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Convert a tree to forest of even nodes

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  • Difficulty Level : Hard
  • Last Updated : 28 Jun, 2021

Given a tree of n even nodes. The task is to find the maximum number of edges to be removed from the given tree to obtain forest of trees having even number of nodes. This problem is always solvable as given graph has even nodes.
Examples: 
 

Input : n = 10
Edge 1: 1 3
Edge 2: 1 6
Edge 3: 1 2
Edge 4: 3 4
Edge 5: 6 8
Edge 6: 2 7
Edge 7: 2 5
Edge 8: 4 9
Edge 9: 4 10

Output : 2

By removing 2 edges we can obtain the forest with even node tree.

Dotted line shows removed edges. Any further removal of edge will not satisfy 
the even nodes condition.

Find a subtree with even number of nodes and remove it from rest of tree by removing the edge connecting it. After removal, we are left with tree with even node only because initially we have even number of nodes in the tree and removed subtree has also even node. Repeat the same procedure until we left with the tree that cannot be further decomposed in this manner.
To do this, the idea is to use Depth First Search to traverse the tree. Implement DFS function in such a manner that it will return number of nodes in the subtree whose root is node on which DFS is performed. If the number of nodes is even then remove the edge, else ignore.
Below is implementation of this approach: 
 

C++




// C++ program to find maximum number to be removed
// to convert a tree into forest containing trees of
// even number of nodes
#include<bits/stdc++.h>
#define N 12
using namespace std;
 
// Return the number of nodes of subtree having
// node as a root.
int dfs(vector<int> tree[N], int visit[N],
                      int *ans, int node)
{
    int num = 0, temp = 0;
 
    // Mark node as visited.
    visit[node] = 1;
 
    // Traverse the adjacency list to find non-
    // visited node.
    for (int i = 0; i < tree[node].size(); i++)
    {
        if (visit[tree[node][i]] == 0)
        {
            // Finding number of nodes of the subtree
            // of a subtree.
            temp = dfs(tree, visit, ans, tree[node][i]);
 
            // If nodes are even, increment number of
            // edges to removed.
            // Else leave the node as child of subtree.
            (temp%2)?(num += temp):((*ans)++);
        }
    }
 
    return num+1;
}
 
// Return the maximum number of edge to remove
// to make forest.
int minEdge(vector<int> tree[N], int n)
{
    int visit[n+2];
    int ans = 0;
    memset(visit, 0, sizeof visit);
 
    dfs(tree, visit, &ans, 1);
 
    return ans;
}
 
// Driven Program
int main()
{
    int n = 10;
 
    vector<int> tree[n+2];
    tree[1].push_back(3);
    tree[3].push_back(1);
 
    tree[1].push_back(6);
    tree[6].push_back(1);
 
    tree[1].push_back(2);
    tree[2].push_back(1);
 
    tree[3].push_back(4);
    tree[4].push_back(3);
 
    tree[6].push_back(8);
    tree[8].push_back(6);
 
    tree[2].push_back(7);
    tree[7].push_back(2);
 
    tree[2].push_back(5);
    tree[5].push_back(2);
 
    tree[4].push_back(9);
    tree[9].push_back(4);
 
    tree[4].push_back(10);
    tree[10].push_back(4);
 
    cout << minEdge(tree, n) << endl;
    return 0;
}

Java




// Java program to find maximum number to be removed
// to convert a tree into forest containing trees of
// even number of nodes
import java.util.*;
 
class GFG
{
    static int N = 12,ans;
     
    static Vector<Vector<Integer>> tree=new Vector<Vector<Integer>>();
     
    // Return the number of nodes of subtree having
    // node as a root.
    static int dfs( int visit[], int node)
    {
        int num = 0, temp = 0;
     
        // Mark node as visited.
        visit[node] = 1;
     
        // Traverse the adjacency list to find non-
        // visited node.
        for (int i = 0; i < tree.get(node).size(); i++)
        {
            if (visit[tree.get(node).get(i)] == 0)
            {
                // Finding number of nodes of the subtree
                // of a subtree.
                temp = dfs( visit, tree.get(node).get(i));
     
                // If nodes are even, increment number of
                // edges to removed.
                // Else leave the node as child of subtree.
                if(temp%2!=0)
                num += temp;
                else
                ans++;
            }
        }
     
        return num+1;
    }
     
    // Return the maximum number of edge to remove
    // to make forest.
    static int minEdge( int n)
    {
        int visit[] = new int[n+2];
        ans = 0;
     
        dfs( visit, 1);
     
        return ans;
    }
     
    // Driven Program
    public static void main(String args[])
    {
        int n = 10;
         
        //set the size of vector
        for(int i = 0; i < n + 2;i++)
        tree.add(new Vector<Integer>());
     
        tree.get(1).add(3);
        tree.get(3).add(1);
     
        tree.get(1).add(6);
        tree.get(6).add(1);
     
        tree.get(1).add(2);
        tree.get(2).add(1);
     
        tree.get(3).add(4);
        tree.get(4).add(3);
     
        tree.get(6).add(8);
        tree.get(8).add(6);
     
        tree.get(2).add(7);
        tree.get(7).add(2);
     
        tree.get(2).add(5);
        tree.get(5).add(2);
     
        tree.get(4).add(9);
        tree.get(9).add(4);
     
        tree.get(4).add(10);
        tree.get(10).add(4);
     
        System.out.println( minEdge( n));
    }
}
 
// This code is contributed by Arnab Kundu

Python3




# Python3 program to find maximum
# number to be removed to convert
# a tree into forest containing trees
# of even number of nodes
 
