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]
]

Solution: 1. Use queue. In order to seperate the levels, use 'NULL' as the end indicator of one level.
2. DFS.

/**
 * 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<vector<int> > levelOrder(TreeNode *root) {
        vector<vector<int> >res;
        levelOrderRe(root, 0, res);
        return res;
    }
    
    void levelOrderRe(TreeNode* root, int level, vector<vector<int> >& res)
    {
        if(!root) return;
        if(res.size() <= level) {
            res.push_back(vector<int>());
        }
        res[level].push_back(root->val);
        levelOrderRe(root->left, level+1, res);
        levelOrderRe(root->right, level+1, res);
    }
};