[LeetCode] 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]]问题描述：给定一个二叉树，返回它的层次遍历，要求相邻的两层的访问次序不一样，一层是从左到右，另外一层是从右到左。之前做的关于层次遍历的采用的是队列，这样，每层都是同一个方向访问。在这里要求是相邻的两层的访问次序不一样。是否可以采用栈呢？比如这个例子，先访问根节点，然后入栈，然后访问根节点的左孩子和右孩子，接着也入栈，再逐个出栈并访问右孩子，再访问左孩子，也依次入栈。代码中使用了两个栈，一个栈的内容对应了一层，交替访问两个栈。
class Solution {public:    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        if(root == NULL)            return vector<vector<int> >(0, vector<int>());        stack<TreeNode *> sta1, sta2;        vector<vector<int> > vec;        vector<int> ivec;        TreeNode *pnode = NULL;        sta1.push(root);        ivec.push_back(root->val);        vec.push_back(ivec);        ivec.clear();        while(!sta1.empty()) {            while(!sta1.empty()) {                pnode = sta1.top();                sta1.pop();                if(pnode->right) {                    sta2.push(pnode->right);                    ivec.push_back(pnode->right->val);                }                if(pnode->left) {                    sta2.push(pnode->left);                    ivec.push_back(pnode->left->val);                }            }            if(!ivec.empty()) {                vec.push_back(ivec);                ivec.clear();            }            while(!sta2.empty()) {                pnode = sta2.top();                sta2.pop();                if(pnode->left) {                    sta1.push(pnode->left);                    ivec.push_back(pnode->left->val);                }                if(pnode->right) {                    sta1.push(pnode->right);                    ivec.push_back(pnode->right->val);                }            }            if(!ivec.empty()) {                vec.push_back(ivec);                ivec.clear();            }        }        return vec;    }};