# Return the number of nodes of 
# subtree having node as a root.
def dfs(tree, visit, ans, node):
    num = 0
    temp = 0
 
    # Mark node as visited.
    visit[node] = 1
 
    # Traverse the adjacency list 
    # to find non-visited node.
    for i in range(len(tree[node])):
        if (visit[tree[node][i]] == 0):
             
            # Finding number of nodes of
            # the subtree of a subtree.
            temp = dfs(tree, visit, ans,
                          tree[node][i])
 
            # If nodes are even, increment
            # number of edges to removed.
            # Else leave the node as child
            # of subtree.
            if(temp % 2):
                num += temp
            else:
                ans[0] += 1
 
    return num + 1
 
# Return the maximum number of
# edge to remove to make forest.
def minEdge(tree, n):
    visit = [0] * (n + 2)
    ans = [0]
    dfs(tree, visit, ans, 1)
 
    return ans[0]
 
# Driver Code
N = 12
n = 10
 
tree = [[] for i in range(n + 2)]
tree[1].append(3)
tree[3].append(1)
 
tree[1].append(6)
tree[6].append(1)
 
tree[1].append(2)
tree[2].append(1)
 
tree[3].append(4)
tree[4].append(3)
 
tree[6].append(8)
tree[8].append(6)
 
tree[2].append(7)
tree[7].append(2)
 
tree[2].append(5)
tree[5].append(2)
 
tree[4].append(9)
tree[9].append(4)
 
tree[4].append(10)
tree[10].append(4)
 
print(minEdge(tree, n))
 
# This code is contributed by pranchalK

C#




// C# program to find maximum number
// to be removed to convert a tree into
// forest containing trees of even number of nodes
using System;
using System.Collections.Generic;
 
class GFG
{
    static int N = 12, ans;
     
    static List<List<int>> tree = new List<List<int>>();
     
    // Return the number of nodes of
    // subtree having node as a root.
    static int dfs(int []visit, int node)
    {
        int num = 0, temp = 0;
     
        // Mark node as visited.
        visit[node] = 1;
     
        // Traverse the adjacency list to
        // find non-visited node.
        for (int i = 0; i < tree[node].Count; i++)
        {
            if (visit[tree[node][i]] == 0)
            {
                // Finding number of nodes of the
                // subtree of a subtree.
                temp = dfs(visit, tree[node][i]);
     
                // If nodes are even, increment number of
                // edges to removed.
                // Else leave the node as child of subtree.
                if(temp % 2 != 0)
                    num += temp;
                else
                    ans++;
            }
        }
        return num + 1;
    }
     
    // Return the maximum number of edge
    // to remove to make forest.
    static int minEdge(int n)
    {
        int []visit = new int[n + 2];
        ans = 0;
     
        dfs(visit, 1);
     
        return ans;
    }
     
    // Driver Code
    public static void Main(String []args)
    {
        int n = 10;
         
        //set the size of vector
        for(int i = 0; i < n + 2;i++)
        tree.Add(new List<int>());
     
        tree[1].Add(3);
        tree[3].Add(1);
     
        tree[1].Add(6);
        tree[6].Add(1);
     
        tree[1].Add(2);
        tree[2].Add(1);
     
        tree[3].Add(4);
        tree[4].Add(3);
     
        tree[6].Add(8);
        tree[8].Add(6);
     
        tree[2].Add(7);
        tree[7].Add(2);
     
        tree[2].Add(5);
        tree[5].Add(2);
     
        tree[4].Add(9);
        tree[9].Add(4);
     
        tree[4].Add(10);
        tree[10].Add(4);
     
        Console.WriteLine(minEdge(n));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// JavaScript program to find maximum number
// to be removed to convert a tree into
// forest containing trees of even number of nodes
var N = 12, ans;
 
var tree = Array();
 
// Return the number of nodes of
// subtree having node as a root.
function dfs(visit, node)
{
    var num = 0, temp = 0;
 
    // Mark node as visited.
    visit[node] = 1;
 
    // Traverse the adjacency list to
    // find non-visited node.
    for (var i = 0; i < tree[node].length; i++)
    {
        if (visit[tree[node][i]] == 0)
        {
            // Finding number of nodes of the
            // subtree of a subtree.
            temp = dfs(visit, tree[node][i]);
 
            // If nodes are even, increment number of
            // edges to removed.
            // Else leave the node as child of subtree.
            if(temp % 2 != 0)
                num += temp;
            else
                ans++;
        }
    }
    return num + 1;
}
 
// Return the maximum number of edge
// to remove to make forest.
function minEdge(n)
{
    var visit = Array(n+2).fill(0);
    ans = 0;
 
    dfs(visit, 1);
 
    return ans;
}
 
// Driver Code
var n = 10;
 
//set the size of vector
for(var i = 0; i < n + 2;i++)
    tree.push(new Array());
tree[1].push(3);
tree[3].push(1);
tree[1].push(6);
tree[6].push(1);
tree[1].push(2);
tree[2].push(1);
tree[3].push(4);
tree[4].push(3);
tree[6].push(8);
tree[8].push(6);
tree[2].push(7);
tree[7].push(2);
tree[2].push(5);
tree[5].push(2);
tree[4].push(9);
tree[9].push(4);
tree[4].push(10);
tree[10].push(4);
document.write(minEdge(n));
 
 
</script>

Output:  

2

Time Complexity: O(n).
Reference: 
http://stackoverflow.com/questions/12043252/obtain-forest-out-of-tree-with-even-number-of-nodes 
This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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