PTA 03-树2 List Leaves (25分)
题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/666
5-4 List Leaves (25分)
Given a tree, you are supposed to list all the leaves in the order of top down, and left to right.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer NN (\le 10≤10) which is the total number of nodes in the tree -- and hence the nodes are numbered from 0 to N-1N−1. Then NNlines follow, each corresponds to a node, 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 one line all the leaves' indices in the order of top down, and left to right. 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:
4 1 5
大部分代码都是在做数据结构,其它的,就是简单的层序遍历
/*
评测结果
时间 结果 得分 题目 编译器 用时(ms) 内存(MB) 用户
2017-07-08 15:59 答案正确 25 5-4 gcc 2 1
测试点结果
测试点 结果 得分/满分 用时(ms) 内存(MB)
测试点1 答案正确 13/13 1 1
测试点2 答案正确 5/5 2 1
测试点3 答案正确 1/1 1 1
测试点4 答案正确 5/5 2 1
测试点5 答案正确 1/1 1 1
*/
#include<stdio.h>
#include<stdlib.h>
#define MAXLEN 10
struct treenode{
int lc;
int rc;
int isLeaf;
int father;
}; typedef struct tree{
struct treenode nodes[MAXLEN];
int root;
int length;
}*ptrTree; typedef struct queue{
int data[MAXLEN];
int front;
int rear;
}*Que; Que CreateQueue()
{
Que temp;
temp=(Que)malloc(sizeof(struct queue));
temp->front=0;
temp->rear=0;
return temp;
} void EnQueue(Que Q,int item)
{
if((Q->rear+1)%MAXLEN == Q->front)
printf("Queue is full!");
Q->rear=(Q->rear+1)%MAXLEN;
Q->data[Q->rear]=item;
} int DeQueue(Que Q)
{
if(Q->front==Q->rear){
printf("Queue is Empty!");
return -1;
}
Q->front=(Q->front+1)%MAXLEN;
return Q->data[Q->front];
} int IsQueueEmpty(Que Q)
{
return Q->front==Q->rear;
} ptrTree CreateTree()
{
int i;
ptrTree temp;
temp=(ptrTree)malloc(sizeof(struct tree));
for(i=0;i<MAXLEN;i++)
temp->nodes[i].father=-1;
return temp; } void DestroyQueue(Que Q)
{
free(Q);
} void DestroyTree(ptrTree T)
{
free(T);
} void Input(ptrTree T)
{
int i,k,len;
scanf("%d",&len);
getchar();//skip a \n
T->length=len;
for(i=0;i<len;i++)
{
T->nodes[i].lc=getchar()-'0';
getchar();//skip a space
T->nodes[i].rc=getchar()-'0';
getchar();//skip a \n if(T->nodes[i].lc>=0)
T->nodes[T->nodes[i].lc].father=i;
if(T->nodes[i].rc>=0)
T->nodes[T->nodes[i].rc].father=i; if(T->nodes[i].lc<0 && T->nodes[i].rc<0)
T->nodes[i].isLeaf=1;
} for(i=0;i<len;i++){
if(T->nodes[i].father<0){
T->root=i;
break;
}
}
} void Process(ptrTree T,Que Q)
{
EnQueue(Q,T->root);
while(!IsQueueEmpty(Q))
{
int i;
i=DeQueue(Q);
if(T->nodes[i].isLeaf==1)
printf("%d",i);
else {
if(T->nodes[i].lc>=0)
EnQueue(Q,T->nodes[i].lc);
if(T->nodes[i].rc>=0)
EnQueue(Q,T->nodes[i].rc);
}
if(T->nodes[i].isLeaf==1 && !IsQueueEmpty(Q))
printf(" ");
}
} int main()
{
Que Q=CreateQueue();
ptrTree T=CreateTree();
Input(T);
Process(T,Q);
return 0;
}
PTA 03-树2 List Leaves (25分)的更多相关文章
- PTA甲级1094 The Largest Generation (25分)
PTA甲级1094 The Largest Generation (25分) A family hierarchy is usually presented by a pedigree tree wh ...
