Binary Tree Zigzag Level Order Traversal

题目大意：按从左到右，然后从右到左层次遍历二叉树

解题思路：用两个栈来保存每层的子节点。

用left2right表示从左到右的栈

right2left表示从右到左的栈。

当left2right 不为空时，顺序弹出left2right 中的元素，并把它的左节点和右节点顺序压入right2left栈中

否则当right2left不为空时，顺序弹出right2left中的元素,并把它的右节点和左节点顺序压入left2right 栈中

/**
 * 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> > result;
        if(root == NULL) {
            return result;
        }
        stack<TreeNode *> left2right;
        stack<TreeNode *> right2left;
        left2right.push(root);
        while(!left2right.empty() || !right2left.empty()) {
            TreeNode *tmp = NULL;
            vector<int> vs;
            if(!left2right.empty()) {
                while(!left2right.empty()) {
                    tmp = left2right.top();
                    left2right.pop();
                    vs.push_back(tmp->val);
                    if(tmp->left != NULL) {
                        right2left.push(tmp->left);
                    }
                    if(tmp->right != NULL) {
                        right2left.push(tmp->right);
                    }
                }
            } else {
                while(!right2left.empty()) {
                    tmp = right2left.top();
                    right2left.pop();
                    vs.push_back(tmp->val);
                    if(tmp->right != NULL) {
                        left2right.push(tmp->right);
                    }
                    if(tmp->left != NULL) {
                        left2right.push(tmp->left);
                    }
                }
            }
            result.push_back(vs);
        }
        return result;
    }
};