LC 971. Flip Binary Tree To Match Preorder Traversal
Given a binary tree with N nodes, each node has a different value from {1, ..., N}.
A node in this binary tree can be flipped by swapping the left child and the right child of that node.
Consider the sequence of N values reported by a preorder traversal starting from the root. Call such a sequence of N values the voyage of the tree.
(Recall that a preorder traversal of a node means we report the current node's value, then preorder-traverse the left child, then preorder-traverse the right child.)
Our goal is to flip the least number of nodes in the tree so that the voyage of the tree matches the voyage we are given.
If we can do so, then return a list of the values of all nodes flipped. You may return the answer in any order.
If we cannot do so, then return the list [-1].
Runtime: 4 ms, faster than 99.80% of C++ online submissions for Flip Binary Tree To Match Preorder Traversal.
//
// Created by yuxi on 2019-01-18.
// #include <vector>
#include <unordered_map>
using namespace std; /**
* 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 {
private:
unordered_map<int,int> mp;
public:
vector<int> flipMatchVoyage(TreeNode* root, vector<int>& voyage) {
for(int i=; i<voyage.size(); i++) mp[voyage[i]] = i;
vector<int> ret;
bool flag = helper(ret, root, );
if(!flag) return {-};
return ret;
}
bool helper(vector<int>& ret, TreeNode* root, int idx){
if(!root) return true;
if(mp[root->val] != idx) return false;
int left = root->left ? mp[root->left->val] : -;
int right = root->right ? mp[root->right->val] : -;
if((left >= && left < mp[root->val]) || (right >= && right < mp[root->val])) return false;
if(left == - && right == -) return true;
else if(left == - && right != -) {
if(right != idx+) return false;
return helper(ret, root->right, mp[root->right->val]);
}
else if(left != - && right == -) {
if(left != idx+) return false;
return helper(ret, root->left, mp[root->left->val]);
}
else if(left > right) {
if(right != idx+) return false;
ret.push_back(root->val);
return helper(ret, root->right, idx+) && helper(ret, root->left, mp[root->left->val]);
}
else {
if(left != idx+) return false;
return helper(ret, root->left, idx+) && helper(ret, root->right, mp[root->right->val]);
}
}
};
LC 971. Flip Binary Tree To Match Preorder Traversal的更多相关文章
- LeetCode 971. Flip Binary Tree To Match Preorder Traversal
原题链接在这里:https://leetcode.com/problems/flip-binary-tree-to-match-preorder-traversal/ 题目: Given a bina ...
- 【LeetCode】971. Flip Binary Tree To Match Preorder Traversal 解题报告(C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 前序遍历 日期 题目地址:https://leetc ...
- [Swift]LeetCode971.翻转二叉树以匹配先序遍历 | Flip Binary Tree To Match Preorder Traversal
Given a binary tree with N nodes, each node has a different value from {1, ..., N}. A node in this b ...
- 【leetcode】Binary Tree Zigzag Level Order Traversal
Binary Tree Zigzag Level Order Traversal Given a binary tree, return the zigzag level order traversa ...
- 37. Binary Tree Zigzag Level Order Traversal && Binary Tree Inorder Traversal
Binary Tree Zigzag Level Order Traversal Given a binary tree, return the zigzag level order traversa ...
- Binary Tree Zigzag Level Order Traversal (LeetCode) 层序遍历二叉树
题目描述: Binary Tree Zigzag Level Order Traversal AC Rate: 399/1474 My Submissions Given a binary tree, ...
- 剑指offer从上往下打印二叉树 、leetcode102. Binary Tree Level Order Traversal(即剑指把二叉树打印成多行、层序打印)、107. Binary Tree Level Order Traversal II 、103. Binary Tree Zigzag Level Order Traversal(剑指之字型打印)
从上往下打印二叉树这个是不分行的,用一个队列就可以实现 class Solution { public: vector<int> PrintFromTopToBottom(TreeNode ...
- 【LeetCode】103. Binary Tree Zigzag Level Order Traversal
Binary Tree Zigzag Level Order Traversal Given a binary tree, return the zigzag level order traversa ...
- [LeetCode] Binary Tree Level Order Traversal 与 Binary Tree Zigzag Level Order Traversal,两种按层次遍历树的方式,分别两个队列,两个栈实现
Binary Tree Level Order Traversal Given a binary tree, return the level order traversal of its nodes ...
随机推荐
- CHD-5.3.6集群上Flume安装
Flume is a distributed, reliable, and available service for efficiently collecting, aggregating, and ...
- 第七章、Ajango自带auth模块
目录 第七章.Ajango自带auth模块 一.什么是auth auth是django自带的用户认证模块 二.auth模块的常用方法 三.拓展默认的auth_user表 第七章.Ajango自带aut ...
- 浅谈String、StringBuffer与StringBuilder
浅谈String.StringBuffer与StringBuilder 先详细介绍一下String.StringBuffer与StringBuilder String: 官方对String的说明: ...
- 如何处理不能新建word、excel、PPT的情况?
Office系列办公软件是大家都非常喜欢使用的软件,但是有些朋友反映在使用电脑时,在桌面右键菜单新建选项里没有Word.Excel或PPT,非常的耽误工作. 下面就为大家介绍一下桌面右键菜单新建选项里 ...
- redis和memcacahe、mongoDB的区别
都是非关系型数据库,性能都非常高,但是mongoDB和memcache.redis是不同的两种类型.后两者主要用于数据的缓存,前者主要用在查询和储存大数据方面,是最接近数据库的文档型的非关系数据库. ...
- web开发: css高级与盒模型
一.组合选择器 二.复制选择器优先级 三.伪类选择器 四.盒模型 五.盒模型显示区域 六.盒模型布局 一.组合选择器 <!DOCTYPE html> <html> <he ...
- Hadoop_12_Hadoop 中的RPC框架演示
Hadoop中自己提供了一个RPC的框架.集群中各节点的通讯都使用了那个框架 1.服务端 1.1.业务接口:ClientNamenodeProtocol package cn.bigdata.hdfs ...
- 从程序员之死看 IT 人士如何摆脱低情商诅咒
(1) IT公司的创业者苏享茂忽然跳楼自杀了,自杀前,他留下几万字的文字记录.遗书,并且在自己开发的软件界面上,设置了弹出页面,控诉是恶毒前妻逼死了自己. 生命戛然而止,留给亲人痛苦,留给世人震惊. ...
- MySQL 5.6, 5.7, 8.0版本的新特性汇总大全
转载:http://blog.itpub.net/15498/viewspace-2650661/ MySQL 5.6 1).支持GTID复制 2).支持无损复制 3).支持延迟复制 4).支持基于库 ...
- [NOI2008]假面舞会——数论+dfs找环
原题戳这里 思路 分三种情况讨论: 1.有环 那显然是对于环长取个\(gcd\) 2.有类环 也就是这种情况 1→2→3→4→5→6→7,1→8→9→7 假设第一条链的长度为\(l_1\),第二条为\ ...