103. Binary Tree Zigzag Level Order Traversal

题目：Binary Tree Zigzag Level Order Traversal

原题描述

原题链接：https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/#/description

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,null,null,15,7],

return its zigzag level order traversal as:
[[3],[20,9], [15,7]]

给出一颗二叉树，返回对二叉树进行Z字形层序遍历的结果。每一层的数据作为一个数组输出。
Z字形层序遍历就是，比如第一层层序遍历的顺序是从左到右，第二层就从右到左，第三层从左到右，依次循环往复。
例：二叉树如上图，则遍历结果是[[3],[20,9], [15,7]]

思路

http://blog.csdn.net/gcs6564157/article/details/62451157
和这一题的思路很像，只不过是需要每隔一层就反转一下数组。可以用一个cnt记录当前层数，初始第0层，每遍历到新的一层cnt+1，当cnt & 1为真时就反转数组。

参考代码

/** * Definition for a binary tree node. * 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>> ans;        if(!root) return ans;        vector<int> temp;        queue<TreeNode*> q;        q.push(root);        int cnt = 0;        TreeNode* firstLeft = NULL;        bool findFirst = false;        while(!q.empty()) {            TreeNode* front = q.front();            q.pop();            if(front == firstLeft) {                if(cnt & 1) {                    reverse(temp.begin(), temp.end());                }                ans.push_back(temp);                temp.clear();                firstLeft = NULL;                findFirst = false;                cnt++;            }            temp.push_back(front->val);            if(front->left) {                if(!findFirst) {                    firstLeft = front->left;                    findFirst = true;                }                q.push(front->left);            }            if(front->right) {                if(!findFirst) {                    firstLeft = front->right;                    findFirst = true;                }                q.push(front->right);            }        }        if(cnt & 1) {            reverse(temp.begin(), temp.end());        }        ans.push_back(temp);        return ans;    }};