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.

/**
 * 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> > zigzagLevelOrder(TreeNode *root) {
        vector<vector<int> > res;
        if(!root) return res;
        queue<TreeNode*> q;
        q.push(root);
        q.push(NULL); // NULL is used to gap between levels
        vector<int> level;
        bool left2right = true;
        while(true) {
            TreeNode* node = q.front(); q.pop();
            if(node) {
                level.push_back(node->val);
                if(node->left) q.push(node->left);
                if(node->right) q.push(node->right);
            }
            else {
                if(!left2right) {
                    reverse(level.begin(), level.end());
                }
                left2right = !left2right;
                res.push_back(level);
                level.clear();
                if(q.empty()) break;
                q.push(NULL);
            }
        }
    }
};