# Description

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,#,#,15,7},
3
/
9 20
/
15 7
return its level order traversal as:
[
,
[9,20],
[15,7]
]

The original problem is here.

The original code is here.

# My Solution

I solve this problem in C++, as below:

``````/*
*Binary Tree Level Order Traversal
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
*/
#include<iostream>
#include<stack>
#include<vector>
#include<stdlib.h>
using namespace std;

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<int> vec;
vector<vector<int>> res;
TreeNode* node;
vector<TreeNode*> currentLevel, nextLevel;
if(root==NULL)
return res;
if(root!=NULL)
currentLevel.push_back(root);
do{
for(int i=0;i<currentLevel.size();i++){
node = currentLevel[i];
vec.push_back(node->val);
if(node->left!=NULL)
nextLevel.push_back(node->left);
if(node->right!=NULL)
nextLevel.push_back(node->right);
}
res.push_back(vec);
vec.clear();
currentLevel.clear();
currentLevel = nextLevel;
nextLevel.clear();
}while(!currentLevel.empty());

return res;
}
};
``````

# Note

To solve the problem, a vector is needed to save tree nodes at one level.