用深搜去求出每个点管辖的子树在线段树上所对应的区间，然后进行常规修改、查询即可

#include <bits/stdc++.h>

//#pragma GCC optimize(3, "Ofast", "inline")

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;

const int N = 100010;

int w[N], e[N], ne[N], h[N], idx;
int T, n, ml[N], mr[N];
int father[N];

struct node
{
    int l, r, x;
}tr[N * 4];

void add(int a, int b) // 存要维护的图
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}

void push_down(int u) // 懒标记下放操作
{
    if (tr[u].x != -1)
    {
        tr[u << 1].x = tr[u].x;
        tr[u << 1 | 1].x = tr[u].x;
        tr[u].x = -1;
    }
}

void build(int u, int l, int r)
{
    if (l == r)
        tr[u] = {l, r, -1};
    else
    {
        tr[u] = {l, r, -1};
        int mid = l + r >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
    }
}
int cnt;
void dfs(int u) // 跑一遍深搜去求出每个点的子树在线段树中所对应的区间
{
    ml[u] = ++ cnt;
    for (int i = h[u]; i != -1; i = ne[i])
    {
        int j = e[i];
        dfs(j);
    }
    mr[u] = cnt;
}

void modify(int u, int l, int r, int k)
{
    if (tr[u].l >= l && tr[u].r <= r)
    {
        tr[u].x = k;
    }
    else
    {
        push_down(u);
        int mid = tr[u].l + tr[u].r >> 1;
        if (l <= mid) modify(u << 1, l, r, k);
        if (r > mid) modify(u << 1 | 1, l, r, k);
    }
}

int query(int u, int x)
{
    if (tr[u].l == x && tr[u].r == x)
        return tr[u].x;
    else
    {
        push_down(u);
        int mid = tr[u].l + tr[u].r >> 1;
        int res;
        if (x <= mid) res = query(u << 1, x);
        if (x > mid) res = query(u << 1 | 1, x); 
        return res;
    }
}

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int res = 0;
    cin >> T;
    while (T --)
    {
        cout << "Case #" << ++ res << ":\n";
        memset(h, -1, sizeof(h));
        memset(father, 0, sizeof(father));
        idx = 0;
        cin >> n;

        for (int i = 1; i < n; i ++)
        {
            int a, b;
            cin >> a >> b;
            add(b, a);
            father[a] = b;
        }

        int root = 1;
        while (father[root]) root ++; // 找到根节点

        cnt = 0;
        dfs(root);

        build(1, 1, n);

        int q;
        cin >> q;

        while (q --)
        {
            char ch;
            cin >> ch;

            if (ch == 'C')
            {
                int x;
                cin >> x;
                cout << query(1, ml[x]) << '\n';
            }
            else
            {
                int x, y;
                cin >> x >> y;
                modify(1, ml[x], mr[x], y);
            }
        }
    }

    return 0;
}