03-树2 List Leaves（25 分）

题目来源：中国大学MOOC-陈越、何钦铭-数据结构-2018春
作者: 陈越
单位: 浙江大学

问题描述：
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

解答：
目标是找出所有叶子节点，按序输出。（要求从上到下，从左到右顺序，可以用BFS做）
一开始看到题目有点懵，感觉重要的是找到树根，其他就简单了，BFS都是套路。树根怎么找呢？可以用一个0到N-1数组表示,注意输入为字符类型，若要数值型数据要减‘0‘以输入左节点为例：

        char temp;
        cin>>temp;
        if(temp!='-')
        {
            tree[i].leftChild=temp-'0';
            isHead[temp-'0']=false;
        }

找到根节点后利用队列做BFS即可

#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;

struct node
{
    int leftChild;
    int rightChild;
    node()
    {
        leftChild=-1;
        rightChild=-1;
    }
};
node tree[10];
bool isHead[10];
int main()
{
    int M;
    cin>>M;
    fill(isHead,isHead+10,true);
    for(int i=0;i<M;i++)
    {
        char temp;
        cin>>temp;
        if(temp!='-')
        {
            tree[i].leftChild=temp-'0';
            isHead[temp-'0']=false;
        }
        cin>>temp;
        if(temp!='-')
        {
            tree[i].rightChild=temp-'0';
            isHead[temp-'0']=false;
        }
    }
    int head;
    for(int i=0;i<M;i++)
    {
        if(isHead[i]==true)
        {
            head=i;
            break;
        }
    }
    queue<int> bfs;
    queue<int> result;
    bfs.push(head);
    while(!bfs.empty())
    {
        int frontNode=bfs.front();
        bfs.pop();
        if(tree[frontNode].leftChild==-1&&tree[frontNode].rightChild==-1)
            result.push(frontNode);
        if(tree[frontNode].leftChild!=-1)
            bfs.push(tree[frontNode].leftChild);
        if(tree[frontNode].rightChild!=-1)
            bfs.push(tree[frontNode].rightChild);
    }
    cout<<result.front();
    result.pop();
    while(!result.empty())
    {
        cout<<" "<<result.front();
        result.pop();
    }
    return 0;
}