Binary Tree Zigzag Level Order Traversal

题目：

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).

结点：

class TreeNode {
public:
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {} 
};

分析：

基本题，5分钟，1次过。

解法：

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> > res;
        if (root == NULL) {
            return res;
        }
        queue<TreeNode*> Q;
        Q.push(root);
        bool needReverse = false;
        while (!Q.empty() ) {
            int size = Q.size();
            vector<int> level;
            for (int i = 0; i < size; i++) {
                TreeNode* temp = Q.front();
                Q.pop();
                level.push_back(temp->val);
                if (temp->left != NULL) {
                    Q.push(temp->left);
                }
                if (temp->right != NULL) {
                    Q.push(temp->right);
                }
            }
            if (needReverse) {
                reverse(level.begin(), level.end() );
            }
            res.push_back(level);
            needReverse = !needReverse;
        }
        return res;
    }
};