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429. N 叉树的层序遍历(BFS)

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429. N 叉树的层序遍历

给定一个 N 叉树,返回其节点值的层序遍历。(即从左到右,逐层遍历)。

树的序列化输入是用层序遍历,每组子节点都由 null 值分隔(参见示例)。

 

示例 1:

输入:root = [1,null,3,2,4,null,5,6]
输出:[[1],[3,2,4],[5,6]]

示例 2:

输入:root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
输出:[[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]

 

提示:

 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         vector<vector<int>> ans;
25         if (root == nullptr) {
26             return ans;
27         }
28         queue<Node *> q;
29         q.push(root);
30         while (!q.empty()) {
31             int size = q.size();
32             vector<int> vec;
33             for (int i = 0; i < size; i++) {
34                 Node *tmp = q.front();
35                 q.pop();
36                 if (tmp == nullptr) {
37                     continue;
38                 }
39                 vec.push_back(tmp->val);
40                 for (auto &node : tmp->children) {
41                     if (node != nullptr) {
42                         q.push(node);
43                     }
44                 }
45             }
46             ans.push_back(vec);
47         }
48         return ans;
49     }
50 };

 

标签:Node,BFS,val,层序,429,vector,null,root,children
来源: https://www.cnblogs.com/MGFangel/p/16214889.html