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;
}
};