ocpc/p/12821362.html