树的遍历问题，本题是普通的树，不是二叉树。

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

const int N = 1010;

int n, m;
bool st[N];                             //判断一个节点是否是子节点
vector<int> g[N];                       //储存每个父节点的子节点

int main()
{
    cin >> n >> m;

    for (int i = 0; i < m; i ++ )
    {
        int father, son;
        cin >> father >> son;

        st[son] = true;
        g[father].push_back(son);
    }

    for (int i = 1; i <= 1000; i ++ )   //找根节点
        if (g[i].size() && !st[i])      //该节点存在且没有父节点,即是根节点
        {
                cout << i << endl;
                break;
        }

    int p = 1;
    for (int i = 2; i <= 1000; i ++ )   //遍历找子节点最多的节点
        if (g[i].size() > g[p].size())
            p = i;

    cout << p << endl;

    sort(g[p].begin(), g[p].end());     //将儿子从小到大排序输出

    for (auto x : g[p]) cout << x << ' ';
    cout << endl;

    return 0;
}