HackerEarth Rooted trees problem solution

In this HackerEarth Rooted trees problem solution, You are given a rooted tree of N nodes. Each node i contains a value ai. Initially, all values of the node are 0, your task is to process Q queries of the following two types:

0 v x for every node u in a subtree of v if au < Y add x to au
1 v print the value au

The tree is rooted at 1.

HackerEarth Rooted trees problem solution.

import java.io.*;
import java.util.*;

public class MainJava{
    static ArrayList<Integer>[] adj;
    static int[] vals, par;
    static int Y;
    static void dfs(int u ,int p){
        par[u]=p;
        for(int v:adj[u]){
            if(v!=p){
                dfs(v,u);
            }
        }
    }
    static void Update(int u, int p,int x) {
        if (vals[u] >= Y)
            return;
        vals[u] += x;
        for (int v : adj[u]){
            if(v!=p)
                Update(v,u, x);
        }
    }

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t = sc.nextInt();
        while(t-->0){
            int n = sc.nextInt();
            int q = sc.nextInt();
            Y = sc.nextInt();
            adj = new ArrayList[n];
            for (int i = 0; i < n; i++)
                adj[i] = new ArrayList<>();
    
            for (int i = 1; i < n; i++) {
                int u = sc.nextInt() - 1;
                int v = sc.nextInt() - 1;
                adj[u].add(v);
                adj[v].add(u);
            }
            vals = new int[n];
            par = new int[n];
            dfs(0,-1);
    
            while (q-- > 0) {
    
                if (sc.nextInt() == 0) {
                    int v = sc.nextInt() - 1;
                    int x = sc.nextInt();
                    Update(v,par[v], x);
                } else {
                    pw.println(vals[sc.nextInt() - 1]);
                }
            }
        }


        pw.flush();
    }


    static class Scanner {
        BufferedReader br;
        StringTokenizer st;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(FileReader f) {
            br = new BufferedReader(f);
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        public int[] nextIntArr(int n) throws IOException {
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = Integer.parseInt(next());
            }
            return arr;
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }

}

Second solution

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

const int N = 1e5 + 14;
int n, y, q, st[N], en[N], a[N], ver[N];
vector<int> g[N];
set<int> ley;

void dfs(int v = 0, int p = -1) {
    static int time;
    if (p == -1)
        time = 0;
    ver[time] = v;
    st[v] = time++;
    for (auto u : g[v])
        if (u != p)
            dfs(u, v);
    en[v] = time;
}

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
        cin >> n >> q >> y;
        fill(g, g + n, vector<int>());
        fill(a, a + n, 0);
        for (int i = 0; i < n - 1; ++i) {
            int v, u;
            cin >> v >> u;
            --v;
            --u;
            g[v].push_back(u);
            g[u].push_back(v);
        }
        ley.clear();
        for (int i = 0; i < n; ++i)
            ley.insert(ley.begin(), i);
        dfs();
        while (q--) {
            int t, v;
            cin >> t >> v;
            --v;
            if (t == 0) {
                int x;
                cin >> x;
                auto it = ley.lower_bound(st[v]);
                while (it != ley.end() && *it < en[v]) {
                    int u = ver[*it];
                    a[u] += x;
                    if (a[u] >= y)
                        it = ley.erase(it);
                    else
                        ++it;
                }
            }
            else
                cout << a[v] << 'n';
        }
    }
}

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