103. Binary Tree Zigzag Level Order Traversal

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题目

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],

    3   / \  9  20    /  \   15   7

return its zigzag level order traversal as:

[  [3],  [20,9],  [15,7]]

翻译

给定一个二叉树，返回其节点值的Z形层次顺序遍历。（即从左到右，然后从右到左，进行下一个级别的交替）。

分析

类似于上一题的递归遍历，此题设置一个level

level=偶数,下一层从右到左

level=奇数，下一层从左到右

犯的错误

没有想清楚这个level作为递归的key，判断条件应该是它本轮的值

一开始弄反了

/** * 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>> res;    void buildOrder(TreeNode* root, int level){        if(root==NULL)   return;        if(res.size()==level)   res.push_back({});        res[level].push_back(root->val);        if(level%2==1){            buildOrder(root->left,level+1);            buildOrder(root->right,level+1);        }        else{            buildOrder(root->right,level+1);            buildOrder(root->left,level+1);        }    }    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {        buildOrder(root,0);        return res;    }};

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