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Forest of a Tree in Data Structure

A is a data structure that is composed of nodes in a tree-like structure. Each node holds a value and a list of references to other nodes, called children. The root node is the starting point of the tree, and each node can have any number of children.

A tree is a data structure that is composed of nodes connected by edges. Each node holds a value and a list of references to other nodes, called children.

 The root node is the starting point of the tree, and each node can have any number of children. A tree is a subset of a forest, and a forest is a group of trees. In data structure, a forest is a set of trees. All the trees in a forest share the same root node.

Each tree in the forest can have different properties, and the nodes in each tree can be arranged in any way. A forest can be used to represent complex relationships between objects, and it is an efficient way to store hierarchical data.

For example, a forest can be used to represent a family tree. Each node in the forest can represent a family member, and a child node can represent a child of the family member. The forest can also be used to represent an organizational structure, with each node representing an employee or department in the organization.

By connecting the nodes with edges, the structure of the organization can be represented. In addition, a forest can also be used to store data in a graph structure. By connecting the nodes with edges, the relationships between the nodes can be represented.

This structure is useful when dealing with complex data such as social networks, where the relationships between different people can be represented. Overall, a forest is a powerful data structure that can be used to represent complex relationships between objects and store hierarchical data. It is an efficient way to store and visualize data, and it can be used in a variety of applications.

Tree is an important concept in data structure. A tree is a hierarchical data structure, consisting of nodes connected by edges. Each node has a parent node, which is the node above it in the hierarchy, and zero or more child nodes, which are the nodes below it. The root node is the topmost node in the tree. Trees are used to represent hierarchical structures, such as family trees, organizational structures, file systems, and networks.

They are also used in sorting algorithms, such as quicksort and heapsort. Trees are also used to represent the relationships between objects in a database. A forest of trees is a collection of trees.

Each tree in the forest is a separate entity, with its own structure and nodes. The trees in a forest can be related to each other, such as in a family tree, or they can be completely unrelated.

A forest of trees is a powerful data structure, as it allows for the efficient storage and manipulation of data. It is often used for storing and manipulating large data sets.

For example, a forest of trees can be used to represent a graph, with each tree representing a path from one node to another. The forest of trees is also used in algorithms such as dijkstra’s algorithm, Kruskal’s algorithm, and Prim’s algorithm. These algorithms are used to find the shortest path between two nodes in a graph.

The forest of trees is also used in artificial intelligence algorithms, such as Monte Carlo tree search.

This algorithm is used to find the best move in a game. The forest of trees is an important concept in data structure. It is used to efficiently store and manipulate data, as well as to solve complex problems. dijkstra’s algorithm, Kruskal’s algorithm, Prim’s algorithm, and Monte Carlo tree search are just a few of the algorithms that rely on forests of trees.

A tree is an undirected, linked, and acyclic graph according to graph theory. To put it another way, a tree is a linked graph without even a single cycle.

A graphical representation of hierarchical organisation is a tree.

The nodes and branches of a tree, respectively, are its constituent parts.

There are (n-1) edges in a tree with n vertices.

Applications of a Forest Data Structure:

  • Big Data Web Scrapers

In a website's organisational structure, which resembles a tree, the main page serves as the root node and the subsequent hyperlinks from that page serve as the nodes for the rest of the Tree. Online scrapers present the data they have gathered as a forest of trees when they have gathered it from multiple related websites.

  • Social Networking related Websites

In order to explain their data, social networking services (such as Facebook, LinkedIn, Twitter, etc.) use tree and graph data structures. By working to add two people as friends, you create a forest of two people.

  • Operating System Storage

If your operating system was Windows-based, you would be able to access several discs in the system, including C drive (C:), D drive (D:), etc. A forest can be compared to the complete storage, while each drive can be thought of as a separate tree.

  • Big Data Web Scrapers

In a website's organisational structure, which resembles a tree, the main page serves as the root node and the subsequent hyperlinks from that page serve as the nodes for the rest of the Tree. Online scrapers present the data they have gathered as a forest of trees when they have gathered it from multiple related websites.

Converting a tree to a forest of even nodes:

A tree with n even nodes is given. To create a forest of trees with an even number of nodes, the challenge is to determine the greatest number of edges that may be eliminated from the given tree. As the provided graph includes even nodes, this problem can always be solved.

Remove an even-numbered subtree from the rest of the tree by cutting the edge that connects it. Only because the original number of nodes in the tree was even, and the deleted subtree likewise included an even node, are we left with a tree with an even number of nodes. Continue this process until we are left with a tree that cannot be disassembled any more in this way.

To do this, the plan is to navigate the tree using Depth First Search. The DFS function should be implemented so that it returns the number of nodes in the subtree whose root is the node on which DFS is executed. Remove the edge if there are an equal number of nodes; else, ignore.

