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]
]

這道二叉樹的之字形層序遍歷是之前那道[LeetCode] Binary Tree Level Order Traversal 二叉樹層序遍歷的變形，不同之處在於一行是從左到右遍歷，下一行是從右往左遍歷，交叉往返的之字形的層序遍歷。根據其特點我們用到棧的後進先出的特點，這道題我們維護兩個棧，相鄰兩行分別存到兩個棧中，進棧的順序也不相同，一個棧是先進左子結點然後右子節點，另一個棧是先進右子節點然後左子結點，這樣出棧的順序就是我們想要的之字形了，程式碼如下：

/**
 * 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;
        stack<TreeNode*> s1;
        stack<TreeNode*> s2;
        s1.push(root);
        vector<int> out
 
;
        while (!s1.empty() || !s2.empty()) {
            while (!s1.empty()) {
                TreeNode *cur = s1.top();
                s1.pop();
                out.push_back(cur->val);
                if (cur->left) s2.push(cur->left);
                if (cur->right) s2.push(cur->right);
            } 
            if (!out.empty()) res.push_back(out);
            out.clear();
            while (!s2.empty()) {
                TreeNode *cur = s2.top();
                s2.pop();
                out.push_back(cur->val);
                if (cur->right) s1.push(cur->right);
                if (cur->left) s1.push(cur->left);
            }
            if (!out.empty()) res.push_back(out);
            out.clear();
        }
        return res;
    }
};

比如對於題幹中的那個例子：

    3
   / \
  9  20
    /  \
   15   7

我們來看每一層兩個棧s1, s2的情況：

s1:　　3

s2:

s1:　　

s2:　　9　　20

s1:　　7　　15

s2: