lintcode 二叉树的锯齿形层次遍历

给出一棵二叉树，返回其节点值的锯齿形层次遍历（先从左往右，下一层再从右往左，层与层之间交替进行）
样例
给出一棵二叉树 {3,9,20,#,#,15,7},
3
/
9 20
/
15 7
返回其锯齿形的层次遍历为：
[
[3],
[20,9],
[15,7]
]
题目链接：http://www.lintcode.com/zh-cn/problem/binary-tree-zigzag-level-order-traversal/#
和层次遍历的道理相同，只不过我们需要设置一个bool变量，来判断在该行是正序遍历还是倒序遍历，维护一个双向队列。

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
 

class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: A list of lists of integer include 
     *          the zigzag level order traversal of its nodes' values 
     */
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode *root) {
        // write your code here
        vector<vector<int>> result;
        if(root == NULL) return result;
        deque<TreeNode *> q;
        q.push_back(root);
        bool d = true;
        int len;
        while(!q.empty()) {
            vector<int> res;
            len = q.size();
            TreeNode *temp;
            while (len--) {
                if (d) {
                    temp = q.front();
                    q.pop_front();
                    res.push_back(temp->val);
                    if (temp->left) q.push_back(temp->left);
                    if (temp->right) q.push_back(temp->right);
                }
                else {
                    temp = q.back();
                    q.pop_back();
                    res.push_back(temp->val);
                    if (temp->right) q.push_front(temp->right);
                    if (temp->left) q.push_front(temp->left);
                }
            }
            result.push_back(res);
            d = !d;
        }
        return result;
    }
};