leetcode笔记：Binary Tree Zigzag Level Order Traversal

一. 题目描述

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

二. 题目分析

BFS，这里用一个bool记录是从左到右还是从右到左读取数据，每一层遍历结束就翻转一下。

三. 示例代码

#include <iostream>#include <vector>using namespace std;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> > result;        Traverse(root, 1, result, true);        return result;    }private:    void Traverse(TreeNode *root, size_t level, vector<vector<int> > & result, bool leftToRight)    {        if (root == NULL) return;        if (level > result.size())            result.push_back(vector<int>());        if (leftToRight)            result[level - 1].push_back(root->val);        else            result[level - 1].insert(result[level - 1].begin(), root->val);        Traverse(root->left, level + 1, result, !leftToRight);        Traverse(root->right, level + 1, result, !leftToRight);    }};

四. 小结

比Binary Tree Level Order Traversal 和Binary Tree Level Order Traversal II 稍微难一些，需要注意更多细节。