Binary Tree Zigzag Level Order

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。