Invert a Binary Tree

pat-1102

import java.util.Arrays;
import java.util.Queue;
import java.util.Scanner;
import java.util.concurrent.LinkedBlockingQueue; /**
* @Author WaleGarrett
* @Date 2020/9/5 20:04
*/
public class PAT_1102 {
static InvertTree[] tree;
static boolean[] flag;
public static void main(String[] args) {
Scanner scanner=new Scanner(System.in);
int n=scanner.nextInt();
tree=new InvertTree[n];
flag=new boolean[n];
Arrays.fill(flag,false);
scanner.nextLine();
for(int i=0;i<n;i++){
String a=scanner.next(),b=scanner.next();
int na,nb;
if(a.equals("-")){
na=-1; }else {
na=Integer.parseInt(a);
flag[na]=true;
}
if(b.equals("-")){
nb=-1;
}else {
nb=Integer.parseInt(b);
flag[nb]=true;
}
tree[i]=new InvertTree();
tree[i].left=na;
tree[i].right=nb;
}
int head=-1;
for(int i=0;i<n;i++){
if(!flag[i]){
head=i;
break;
}
}
System.out.println(levelOrder(head,"").trim());
System.out.println(inOrder(head,"").trim());
}
public static String levelOrder(int head,String result){
Queue<Integer> que=new LinkedBlockingQueue<>();
que.add(head);
while(!que.isEmpty()){
int now=que.poll();
result=result+now+" ";
if(tree[now].right!=-1) que.add(tree[now].right);
if(tree[now].left!=-1) que.add(tree[now].left);
}
return result;
}
public static String inOrder(int head,String result){
if(head==-1)
return result;
result=inOrder(tree[head].right,result);
result=result+head+" ";
result=inOrder(tree[head].left,result);
return result;
}
}
class InvertTree{
int left,right,value;
}

PAT-1102(Invert a Binary Tree)+二叉树的镜像+层次遍历+中序遍历+已知树的结构构树的更多相关文章

  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 1102 Invert a Binary Tree

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

  3. 【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 ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. 1102 Invert a Binary Tree (25 分)(二叉树遍历)

    二叉树有N个结点,给出每个结点的左右孩子结点的编号,把二叉树反转(左右孩子交换  所以是后序遍历交换) 输出反转后二叉树的层序遍历和中序遍历 #include<bits/stdc++.h> ...

  8. 1102. Invert a Binary Tree (25)

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

  9. 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 ( ...

随机推荐

  1. 【hdu 4859】海岸线(图论--网络流最小割)

    题意:有一个区域,有'.'的陆地,'D'的深海域,'E'的浅海域.其中浅海域可以填充为陆地.这里的陆地区域不联通,并且整个地图都处在海洋之中.问填充一定浅海域之后所有岛屿的最长的海岸线之和. 解法:最 ...

  2. AtCoder Beginner Contest 188 E - Peddler (树)

    题意:有\(n\)个点,\(m\)条单向边,保证每条边的起点小于终点,每个点都有权值,找到联通的点的两个点的最大差值. 题解:因为题目说了起点小于终点,所以我们可以反向存边,然后维护连通边的前缀最小值 ...

  3. Codeforces Round #577 (Div. 2) C. Maximum Median (模拟,中位数)

    题意:给你一个长度为奇数\(n\)的序列.你可以对任意元素加上\(k\)次\(1\),求操作后的中位数最大. 题解:先对序列进行排序,然后对中位数相加,如果中位数和后面的元素相等,就对后面所有和当前中 ...

  4. 郁闷的出纳员 HYSBZ - 1503

    OIER公司是一家大型专业化软件公司,有着数以万计的员工.作为一名出纳员,我的任务之一便是统计每位员工的 工资.这本来是一份不错的工作,但是令人郁闷的是,我们的老板反复无常,经常调整员工的工资.如果他 ...

  5. 爬虫——urllib.request包

    一.引用包 import urllib.request 二.常用方法 (1)urllib.request.urlretrieve(网址,本地文件存储地址):直接下载网页到本地 urllib.reque ...

  6. CODE_TEST-- gtest

    gtest 提供了一套优秀的 C++ 单元测试解决方案,简单易用,功能完善,非常适合在项目中使用以保证代码质量. https://blog.csdn.net/jcjc918/article/detai ...

  7. Leetcode(1)-两数之和

    给定一个整数数组和一个目标值,找出数组中和为目标值的两个数. 你可以假设每个输入只对应一种答案,且同样的元素不能被重复利用. 示例: 给定 nums = [2, 7, 11, 15], target ...

  8. js optional chaining operator

    js optional chaining operator js 可选链 可选链操作符( ?. )允许读取位于连接对象链深处的属性的值,而不必明确验证链中的每个引用是否有效. ?. 操作符的功能类似于 ...

  9. Vue dynamic component All In One

    Vue dynamic component All In One Vue 动态组件 vue 2.x https://vuejs.org/v2/guide/components-dynamic-asyn ...

  10. VSCode Plugin & Auto File Header Comments Generator

    VSCode Plugin & Auto File Header Comments Generator Xcode SwiftUI // // ContentView.swift // Mem ...