Reorder Routes to Make All Paths Lead to the City Zero

There are n cities numbered from 0 to n-1 and n-1 roads such that there is only one way to travel between two different cities (this network form a tree). Last year, The ministry of transport decided to orient the roads in one direction because they are too narrow.
Roads are represented by connections where connections[i] = [a, b] represents a road from the city a to b.
This year, there will be a big event in the capital (city 0), and many people want to travel to this city.
Your task consists of reorienting some roads such that each city can visit the city 0. Return the minimum number of edges changed.
It's guaranteed that each city can reach the city 0 after reorder.

Example 1:

Input: n = 5, connections ={{1,0},{1,2},{3,2},{3,4}}
Output: 2

Approach

Java

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

public class ReorderRoutes {
    public static void main(String[] args) {
        int n = 5;
        int[][] connections = { { 1, 0 }, { 1, 2 }, { 3, 2 }, { 3, 4 } };
        System.out.println(minReorder(n, connections));
    }

    static int minReorder(int n, int[][] connections) {
        List<List<Integer>> arr = new ArrayList<List<Integer>>();
        List<List<Integer>> tr = new ArrayList<List<Integer>>();

        // initialization the list
        for (int i = 0; i < connections.length+1; i++) {
            arr.add(new ArrayList<Integer>());
            tr.add(new ArrayList<Integer>());
        }

        int vis[] = new int[n];

        // create the transpose graph
        // and graph
        for (int i = 0; i < connections.length; i++) {
            int a = connections[i][0];
            int b = connections[i][1];
            arr.get(a).add(b);
            tr.get(b).add(a);
        }
        Queue<Integer> q = new LinkedList<>();
        q.add(0);
        int res = 0;

        // iterate till the queue is not
        // empty
        while (!q.isEmpty()) {
            int curr = q.poll();
            vis[curr] = 1;
            for (int child : arr.get(curr)) {
                if (vis[child] == 0) {
                    res++;
                    q.add(child);
                }
            }
            for (int child : tr.get(curr)) {
                if (vis[child] == 0) {
                    q.add(child);
                }
            }
        }
        return res;
    }
}

C++

#include <bits/stdc++.h>
using namespace std;

int minReorder(int n, vector<vector<int>>& connections) 
{
    vector<vector<int>> arr(n),tr(n);
    vector<int> vis(n);

    //create the transpose graph
    //and graph
      for(int i=0;i<connections.size();i++)
        {
          int a=connections[i][0];
          int b=connections[i][1];
          arr[a].push_back(b);
          tr[b].push_back(a);
       }
      queue<int> q;
        q.push(0);
        int res=0;

        //iterate till the queue is not
        //empty
      while(!q.empty())
        {
            int curr=q.front();
            q.pop();
            vis[curr]=1;
            for(int child:arr[curr])
            {
                if(vis[child]==0)
                {
                    res++;
                    q.push(child);
                }
            }
        for(int child:tr[curr])
        {
            if(vis[child]==0)
            {
                q.push(child);
            }
        }
        }
        return res;
}

int main()
{
    int  n = 5;
    vector<vector<int>>  connections ={{1,0},{1,2},{3,2},{3,4}};
    cout<<minReorder(n,connections);
    return 0;
}