【41】102. Binary Tree Level Order Traversal

102. Binary Tree Level Order Traversal

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  • Total Accepted: 151545
  • Total Submissions: 405172
  • Difficulty: Medium
  • Contributors: Admin

 

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

 

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

 

 

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<vector<int>> levelOrder(TreeNode* root) {
13         vector<vector<int>> res;
14         if(!root) return res;
15         queue<TreeNode*> q;
16         q.push(root);
17         while(!q.empty()){
18             int size = q.size();
19             vector<int> level;
20             //level.clear();
21             for(int i = 0; i < size; i++){
22                 //vector<int> level;
23                 TreeNode* tmp = q.front();
24                 q.pop();
25                 level.push_back(tmp -> val);
26                 if(tmp -> left){
27                     q.push(tmp -> left);
28                 }
29                 if(tmp -> right){
30                     q.push(tmp -> right);
31                 }
32             }
33             res.push_back(level);
34         }
35         return res;
36     }
37 };

 

posted @ 2017-02-09 09:55  会咬人的兔子  阅读(134)  评论(0)  编辑  收藏  举报