甲级-1004 Counting Leaves (30 分)

题目：

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 0

ID K ID[1] ID[2] ... ID[K]
where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID's of its children. For the sake of simplicity, let us fix the root ID to be 01.

The input ends with N being 0. That case must NOT be processed.

Output Specification:

For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.

The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output 0 1 in a line.

Sample Input:

2 1
01 1 02

Sample Output:

0 1

解题思路：

用左孩子右兄弟的形式构建二叉树。
输出的时候使用栈或队列，依次遍历每个层次，同时统计每层的叶子结点数量。
缺点是为了快速访问到每个结点使用 vector 保存结点，在此假定结点的ID范围为01~N，如超出此范围则代码不适用

代码：

编译器：C++(g++)

#include 
#include 
#include 
using namespace std;

struct Node{
    int id;
    Node *left;
    Node *right;
    Node(int n=0):id(n),left(NULL),right(NULL){}
};
int main()
{
    int n,m;
    cin>>n>>m;
    if(0==n)
    {
        return 0;
    }
    vector allNode(n,0);
    for(int i=0;i!=n;++i)
    {
        allNode[i].id=i+1;
    }
    for(int i=0;i!=m;++i)
    {
        int pre,num,cur;
        cin>>pre>>num;
        for(int j=0;j!=num;++j)
        {
            cin>>cur;
            if(0==j)
            {
                allNode[pre-1].left=&(allNode[cur-1]);
            }
            else
            {
                allNode[pre-1].right=&(allNode[cur-1]);
            }
            pre=cur;
        }
    }
    stack s1,s2;
    s1.push(allNode[0]);
    while(!s1.empty())
    {
        if(s1.top().id!=allNode[0].id)
        {
            cout<<" ";
        }
        int count=0;
        Node t;
        while(!s1.empty())
        {
            t=s1.top();
            s1.pop();
            if(NULL==t.left)
            {
                ++count;
            }
            else
            {
                s2.push(*(t.left));
            }
            if(t.right)
            {
                s1.push(*(t.right));
            }
        }
        swap(s1,s2);
        cout<