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]
]

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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
//和前一题，lever order traversal相似，不同的是需要一个isReverse的boolean型变量来记录是否要反转当前的层的节点。
//翻转的话就insert到首位，不翻转的话就push_back
//lastNode用来记录当前层的最后一个节点，用来确定是否遍历到了新的层。
class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        queue<TreeNode *> q;
        vector<vector<int > > re;
        vector<int> thisRe;
        bool isReverse = false;
        if(root == NULL){
            return re;
        }
        q.push(root);
        TreeNode * lastNode = root;
        while(!q.empty()){
            TreeNode * p = q.front();
            q.pop();
            if(isReverse == false){
                thisRe.push_back(p->val);
            }
            else{
                thisRe.insert(thisRe.begin(), p->val);
            }
            if(p->left){
                q.push(p->left);
            }
            if(p->right){
                q.push(p->right);
            }
            if(p == lastNode){
                re.push_back(thisRe);
                thisRe.clear();
                isReverse = isReverse == true?false:true;
                if(!q.empty()){
                    lastNode = q.back();    
                }
            }
        }
        return re;
    }
};