03-树2 List Leaves (25分)

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本文介绍了一种使用队列实现的层次遍历方法来按从上到下、从左到右的顺序输出二叉树的所有叶子节点的索引。输入包括节点总数及各节点的左右子节点信息，输出则为叶子节点的索引列表。

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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 NN 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

思路：这道题跟上道题二叉树的构建类似，不同的是这个题节点的数据就是它在数组中的下标，很明显必须采用层次遍历的方式才能从上到下从左到右输出叶子节点。这里采用队列的方式实现层次遍历。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<queue>
using namespace std;
#define MaxTree 10
#define Null -1
typedef int Tree;
struct TreeNode{
    Tree Left;
    Tree Right;
}T[MaxTree];
int N,check[MaxTree]={0},count=0;
queue<int> q;

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

void countleaves(Tree Root){        //每碰到一个节点将它的左右孩子入队（若有的话），然后依次从队头取出判断是否为叶子节点
    Tree temp;
    if(Root==Null)  return;
    q.push(Root);               
    while(!q.empty()){
        temp=q.front();
        q.pop();
        if(T[temp].Left==Null&&T[temp].Right==Null){    //如果没有左右孩子即为叶子节点，则输出
            if(count++!=0){     //不是第一个节点的话前面输出空格
                putchar(' ');
            }
            printf("%d",temp);          
        }
        if(T[temp].Left!=Null)      q.push(T[temp].Left);
        if(T[temp].Right!=Null)     q.push(T[temp].Right);
    }
}
int main(){
    Tree R;
    R=BuildTree(T);
    countleaves(R);
    return 0;
}