```/题目要求：①判断是否是树，如果是，②输出最深的根
已知N个点，N-1条边，若只有一个连通块，则是树
①并查集②dfs/

include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

using namespace std;

const int N = 10010, M = N * 2;

int n;
int h[N], e[M], ne[M], idx;//无向图，e,ne数组要开二倍
int p[N];

int find(int x)//判断联通图
{
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}

void add(int a, int b)
{
e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
}

int dfs(int u, int father)//避免重复找，找最大深度
{
int depth = 0;
for (int i = h[u]; ~i; i = ne[i])
{
int j = e[i];
if (j == father) continue;
depth = max(depth, dfs(j, u) + 1);
}
return depth;
}

int main()
{
cin >> n;

memset(h, -1, sizeof h);
for (int i = 1; i <= n; i ++ ) p[i] = i;

int k = n;
for (int i = 0; i < n - 1; i ++ )
{
    int a, b;
    cin >> a >> b;
    if (find(a) != find(b))
    {
        k -- ;//更新联通块数量
        p[find(a)] = find(b);
    }
    add(a, b), add(b, a);
}

if (k > 1) printf("Error: %d components", k);
else
{
    vector<int> nodes;
    int max_depth = -1;

    for (int i = 1; i <= n; i ++ )
    {
        int depth = dfs(i, -1);//遍历每一个点的最大深度
        if (depth > max_depth)
        {
            max_depth = depth;
            nodes.clear();//把非最长路径上的点清空
            nodes.push_back(i);
        }
        else if (depth == max_depth)
            nodes.push_back(i);
    }

    for (auto v : nodes) cout << v << endl;
}

return 0;

}```