Implementation of the above approach in java language:

// A Java programme to determine the maximum number of nodes to remove to turn a tree into a forest of trees with an even number of nodes
import java. util. *;
 
class ForestDemo
{
    static int N = 12, ans;
     
    static Vector < Vector < Integer > > tree = new Vector < Vector < Integer > > ( );
     
    // Retrieve the number of subtree nodes that have node as their root.
    static int dfsearch ( int visited [ ], int node)
    {
        int num = 0, temp = 0;
     
        // Mark the below node as Visited.
        visited [ node] = 1;
     
        // To identify a node that hasn't been visited, search the adjacency list.
        for (int i = 0; i < tree. get (node). Size ( ); i ++)
        {
            if (visited [ tree. get (node). get (i)] == 0)
            {
                // Counting the nodes in a subtree of a subtree.
                tempNd = dfsearch ( visited, tree. get (node). get (i));
     
                // Increase the number of edges to be deleted if the number of nodes is even. 
// Otherwise, leave the node as a subtree child.
                if (tempNd % 2 != 0)
                num += tempNd;
                else
                ans ++;
            }
        }
     
        return num + 1;
    }
     
    // Give back the most edges that need to be cut out to create a forest.
    static int getMinEdge ( int n)
    {
        int visited [ ] = new int [ n + 2];
        ans = 0;
     
        dfsearch ( visit, 1);
     
        return ans;
    }
     
    // Driven Program for checking the above functionalities
    public static void main (String args [ ])
    {
        int n = 10;
         
        // determine the vector's size
        for (int i = 0; i < n + 2; i ++)
        tree. add (new Vector < Integer > ( ));
     
        tree. get (10). add (30);
        tree. get (30). add (10);
     
        tree. get (10). add (60);
        tree. get (60). add (10);
     
        tree. get (10). add (20);
        tree. get (20). add (10);
     
        tree. get (30). add (40);
        tree. get (40). add (30);
     
        tree. get (60). add (80);
        tree. get (80). add (60);
     
        tree. get (20). add (70);
        tree. get (70). add (20);
     
        tree. get (20). add (50);
        tree. get (50). add (20);
     
        tree. get (40). add (90);
        tree. get (90). add (40);
     
        tree. get (40). add (100);
        tree. get (100). add (40);
     
        System. out. println (getMinEdge ( n));
    }
}

OUTPUT:

20

For the above approach the time and space complexity will be O (N).

Implementation of the above approach in C++ language:

// A C++ programme to determine the maximum number of nodes to remove to turn a tree an even number of nodes in a grove of trees


# include < bits / stdc ++. h >
# define N 12
using namespace std;
 
// The number of nodes in the subtree that has the root node as its value.
int dfsearch (vector < int > tree [ N], int visit [ N],
                      int * ans, int node)
{
    int num = 0, temp = 0;
 
    // make the value for the visited node as visited.
    visit [ node] = 1;
 
    // now traverse the remaining adjacency list for finding the non visited nodes in the list.


    for (int i = 0; i < tree [ node]. size ( ); I ++)
    {
        if (visit [tree [node] [ i]] == 0)
        {
            // Get the number of nodes present in the sub tree.
            temp = dfsearch (tree, visit, ans, tree [ node] [ i]);
 
            // If the number of nodes present in the sub tree is even, then increment the value of the edges that are to be removed.
            // If the number of nodes present in the sub tree is odd, then leave the nodes as the child nodes.
            (temp % 2) ? (num += temp) : (( * ans) ++);
        }
    }
 
    return num + 1;
}
 
// Give back the most edges that need to be cut out to create a forest.
int minEdgeCount (vector < int > tree [ N], int n)
{
    int visit [ n + 2];
    int ans = 0;
    memset (visit, 0, sizeof visit);
 
    dfsearch (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 << minEdgeCount (tree, n) << endl;
    return 0;
}

OUTPUT:

20

For the above approach the time and space complexity will be O (N).

Implementation of the above approach in python language:

# A Python 3 programme to determine the maximum number of nodes to remove to turn a tree into a forest of trees with an even number of nodes
 
# Retrieve the number of subtree nodes that have node as their root.
def dfsearch (tree, visit, ans, node):
    num = 0
    temp = 0
 
    # make the value for the visited node as visited..
    visit [ node] = 1
 
    # To identify a node that hasn't been visited, search the adjacency list.
    for i in range (len (tree [ node])):
        if (visit [ tree [ node] [ i] ] == 0):
             
            # Counting the nodes in a subtree of a subtree.


            temp = dfsearch (tree, visit, ans,
                          tree [ node] [ i])
 
            # If the number of nodes present in the sub tree is even, then increment the value of the edges that are to be removed.
            # If the number of nodes present in the sub tree is odd, leave the node as child of subtree.
            if (temp % 2):
                num += temp
            else:
                ans [ 0] += 1
 
    return num + 1
 
# Give back the most edges that need to be cut out to create a forest.
def minEdgeCount (tree, n):
    visit = [ 0] * (n + 2)
    ans = [ 0]
    dfsearch (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 (minEdgeCount (tree, n))

OUTPUT:

2

For the above approach the time and space complexity will be O (N).