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
#include <stdio.h>
#define maxsize 10
#define Null -1
struct TreeNode{
int left;
int right;
}T[maxsize];
typedef struct TreeNode TN;
int BuildTree(TN T[]);//此处改为TN *T完全没问题,说明T[]本质是指针
int ListLeaves(int r);
int IsLeaves(int a);
int main()
{
int r;
r = BuildTree(T); //结合这行说明T本质是地址
ListLeaves(r);
return 0;
}
int BuildTree(TN T[])
{
int i,N;
int Root = -1;
char l,r;
scanf("%d\n",&N);//这里的\n坑惨了,如果没有,后面%c会把回车当char读进去
//从而出现一些奇怪的bug,所以要么在第一个scanf里面后面
//加\n,或者在后面的scanf里面前面加\n;
if(N){
int check[maxsize]={0};
for(i=0;i<N;i++){
scanf("%c %c\n",&l,&r);
if(l!='-'){
T[i].left=l-'0';
check[T[i].left]=1;
}
else {
T[i].left=Null;
}
if(r!='-'){
T[i].right=r-'0';
check[T[i].right]=1;
}
else {
T[i].right=Null;
}
}
for(i=0;i<N;i++){
if(check[i]==0){
Root = i;
break;
}
}
}
return Root;
}
int ListLeaves(int r)
{
int Queue[maxsize];//建立队列
int head = 0;
int rear = 0;
int flag = 0;
Queue[rear]=r;
rear++;
while(rear-head){
if(IsLeaves(Queue[head])){
if(flag){
printf(" ");
}
printf("%d",Queue[head]);
flag = 1;
}
else {
if(T[Queue[head]].left!=Null){
Queue[rear]=T[Queue[head]].left;
rear++;
}
if(T[Queue[head]].right!=Null){
Queue[rear]=T[Queue[head]].right;
rear++;
}
}
head++;
}
return 0;
}
int IsLeaves(int a)
{
int flag = 0;
if(T[a].left==Null&&T[a].right==Null){
flag = 1;
}
return flag;
}