根据树的性质进行两个深度优先搜索就可以了,要注意数据的范围。
代码如下:

#include <iostream>
#include <vector>
using namespace std;

const int MAX_N = 1e6+10;
const long long  MOD = 1e9+7;
int n;
vector<int> nd[MAX_N]; // 连接的顶点
long long ans[MAX_N];

inline long long fastpow(long long x, int k)
{
    long long res = 1, t = x;
    while(k) {
        if(k & 1) res  = res * t % MOD;
        t = t * t % MOD;
        k >>= 1;
    }
    return res;
}
inline long long inv(long long x){
    return fastpow(x, MOD - 2);
}

void dfs1(int p, int cur){
    int size = nd[cur].size(), son;
    for(int i = 0; i < size; ++i){
        son = nd[cur][i];
        if(son == p) continue;
        dfs1(cur, son);
        ans[cur] = (ans[cur]*(ans[son]+1)) % MOD;
    }
}

void dfs2(int p, int cur){
    int size = nd[cur].size(), son;
    long long tm = 1;
    for(int i = 0; i < size; ++i){
        son = nd[cur][i];
        if(son == p) continue;
        if((ans[son] + 1) % MOD != 0){
            tm = tm*(ans[son] + 1) % MOD;         
            ans[son] = ans[son]*((ans[cur]*inv(ans[son] + 1) + 1) % MOD) % MOD;
        }else{
            if(tm != 0){
                for(int j = i + 1; j < size; ++j){
                    if(tm == 0) break;
                    tm = tm*(ans[nd[cur][j]] + 1) % MOD;
                }
                ans[son] = ans[son]*(tm + 1) % MOD;
                tm = 0;
            }         
        }
        dfs2(cur, son);
    }
}

int main(){
    scanf("%lld",&n);
    int x, y;
    for(int i = 1; i < n; ++i){
        scanf("%d%d",&x, &y);
        nd[y].push_back(x);
        nd[x].push_back(y);
    }
    for(int i = 1; i <= n; ++i) ans[i] = 1;
    dfs1(0, 1);
    dfs2(0, 1);   
    for(int i = 1; i <= n; ++i) printf("%lld\n",ans[i]);
}