PTA 7-4 small generation (DFS)

https://pintia.cn/problem-sets/1259041418971803648/problems/1259047109883162624
（ Title link above ）
The question ：
Given the genealogy of a large family , Please give a list of the youngest generation .
Answer key ：
Node design ：

struct tnode{
    
	int parent;
	int num;
	int level;// Record seniority , Which layer of the tree 
}node[100005];

Ideas ： For each node , Search deeply until its ancestor root node returns , When tracing back, the number of layers of each layer is equal to the number of layers of the previous layer plus one . If you find a node level Not for 0, And go straight back to , When backtracking, add from the level of this node .
AC Code ：

#include <iostream>
using namespace std;
int n,lev;
struct tnode{
    
	int parent;
	int num;
	int level;
}node[100005];
void dfs(int now){
    
	if(node[now].parent==-1){
    
		node[now].level=1;
		return;
	}
	if(node[now].level!=0){
    
		return;
	}
	int next=node[now].parent;
	dfs(next);
	node[now].level=node[next].level+1;
}
int main(){
    
	cin>>n;
	for(int i=1;i<=n;i++){
    
		cin>>node[i].parent;
		node[i].num=i;
	}
	int maxlev=-1;
	for(int i=1;i<=n;i++){
    
		dfs(i);
		lev=node[i].level;
		maxlev=max(maxlev,lev);
	}
	int flag=0;
	cout<<maxlev<<endl;
	for(int i=1;i<=n;i++){
    
		if(node[i].level==maxlev){
    
			if(!flag){
    
				cout<<i; flag=1;
			}
			else cout<<" "<<i;
		}
	}
	cout<<endl;
	return 0;
}