Binary Tree Zigzag Level Order Traversal

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left
to right, then right to left for the next level and alternate between).
For example:
Given binary tree {3,9,20,#,#,15,7},
　　3
　 /
9 20
　　/
　 15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]

Solution: 1. Queue + reverse.
2. Two stacks.
3. Vector. Contributed by yinlinglin.

 1 /**
 2  * Definition for binary tree
 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> > zigzagLevelOrder(TreeNode *root) {
13         vector<vector<int> > res;
14         if(!root) return res;
15         queue<TreeNode*> q;
16         q.push(root);
17         q.push(NULL); // NULL is used to gap between levels
18         vector<int> level;
19         bool left2right = true;
20         while(true) {
21             TreeNode* node = q.front(); q.pop();
22             if(node) {
23                 level.push_back(node->val);
24                 if(node->left) q.push(node->left);
25                 if(node->right) q.push(node->right);
26             }
27             else {
28                 if(!left2right) {
29                     reverse(level.begin(), level.end());
30                 }
31                 left2right = !left2right;
32                 res.push_back(level);
33                 level.clear();
34                 if(q.empty()) break;
35                 q.push(NULL);
36             }
37         }
38     }
39 };