算法一：经典做法——两遍dfs

任选一点找到距离该点最远的点u

在找到距离u最远的点v则u-v路径就是一条直径

import java.util.*;
class Main{
    private static int ans, maxv, maxd;
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        createAdj(n);
        for(int i = 1; i < n; i++){
            int a = sc.nextInt();
            int b = sc.nextInt();
            int c = sc.nextInt();
            add(a, b, c);
            add(b, a, c);
        }
        dfs(1, -1, 0);
        dfs(maxv, -1, 0);
        System.out.println(maxd);
    }
    private static void dfs(int u, int fa, int d){
        for(int i = h[u]; i != -1; i = ne[i]){
            int j = e[i];
            if(j == fa) continue;
            if(d + w[i] > maxd){
                maxd = d + w[i];
                maxv = j;
            }
            dfs(j, u, d + w[i]);
        }
    }
    private static int[] h, e, w, ne;
    private static int idx;
    private static void createAdj(int n){
        h = new int[n+1];
        e = new int[n*2];
        w = new int[n*2];
        ne = new int[n*2];
        Arrays.fill(h, -1);
        idx = 0;
    }
    private static void add(int a, int b, int c){
        e[idx] = b;
        w[idx] = c;
        ne[idx] = h[a];
        h[a] = idx++;
    }
}

算法二：树形DP——所有挂在节点u中直径的集合

import java.util.*;
class Main{
    private static int ans;
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        createAdj(n);
        for(int i = 1; i < n; i++){
            int a = sc.nextInt();
            int b = sc.nextInt();
            int c = sc.nextInt();
            add(a, b, c);
            add(b, a, c);
        }
        dfs(1, -1);
        System.out.println(ans);
    }
    private static int dfs(int u, int fa){
        int max1 = 0, max2 = 0;                 //表示节点u子节点中深度最长的两条边
        for(int i = h[u]; i != -1; i = ne[i]){
            int j = e[i];
            if(j == fa) continue;
            int d = dfs(j, u) + w[i];
            if(d > max1){
                max2 = max1; max1 = d;
            }else{
                if(d > max2) max2 = d;
            }
        }
        ans = Math.max(ans, max1 + max2);
        return max1;
    }

    private static int[] h, e, w, ne;
    private static int idx;
    private static void createAdj(int n){
          h = new int[n+1];
          e = new int[n*2];
          w = new int[n*2];
          ne = new int[n*2];
          Arrays.fill(h, -1);
          idx = 0;
    }
    private static void add(int a, int b, int c){
        e[idx] = b;
        w[idx] = c;
        ne[idx] = h[a];
        h[a] = idx++;
    }
}