[leetcode]Binary Tree Level Order Traversal

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:

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

confused what "{1,#,2,3}" means > read more on how binary tree is serialized on OJ.

 /**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */ 
class Solution { 
public: 
    vector > levelOrder(TreeNode *root) { 
        // Start typing your C/C++ solution below  
        // DO NOT write int main() function  
        vector> result; 
        if(!root) return result; 
         
        queue q1,q2; 
        q1.push(root); 
         
        TreeNode *cur; 
        vector tmp; 
         
        while(!q1.empty()){ 
            tmp.clear(); 
            while(!q1.empty()){ 
                cur = q1.front(); 
                q1.pop(); 
                 
                tmp.push_back(cur -> val); 
                if(cur -> left) q2.push(cur -> left); 
                if(cur -> right) q2.push(cur -> right); 
            } 
            result.push_back(tmp); 
            swap(q1, q2); 
        } 
         
        return result; 
         
    } 
}; 

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector > levelOrder(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector> result;
        if(!root) return result;
       
        queue q1,q2;
        q1.push(root);
       
        TreeNode *cur;
        vector tmp;
       
        while(!q1.empty()){
            tmp.clear();
            while(!q1.empty()){
                cur = q1.front();
                q1.pop();
               
                tmp.push_back(cur -> val);
                if(cur -> left) q2.push(cur -> left);
                if(cur -> right) q2.push(cur -> right);
            }
            result.push_back(tmp);
            swap(q1, q2);
        }
       
        return result;
       
    }
};