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]]

同上一篇，只是多做了一次reverse： http://blog.csdn.net/anyicheng2015/article/details/42264721

/** * 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> > results;        if(root == NULL) return results;        queue<TreeNode *> onelevel;        onelevel.push(root);        bool shouldReverse = false;        while(!onelevel.empty()) {            // Get a level and get TreeNode of the next level.            vector<int> valOfCurrentLevel;            queue<TreeNode *> nextlevel;            while(!onelevel.empty()) {                TreeNode *front = onelevel.front();                valOfCurrentLevel.push_back(front->val);                if(front->left != NULL)                    nextlevel.push(front->left);                if(front->right != NULL)                    nextlevel.push(front->right);                onelevel.pop();            }            if(shouldReverse)                reverse(valOfCurrentLevel.begin(), valOfCurrentLevel.end());            results.push_back(valOfCurrentLevel);            onelevel = nextlevel;            shouldReverse = !shouldReverse;        }        return results;    }};