LeetCode: Binary Tree Level Order Traversal

Problem: 

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]
]

分层遍历二叉树，用队列模拟。

/**
 * 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< vector > result;
        if (root == NULL)
            return result;
        
        queue q;
        q.push(root);
        int count = 1;
        int level = 0;
        vector tmp(0);
        while(!q.empty())
        {
            tmp.clear();
            level = 0;
            for (int i = 0; i < count; ++i)
            {
                root = q.front();
                q.pop();
                tmp.push_back(root->val);
                if (root->left != NULL)
                {
                    q.push(root->left);
                    ++level;
                }
                if (root->right != NULL)
                {
                    q.push(root->right);
                    ++level;
                }
            }
            count = level;
            result.push_back(tmp);
        }
        return result;
    }
};