#include <bits/stdc++.h> // By Lucky Ox
#define int long long
#define endl "\n"
#define pii pair<int, int>
#define PI atan(1.0) * 4
using namespace std;
using i128 = __int128;
const long long mod = 1000000000000002493, INF = 0x3f3f3f3f3f3f3f3f;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int x, int y) { return x / gcd(x, y) * y; }
int q_pow(int a, int k, int p) { int res = 1; while (k) { if (k & 1) res = res * a % p; k >>= 1; a = a * a % p; } return res; }
int len_10(int x) { int len = 0; while(x) { x /= 10; len ++ ; } return len; }
int P(int x, int p){ return (x % p + p) % p; }
//用__lg()来求一个数二进制下的位数 返回的len 表示这个数是[0, 1, ...., len] 比如10 __lg(10) = 3, 1010 [3, 2, 1, 0]
//__builtin_popcountll(int x) 求二进制下x中1的数量 __buitlin_ctzll(int x) 求二进制下末尾0的个数

const int N = 1e6 + 10;
int n;
vector<int> e[N];
int sz[N];

void dfs1(int u, int fa) {

}
void solve() {
    cin >> n;
    for(int i = 1; i <= n; i ++ ) sz[i] = 1;
    
    map<pii, int> mp;
    for(int i = 1; i <= n - 1; i ++ ) {
        int u, v, w;
        cin >> u >> v >> w;
        e[u].push_back(v), e[v].push_back(u);
        mp[{minmax(u, v)}] = w;
    }
    int ans = 0;
    function<void (int, int)> dfs1 = [&](int u, int fa) {
        for(int v : e[u]) {
            if(v == fa) continue;
            dfs1(v, u);
            sz[u] += sz[v];

            ans +=  mp[minmax(u, v)] * abs(n - sz[v] - sz[v]);
        }
    };
    dfs1(1, 0);
    cout << ans << endl;
}

signed main() {
    ios::sync_with_stdio(0);cin.tie(0);
    cout << fixed << setprecision(10);
    int T = 1;
    //cin >> T;
    while (T -- ) solve();
    return 0;
}   

时间复杂度是Onlogn吧 感觉用不到两秒啊