PTA 03-树2 List Leaves

最新推荐文章于 2021-12-07 19:39:07 发布

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本文介绍了一个程序设计问题，即如何按照从上到下、从左到右的顺序遍历一棵树并打印出所有的叶子节点。文章详细说明了输入格式，包括树的节点数量及其左右子节点信息，并给出了输出格式的要求。

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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.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer $N$ ( $\leq 10$ ) 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

#include <stdio.h>
#include <stdlib.h>
struct Tree {
int data;
int left;
int right;
};
int main()
{
int num,sum;
sum = 0;
int i;
scanf("%d ", &num);
for (i = 0; i < num; i++)
{
  sum += i;
}
struct Tree * T;
T = (struct Tree*)malloc(sizeof(struct Tree)*num);
for (i = 0; i < num; i++)
{
  T[i].data = i;
  int l, r;
  scanf("%c %c \n", &l,&r);

  if (l == 45) T[i].left = -1;
  else T[i].left = l-'0';
  if (r == 45) T[i].right = -1;
  else T[i].right = r-'0';
}
for (i = 0; i < num; i++)
{
  printf("%d %d %d\n", T[i].data, T[i].left, T[i].right);
}
return 0;
}

目前问题：在全部输入后还需再次输入一个任意值程序方能正常进行。