**(leetcode) Binary Tree Zigzag Level Order Traversal (tree)

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]]

思路： 用两个stack来进行Z序遍历即可 

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {    void myPush(stack<TreeNode*> &s, vector<int> &v, TreeNode *root){        if(root==NULL) return;        v.push_back(root->val);        s.push(root);    }public:    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {        vector<vector<int> > ans;        if(NULL==root)  return ans;        stack<TreeNode*> s, tmp;        TreeNode *node;        vector<int> v1; //现将头结点加入        v1.push_back(root->val);        ans.push_back(v1);        int level=2; //level为偶数，这从右到左插入vector                s.push(root);        while(true){            vector<int> v;            while(!s.empty()){                node = s.top();                s.pop();                if(level%2==0){                    myPush(tmp, v, node->right);                    myPush(tmp, v, node->left);                }else{                    myPush(tmp, v, node->left);                        myPush(tmp, v, node->right);                }            }             if(tmp.empty())                return ans;            ans.push_back(v);            s = tmp;            level++;            while(!tmp.empty())                tmp.pop();        }    }};