List Leaves

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 (≤) 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

给定一棵树，您应该按照自上而下和从左到右的顺序列出所有叶子。

#include <stdio.h>
#include <queue>
#include <algorithm>
#define MaxTree 11
#define Null -1 
using namespace std;

int check[MaxTree];
int P[MaxTree];
int num=0;
queue<int> qu;

struct TreeNode
{
    int Left;
    int Right; 
}T[MaxTree];

int BuildTree(struct TreeNode T[])
{
    int N,i;
    char le,ri;
    int Root;
    scanf("%d
",&N);
    if(N)
    {
        for(i=0;i<N;i++)
        {
            check[i]=0;
        }
        for(i=0;i<N;i++)
        {
            scanf("%c %c
",&le,&ri);
            if(le!='-')
            {
                T[i].Left=le-'0';
                check[T[i].Left]=1;
            }
            else T[i].Left=Null;
            if(ri!='-')
            {
                T[i].Right=ri-'0';
                check[T[i].Right]=1;
            }
            else T[i].Right=Null;
        }
        for(i=0;i<N;i++)
        {
            if(!check[i]) break;
        }
        Root = i;
    }
    else 
    {
        Root=Null;
    }
    return Root;
}

void search(int Tree)
{
    if(Tree!=Null)
    {
        qu.push(Tree);
        while(!qu.empty())
        {
            int k;
            k=qu.front();
            if(T[k].Left==Null&&T[k].Right==Null)
                P[num++]=k;
            qu.pop();;
            if(T[k].Left!=Null)
                qu.push(T[k].Left);
            if(T[k].Right!=Null)
                qu.push(T[k].Right);
        }
    }
    else return;
}

int main()
{
    int Tree;
    Tree=BuildTree(T);
    search(Tree);
    int i;
    for(i=0;i<num;i++)
    {
        if(i==0)
            printf("%d",P[i]);
        else printf(" %d",P[i]);
    }
    return 0;
}