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 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, 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
我的答案
 #include <stdio.h>
#include <stdlib.h>
#include <unistd.h> #define MaxTree 10
#define Tree int
#define Null -1 struct TreeNode {
Tree Left;
Tree Right;
} T1[MaxTree]; #define QueueSize 100
struct QNode {
int Data[QueueSize];
int rear;
int front;
};
typedef struct QNode *Queue; int IsEmpty(Queue PtrQ)
{
return (PtrQ->front == PtrQ->rear);
} void AddQ(Queue PtrQ, int item)
{
if((PtrQ->rear+)%QueueSize == PtrQ->front) {
printf("Queue full");
return;
}
PtrQ->rear = (PtrQ->rear+)%QueueSize;
PtrQ->Data[PtrQ->rear] = item;
} int DeleteQ(Queue PtrQ)
{
if(PtrQ->front == PtrQ->rear) {
printf("Queue empty");
return -;
} else {
PtrQ->front = (PtrQ->front+)%QueueSize;
return PtrQ->Data[PtrQ->front];
}
} Tree BuildTree(struct TreeNode T[])
{
char cl, cr;
int N, i, check[MaxTree];
Tree Root = Null;
// scanf("%d\n", &N);
scanf("%d\n", &N);
// printf("N=%d\n", N);
if(N) {
for(i=;i<N;i++)
check[i] = ;
for(i=;i<N;i++) {
scanf("%c %c\n", &cl, &cr);
// printf("cl=%c, cr=%c\n", cl, cr);
if(cl!='-') {
T1[i].Left = cl - '';
check[T1[i].Left] = ;
} else
T1[i].Left = Null; if(cr!='-') {
T1[i].Right = cr - '';
check[T1[i].Right] = ;
} else
T1[i].Right = Null;
}
for(i=;i<N;i++)
if(check[i] != ) break;
Root = i;
}
return Root;
} void PrintLeaves(Tree R) //层序遍历
{
Queue Q;
Tree cur;
int count = ;
if(R==Null) return;
Q = (Queue)malloc(sizeof(struct QNode));
Q->rear = -;
Q->front = -;
AddQ(Q, R);
while(!IsEmpty(Q)) {
cur = DeleteQ(Q);
if((T1[cur].Left == Null) && (T1[cur].Right == Null)) {
if(!count) {
printf("%d", cur);
count++;
} else
printf(" %d", cur);
continue;
}
if(T1[cur].Left!=Null) AddQ(Q, T1[cur].Left);
if(T1[cur].Right!=Null) AddQ(Q, T1[cur].Right);
}
} int main()
{
Tree R;
R = BuildTree(T1); PrintLeaves(R); return ;
}

03-树2 List Leaves(25 分)的更多相关文章

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 浙大数据结构课后习题 练习三 7-4 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. I ...

  7. 7-3 树的同构(25 分) JAVA

    给定两棵树T1和T2.如果T1可以通过若干次左右孩子互换就变成T2,则我们称两棵树是“同构”的. 例如图1给出的两棵树就是同构的,因为我们把其中一棵树的结点A.B.G的左右孩子互换后,就得到另外一棵树 ...

  8. PTA 03-树1 树的同构 (25分)

    题目地址 https://pta.patest.cn/pta/test/15/exam/4/question/711 5-3 树的同构   (25分) 给定两棵树T1和T2.如果T1可以通过若干次左右 ...

  9. PAT 甲级 1021 Deepest Root (25 分)(bfs求树高,又可能存在part数part>2的情况)

    1021 Deepest Root (25 分)   A graph which is connected and acyclic can be considered a tree. The heig ...

  10. PTA 树的同构 (25分)

    PTA 树的同构 (25分) 输入格式: 输入给出2棵二叉树树的信息.对于每棵树,首先在一行中给出一个非负整数N (≤10),即该树的结点数(此时假设结点从0到N−1编号):随后N行,第i行对应编号第 ...

随机推荐

  1. 黄金含量版本——KTV

    呀,进来的都是盆友,首先先给大家拜年了,祝大家新年快乐,万事如意,家和万事兴~! 大家看了标题进来就不能让大家白进来,一定会让大家带着满满的果实. 下面我们就来讨论讨论KTV这个项目: (1)KTV的 ...

  2. 前端每日实战:84# 视频演示如何用纯 CSS 创作一个极品飞车 loader

    效果预览 按下右侧的"点击预览"按钮可以在当前页面预览,点击链接可以全屏预览. https://codepen.io/comehope/pen/MBbEMo 可交互视频 此视频是可 ...

  3. 2018-2019-2 20175126谢文航 实验四《Android开发基础》实验报告

    一.实验封面 课程:Java程序设计 班级:1751 班 姓名:谢文航 学号:20175126 指导教师:娄嘉鹏 实验日期:2019年5月15日 实验时间:--- 实验序号:实验四 实验名称:Andr ...

  4. C# JS 前后端互传数据

    ---恢复内容开始--- 后端: public void ProcessRequest(HttpContext context) { context.Response.ContentType = &q ...

  5. printf ("%*.*s")

    小数点.后“*”表示输出位数,具体的数据来自参数表printf格式字符串中,与宽度控制和精度控制有关的常量都可以换成变量,方法就是使用一个“*”代替那个常量,然后在后面提供变量给“*”. 同样,小数点 ...

  6. jmeter添加自定义扩展函数之if判断

    1,打开eclipse,新建maven工程,在pom中引用jmeter核心jar包,具体请看---https://www.cnblogs.com/guanyf/p/10863033.html---,这 ...

  7. Python笔记(九)_切片、列表生成式

    切片 mylist[:3] 取前3位元素,0可省略不写 mylist[-4:] 取后4位元素,0可省略不写 mylist[2:4] 从第2个开始取,取到第4个,但第4个不取,取的元素值为4-2=2 m ...

  8. Cocos2d-x之Action

    |   版权声明:本文为博主原创文章,未经博主允许不得转载. 在Cocos2d-x中的Node对象可以有动作,特效和动画等动态特性.因此在Node类中定义了这些动态特性,因此精灵,标签,菜单,地图和粒 ...

  9. Ubuntu解压缩rar格式文件

    解压缩rar文件时,出现问题 解决方法: sudo apt-get install unrar

  10. Spark Streaming + Kafka 整合向导之createDirectStream

    启动zk: zkServer.sh start 启动kafka:kafka-server-start.sh $KAFKA_HOME/config/server.properties 创建一个topic ...