- L2-006 树的遍历 (25 分) (根据后序遍历与中序遍历建二叉树)
题目链接:https://pintia.cn/problem-sets/994805046380707840/problems/994805069361299456 L2-006 树的遍历 (25 分 ...
- PTA 04-树5 Root of AVL Tree (25分)
题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/668 5-6 Root of AVL Tree (25分) An AVL tree ...
- PTA 10-排序5 PAT Judge (25分)
题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/677 5-15 PAT Judge (25分) The ranklist of PA ...
- PTA 05-树7 堆中的路径 (25分)
题目地址 https://pta.patest.cn/pta/test/15/exam/4/question/713 5-5 堆中的路径 (25分) 将一系列给定数字插入一个初始为空的小顶堆H[] ...
- 7-4 List Leaves (25分) JAVA
Given a tree, you are supposed to list all the leaves in the order of top down, and left to right. I ...
- PTA 07-图4 哈利·波特的考试 (25分)
哈利·波特要考试了,他需要你的帮助.这门课学的是用魔咒将一种动物变成另一种动物的本事.例如将猫变成老鼠的魔咒是haha,将老鼠变成鱼的魔咒是hehe等等.反方向变化的魔咒就是简单地将原来的魔咒倒过来念 ...
- L2-006 树的遍历 (25 分)
链接:https://pintia.cn/problem-sets/994805046380707840/problems/994805069361299456 题目: 给定一棵二叉树的后序遍历和中序 ...
- 03-树2 List Leaves (25 分)
Given a tree, you are supposed to list all the leaves in the order of top down, and left to right. I ...
随机推荐
- 【LeetCode 337 & 329. memorization DFS】House Robber III
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode ...
- session共享方法
session数据保存在memcached,redis这种内存数据库内 memcache比较简单,使用的场景比较多.redis支持的数据类型多.而且数据具有期限,和session和期限可以配合使用 通 ...
- Java 中 i++和++i的区别
public class Test{ public static void main(String [] args){ int i = 1; int s = ++i; int x= i++; Syst ...
- .NET Core WebAPI Swagger使用
相对于普通的webapi而言,.net core webapi本身并不具备文档的功能,所以可以借助第三方插件:swagger,使用的话很简单. 步骤一. Nuget Packages安装,使用程序包管 ...
- 什么是极坐标? —— 一点微小的想法 What is Polar Coordinate ? - Some Naive Thoughts about It
Can you answer these three questions? The answer seems to be trivial, since we can use our eyes to o ...
- 【转】几种Java序列化方式的实现
0.前言 本文主要对几种常见Java序列化方式进行实现.包括Java原生以流的方法进行的序列化.Json序列化.FastJson序列化.Protobuff序列化. 1.Java原生序列化 Java原生 ...
- 【学习笔记】深入理解js原型和闭包(2)——函数和对象的关系
上文(深入理解jS原型和闭包(1)——一切都是对象)已经提到,函数就是对象的一种,因为通过instanceof函数可以判断. var fn = function () { }; console.log ...
- hdu5792 World is Exploding(多校第五场)树状数组求逆序对 离散化
题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=5792 题目描述:给你n个值,每个值用A[i]表示,然后问你能否找到多少组(a,b,c,d)四个编号,四 ...
- C++ 类中的static成员的初始化和特点
C++ 类中的static成员的初始化和特点 #include <iostream> using namespace std; class Test { public: Test() : ...
- \p{space}|\p{Hex}|\p{Digit}|转译符|\1|\g{-1}|[^ab]|/([^\d\D]+)/
小骆驼 第七章 漫游正则表达式王国 #!/usr/bin/perl use strict; use warnings; $_ = 'ab \ cde f ghijk10.x12ln'; if(/d/) ...