Skip to content
Related Articles

Related Articles

Improve Article

Minimum time required to visit all the special nodes of a Tree

  • Difficulty Level : Hard
  • Last Updated : 22 Jun, 2021

Given an undirected tree consisting of N vertices where some of the nodes are special nodes, the task is to visit all the special nodes from the root node in minimum time. Time for traveling from one node to another node can be assumed as unit time.

A node is special if the path from the root to the node consists of distinct value nodes.

Example: 

Input: N = 7, edges[] = {(0, 1), (0, 2), (1, 4), (1, 5), (2, 3), (2, 6)} 
isSpecial[] = {false, false, true, false, true, true, false} 
Output:
Explanation: 



Input: N = 7, edges[] = {(0, 1), (0, 2), (1, 4), (1, 5), (2, 3), (2, 6)} 
isSpecial[] = {false, false, true, false, false, true, false} 
Output:
Explanation: 

Approach: The idea is to use Depth First Search traversal and traverse the nodes. If any node is having a children which is a special node, add two to the steps required for that node. Also mark that node as a special node such that while moving up the steps are taken into consideration.

Below is the implementation of the above approach: 

C++




// C++ implementation to find
// the minimum time required to
// visit special nodes of a tree
 
#include <bits/stdc++.h>
using namespace std;
 
const int N = 100005;
 
// Time required to collect
vector<int> ans(N, 0);
 
vector<int> flag(N, 0);
 
// Minimum time required to reach
// all the special nodes of tree
void minimumTime(int u, int par,
                 vector<bool>& hasApple,
                 vector<int> adj[])
{
 
    // Condition to check if
    // the vertex has apple
    if (hasApple[u] == true)
        flag[u] = 1;
 
    // Iterate all the
    // adjacent of vertex u.
    for (auto it : adj[u]) {
 
        // if adjacent vertex
        // is it's parent
        if (it != par) {
            minimumTime(it, u, hasApple, adj);
 
            // if any vertex of subtree
            // it contain apple
            if (flag[it] > 0)
                ans[u] += (ans[it] + 2);
 
            // flagbit for node u
            // would be on if any vertex
            // in it's subtree contain apple
            flag[u] |= flag[it];
        }
    }
}
 
// Driver Code
int main()
{
    // Number of the vertex.
    int n = 7;
 
    vector<bool> hasApple{ false, false,
                           true, false,
                           true, true,
                           false };
 
    // Store all the edges,
    // any edge represented
    // by pair of vertex
    vector<pair<int, int> > edges;
 
    // Added all the
    // edge in edges vector.
    edges.push_back(make_pair(0, 1));
    edges.push_back(make_pair(0, 2));
    edges.push_back(make_pair(1, 4));
    edges.push_back(make_pair(1, 5));
    edges.push_back(make_pair(2, 3));
    edges.push_back(make_pair(2, 6));
 
    // Adjacent list
    vector<int> adj[n];
 
    for (int i = 0; i < edges.size(); i++) {
        int source_node = edges[i].first;
 
        int destination_node
            = edges[i].second;
 
        adj[source_node]
            .push_back(destination_node);
 
        adj[destination_node]
            .push_back(source_node);
    }
 
    // Function Call
    minimumTime(0, -1, hasApple, adj);
 
    cout << ans[0];
    return 0;
}

Java




// Java implementation to find
// the minimum time required to
// visit special nodes of a tree
import java.util.*;
 
@SuppressWarnings("unchecked")
class GFG{
 
static class pair
{
    int first, second;
     
    pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
static final int N = 100005;
  
// Time required to collect
static ArrayList ans;
static ArrayList flag;
  
// Minimum time required to reach
// all the special nodes of tree
static void minimumTime(int u, int par,
                        ArrayList hasApple,
                        ArrayList adj[])
{
     
    // Condition to check if
    // the vertex has apple
    if ((boolean)hasApple.get(u) == true)
        flag.set(u, 1);
  
    // Iterate all the
    // adjacent of vertex u.
    for(int it : (ArrayList<Integer>)adj[u])
    {
         
        // If adjacent vertex
        // is it's parent
        if (it != par)
        {
            minimumTime(it, u, hasApple, adj);
  
            // If any vertex of subtree
            // it contain apple
            if ((int)flag.get(it) > 0)
                ans.set(u, (int)ans.get(u) +
                           (int)ans.get(it) + 2 );
  
            // flagbit for node u
            // would be on if any vertex
            // in it's subtree contain apple
            flag.set(u, (int)flag.get(u) |
                        (int)flag.get(it));
        }
    }
}
  
// Driver Code
public static void main(String []args)
{
     
    // Number of the vertex.
    int n = 7;
 
    ans = new ArrayList();
    flag = new ArrayList();
     
    for(int i = 0; i < N; i++)
    {
        ans.add(0);
        flag.add(0);
    }
     
    ArrayList hasApple = new ArrayList();
    hasApple.add(false);
    hasApple.add(false);
    hasApple.add(true);
    hasApple.add(false);
    hasApple.add(true);
    hasApple.add(true);
    hasApple.add(false);
  
    // Store all the edges,
    // any edge represented
    // by pair of vertex
    ArrayList edges = new ArrayList();
  
