题目描述

1004 Counting Leaves (30 分)

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<N<100, the number of nodes in a tree, and M (<N), the number of non-leaf nodes. Then M lines follow, each in the format:

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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:

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2 1
01 1 02



结尾无空行

Sample Output:

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0 1



结尾无空行

参考柳婼的题解

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//
// Created by 张旭辉 on 2021/8/11.
//
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
int nums[100]; // 每一层叶子节点的个数
vector<int> node[100]; //二维数组存放数 
int maxDep = -1;

/**
 * 深度优先搜索
 * @param index 当前的节点的索引
 * @param depth 当前的深度
 */
void dfs(int index,int depth){
   //到达当前节点需要做的事情
   //这里是判断时候有叶子节点，没有就给当前层数的 nums 加 1
    if(node[index].size() == 0){
        nums[depth]++;
        maxDep = max(depth,maxDep);
        return ;
    }
  
  //下面这个循环，其实是dfs的模板，也是回溯的模板
    for(int i = 0 ; i < node[index].size(); i++){
      
       //TODO 如果是回溯，前置操作
        dfs(node[index][i],depth +1);
         //TODO 如果是回溯，后置操作
      
    }
}
int main(){
    fill(nums,nums+100,inf);
    int n,m;
    int curNode ,num;
    int val;
    scanf("%d %d", &n,&m);
     //这里看题目的意思，我反正是没有理解透，看了柳神代码才恍然大悟
    for(int i = 0; i < m; i++){
        scanf("%d %d",&curNode,&num);
        for(int j = 0 ; j < num; j++){
            scanf("%d",&val);
            node[curNode].push_back(val);
        }
    }
  
    dfs(1,0);
    cout<<nums[0];
    for(int i =1; i <= maxDep; i++){
        cout<<" "<<nums[i];
    }
    cout<<endl;
    return 0;
}

总结

DFS 模板(我精通java，会写一点C++)

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void traverse(TreeNode root) {
    for (TreeNode child : root.children)
        // 前序遍历需要的操作
        traverse(child);
        // 后序遍历需要的操作
}