The following is from Max Howell @twitter:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can't invert a binary tree on a whiteboard so fuck off.

Now it's your turn to prove that YOU CAN invert a binary tree!

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=10) which is the total number of nodes in the tree -- and hence the nodes are numbered from 0 to N-1. Then N lines follow, each corresponds to a node from 0 to N-1, and gives the indices of the left and right children of the node. If the child does not exist, a "-" will be put at the position. Any pair of children are separated by a space.

Output Specification:

For each test case, print in the first line the level-order, and then in the second line the in-order traversal sequences of the inverted tree. There must be exactly one space between any adjacent numbers, and no extra space at the end of the line.

Sample Input:

8

1 -

- -

0 -

2 7

- -

- -

5 -

4 6

Sample Output:

3 7 2 6 4 0 5 1

6 5 7 4 3 2 0 1

分析

用vector Left和right来记录每个节点的左孩子节点和右孩子节点,利用根节点不是任何节点的子节点来找到根节点root,在从根节点开始将节点依次插入树中,最后就是基本的层序遍历和中序遍历的实现了。

#include<iostream>
#include<vector>
#include<queue>
#include<string.h>
using namespace std;
vector<int> Left;
vector<int> Right;
int flag=1;
class node{
public:
int value;
node* left=NULL;
node* right=NULL;
node(int val):value(val), left(NULL), right(NULL){
}
};
void CreateTree(node* root){
if(Left[root->value]!=-1){
root->left=new node(Left[root->value]);
CreateTree(root->left);
}
if(Right[root->value]!=-1){
root->right=new node(Right[root->value]);
CreateTree(root->right);
}
}
void inorder(node* root){
if(root->left)
inorder(root->left);
flag++==1?cout<<root->value:cout<<" "<<root->value;
if(root->right)
inorder(root->right);
}
int main(){
int N;
cin>>N;
vector<int> record(N,0);
Left.resize(N);
Right.resize(N);
for(int i=0; i<N; i++){
char l, r;
cin>>l>>r;
if(isdigit(l)){
Right[i]=l-'0';
record[Right[i]]=1;
}
else
Right[i]=-1;
if(isdigit(r)){
Left[i]=r-'0';
record[Left[i]]=1;
}
else
Left[i]=-1;
}
int i;
for(i=0; i<N; i++)
if(record[i]==0)
break;
node* root=new node(i);
CreateTree(root);
queue<node*> q;
q.push(root);
while(q.size()!=0){
node* t=q.front();
q.pop();
flag++==1?cout<<t->value:cout<<" "<<t->value;
if(t->left)
q.push(t->left);
if(t->right)
q.push(t->right);
}
cout<<endl;
flag=1;
inorder(root);
return 0;
}

PAT 1102 Invert a Binary Tree的更多相关文章

  1. PAT 1102 Invert a Binary Tree[比较简单]

    1102 Invert a Binary Tree(25 分) The following is from Max Howell @twitter: Google: 90% of our engine ...

  2. PAT甲级——1102 Invert a Binary Tree (层序遍历+中序遍历)

    本文同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/90577042 1102 Invert a Binary Tree ...

  3. 1102 Invert a Binary Tree——PAT甲级真题

    1102 Invert a Binary Tree The following is from Max Howell @twitter: Google: 90% of our engineers us ...

  4. PAT Advanced 1102 Invert a Binary Tree (25) [树的遍历]

    题目 The following is from Max Howell @twitter: Google: 90% of our engineers use the sofware you wrote ...

  5. 1102. Invert a Binary Tree (25)

    The following is from Max Howell @twitter: Google: 90% of our engineers use the software you wrote ( ...

  6. PAT A1102 Invert a Binary Tree (25 分)——静态树,层序遍历,先序遍历,后序遍历

    The following is from Max Howell @twitter: Google: 90% of our engineers use the software you wrote ( ...

  7. PAT (Advanced Level) 1102. Invert a Binary Tree (25)

    简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #in ...

  8. PAT甲题题解-1102. Invert a Binary Tree (25)-(建树,水题)

    就是把输入给的左孩子右孩子互换一下,然后输出层次遍历和中序遍历. #include <iostream> #include <algorithm> #include <c ...

  9. 【PAT甲级】1102 Invert a Binary Tree (25 分)(层次遍历和中序遍历)

    题意: 输入一个正整数N(<=10),接着输入0~N-1每个结点的左右儿子结点,输出这颗二叉树的反转的层次遍历和中序遍历. AAAAAccepted code: #define HAVE_STR ...

随机推荐

  1. 20170623_oracle备份和恢复_常见问题

    1 为什么需要备份?备份分类? 1)故障.迁移.误操作 2)备份分类: 物理与逻辑角度:物理备份.逻辑备份 备份策略角度:完全备份.增量备份.差异备份 2 使用导入导出进行备份和恢复及其四种模式:其他 ...

  2. luogu2744 量取牛奶

    题目大意 给出一个整数集合$A$,总数$N$,规定一个整数序列$\{a_n\}, \forall i, a_i\in A$满足条件:存在一个正整数序列$\{k_n\}$,使得$\sum_{i=1}^n ...

  3. 使用 StackExchange.Redis 封装属于自己的 RedisHelper

    目录 核心类 ConnectionMultiplexer 字符串(String) 哈希(Hash) 列表(List) 有序集合(sorted set) Key 操作 发布订阅 其他 简介 目前 .NE ...

  4. All Discs Considered(拓扑排序)

    http://poj.org/problem?id=1778 题意:有两个DVD,第一个DVD上有编号为1~n1的安装包,第二个DVD上有编号为n1+1~n1+n2的安装包,给出m组关系(a,b) 表 ...

  5. c++小游戏

    #include <iostream> using namespace std; double shengmingli=2000;//定义主角初始生命力 int gongjili=150; ...

  6. ORA-01075: you are currently logged on

    [root@hear01 ~]# su - oracle[oracle@hear01 ~]$ sqlplus "/as sysdba" SQL*Plus: Release 11.2 ...

  7. UNIX环境高级编程--3

    文件IO 函数lseek: 每个打开文件都有一个与其相关联的“当前文件偏移量”,用来度量从文件开始处计算的字节数.除非指定O_APPEND选项,否则该偏移量被置为0.如果文件描述符指向的是一个管道.F ...

  8. 在Windows2003安装配置Bitvise SSH Server后,不能使用软件内建立的用户登录!

    Google:  I can only log in with an administrator account - attempting to log in with a regular accou ...

  9. 15年用canvas画的

    请恕我当年的工作太轻松,用canvas手打了一个图,技术含量并没有什么,现在看看,甚是怀念_(¦3」∠)_ <!DOCTYPE html> <html> <head&g ...

  10. 从"嘿,今晚..."谈消息安全传输中的技术点

    一.初级阶段:信息裸传 特点:在网络上传递明文 黑客定理一:网络上传递的数据是不安全的,属网络于黑客公共场所,能被截取 结果:传递明文无异于不穿衣服裸奔 改进方案:先加密,再在网络上传输 二.进阶阶段 ...