    // Added all the edge in
    // edges vector.
    edges.add(new pair(0, 1));
    edges.add(new pair(0, 2));
    edges.add(new pair(1, 4));
    edges.add(new pair(1, 5));
    edges.add(new pair(2, 3));
    edges.add(new pair(2, 6));
  
    // Adjacent list
    ArrayList []adj = new ArrayList[n];
 
    for(int i = 0; i < n; i++)
    {
        adj[i] = new ArrayList();
    }
  
    for(int i = 0; i < edges.size(); i++)
    {
        int source_node = ((pair)edges.get(i)).first;
        int destination_node = ((pair)edges.get(i)).second;
  
        adj[source_node].add(destination_node);
        adj[destination_node].add(source_node);
    }
  
    // Function Call
    minimumTime(0, -1, hasApple, adj);
     
    System.out.print(ans.get(0));
}
}
 
// This code is contributed by pratham76

Python3




# Python3 implementation to find
# the minimum time required to
# visit special nodes of a tree
N = 100005
  
# Time required to collect
ans = [0 for i in range(N)]
flag = [0 for i in range(N)]
  
# Minimum time required to reach
# all the special nodes of tree
def minimumTime(u, par, hasApple, adj):
  
    # Condition to check if
    # the vertex has apple
    if (hasApple[u] == True):
        flag[u] = 1
  
    # Iterate all the
    # adjacent of vertex u.
    for it in adj[u]:
  
        # if adjacent vertex
        # is it's parent
        if (it != par):
            minimumTime(it, u, hasApple, adj)
  
            # if any vertex of subtree
            # it contain apple
            if (flag[it] > 0):
                ans[u] += (ans[it] + 2)
  
            # flagbit for node u
            # would be on if any vertex
            # in it's subtree contain apple
            flag[u] |= flag[it]
  
# Driver Code
if __name__=='__main__':
 
    # Number of the vertex.
    n = 7
  
    hasApple = [ False, False, True,
                 False, True, True, False ]
  
    # Store all the edges,
    # any edge represented
    # by pair of vertex
    edges = []
  
    # Added all the
    # edge in edges vector.
    edges.append([0, 1])
    edges.append([0, 2])
    edges.append([1, 4])
    edges.append([1, 5])
    edges.append([2, 3])
    edges.append([2, 6])
  
    # Adjacent list
    adj = [[] for i in range(n)]
     
    for i in range(len(edges)):
        source_node = edges[i][0]
         
        destination_node = edges[i][1]
  
        adj[source_node].append(destination_node)
        adj[destination_node].append(source_node)
     
    # Function Call
    minimumTime(0, -1, hasApple, adj);
     
    print(ans[0])
     
# This code is contributed by rutvik_56

Javascript




<script>
 
    // JavaScript implementation to find
    // the minimum time required to
    // visit special nodes of a tree
     
    let N = 100005;
   
    // Time required to collect
    let ans = [];
    let flag = [];
 
    // Minimum time required to reach
    // all the special nodes of tree
    function minimumTime(u, par, hasApple, adj)
    {
 
        // Condition to check if
        // the vertex has apple
        if (hasApple[u] == true)
            flag[u] = 1;
 
        // Iterate all the
        // adjacent of vertex u.
        for(let it = 0; it < adj[u].length; it++)
        {
 
            // If adjacent vertex
            // is it's parent
            if (adj[u][it] != par)
            {
                minimumTime(adj[u][it], u, hasApple, adj);
 
                // If any vertex of subtree
                // it contain apple
                if (flag[adj[u][it]] > 0)
                    ans[u] = ans[u] + ans[adj[u][it]] + 2 ;
 
                // flagbit for node u
                // would be on if any vertex
                // in it's subtree contain apple
                flag[u] = flag[u] | flag[adj[u][it]];
            }
        }
    }
     
    // Number of the vertex.
    let n = 7;
  
    ans = [];
    flag = [];
      
    for(let i = 0; i < N; i++)
    {
        ans.push(0);
        flag.push(0);
    }
      
    let hasApple = [];
    hasApple.push(false);
    hasApple.push(false);
    hasApple.push(true);
    hasApple.push(false);
    hasApple.push(true);
    hasApple.push(true);
    hasApple.push(false);
   
    // Store all the edges,
    // any edge represented
    // by pair of vertex
    let edges = [];
   
    // Added all the edge in
    // edges vector.
    edges.push([0, 1]);
    edges.push([0, 2]);
    edges.push([1, 4]);
    edges.push([1, 5]);
    edges.push([2, 3]);
    edges.push([2, 6]);
   
    // Adjacent list
    let adj = new Array(n);
  
    for(let i = 0; i < n; i++)
    {
        adj[i] = [];
    }
   
    for(let i = 0; i < edges.length; i++)
    {
        let source_node = edges[i][0];
        let destination_node = edges[i][1];
   
        adj[source_node].push(destination_node);
        adj[destination_node].push(source_node);
    }
   
    // Function Call
    minimumTime(0, -1, hasApple, adj);
      
    document.write(ans[0]);
     
</script>
Output: 
8

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :