Count number of trees in a forest

Last Updated : 14 Apr, 2023

Given n nodes of a forest (collection of trees), find the number of trees in the forest.

Examples :

Input :  edges[] = {0, 1}, {0, 2}, {3, 4}
Output : 2
Explanation : There are 2 trees
                   0       3
                  / \       \
                 1   2       4

Approach :

Apply DFS on every node.
Increment count by one if every connected node is visited from one source.
Again perform DFS traversal if some nodes yet not visited.
Count will give the number of trees in forest.

Implementation:

C++

// CPP program to count number of trees in
// a forest.
#include<bits/stdc++.h>
using namespace std;
 
// A utility function to add an edge in an
// undirected graph.
void addEdge(vector<int> adj[], int u, int v)
{
    adj[u].push_back(v);
    adj[v].push_back(u);
}
 
// A utility function to do DFS of graph
// recursively from a given vertex u.
void DFSUtil(int u, vector<int> adj[],
                    vector<bool> &visited)
{
    visited[u] = true;
    for (int i=0; i<adj[u].size(); i++)
        if (visited[adj[u][i]] == false)
            DFSUtil(adj[u][i], adj, visited);
}
 
// Returns count of tree is the forest
// given as adjacency list.
int countTrees(vector<int> adj[], int V)
{
    vector<bool> visited(V, false);
    int res = 0;
    for (int u=0; u<V; u++)
    {
        if (visited[u] == false)
        {
            DFSUtil(u, adj, visited);
            res++;
        }
    }
    return res;
}
 
// Driver code
int main()
{
    int V = 5;
    vector<int> adj[V];
    addEdge(adj, 0, 1);
    addEdge(adj, 0, 2);
    addEdge(adj, 3, 4);
    cout << countTrees(adj, V);
    return 0;
}

Java

// Java program to count number of trees in a forest.
import java.io.*; 
import java.util.*; 
 
// This class represents a directed graph using adjacency list 
// representation 
class Graph 
{ 
    private int V; // No. of vertices 
 
    // Array of lists for Adjacency List Representation 
    private LinkedList<Integer> adj[]; 
 
    // Constructor 
    Graph(int v) 
    { 
        V = v; 
        adj = new LinkedList[v]; 
        for (int i = 0; i <  v; ++i) 
            adj[i] = new LinkedList(); 
    } 
 
    //Function to add an edge into the graph 
    void addEdge(int v, int w) 
    { 
        adj[v].add(w); // Add w to v's list. 
    } 
 
    // A function used by DFS 
    void DFSUtil(int v,boolean visited[]) 
    { 
        // Mark the current node as visited and print it 
        visited[v] = true; 
         
        // Recur for all the vertices adjacent to this vertex 
        Iterator<Integer> i = adj[v].listIterator(); 
        while (i.hasNext()) 
        { 
            int n = i.next(); 
            if (!visited[n])
            {
                DFSUtil(n,visited); 
            }
        }
    } 
 
    // The function to do DFS traversal. It uses recursive DFSUtil() 
    int countTrees()
    { 
        // Mark all the vertices as not visited(set as 
        // false by default in java) 
        boolean visited[] = new boolean[V]; 
        int res = 0;
         
        // Call the recursive helper function to print DFS traversal 
        // starting from all vertices one by one 
        for (int i = 0; i < V; ++i) 
        {
            if (visited[i] == false)
            { 
                DFSUtil(i, visited); 
                res ++;
            }
        }
        return res;
    } 
 
    // Driver code
    public static void main(String args[]) 
    { 
        Graph g = new Graph(5); 
 
        g.addEdge(0, 1); 
        g.addEdge(0, 2); 
        g.addEdge(3, 4); 
 
        System.out.println(g.countTrees()); 
    } 
} 
 
// This code is contributed by mayankbansal2

Python3

# Python3 program to count number  
# of trees in a forest.
 
# A utility function to add an 
# edge in an undirected graph. 
def addEdge(adj, u, v):
    adj[u].append(v) 
    adj[v].append(u)
 
# A utility function to do DFS of graph 
# recursively from a given vertex u. 
def DFSUtil(u, adj, visited):
    visited[u] = True
    for i in range(len(adj[u])):
        if (visited[adj[u][i]] == False):
            DFSUtil(adj[u][i], adj, visited)
 
# Returns count of tree is the 
# forest given as adjacency list. 
def countTrees(adj, V):
    visited = [False] * V 
    res = 0
    for u in range(V):
        if (visited[u] == False):
            DFSUtil(u, adj, visited) 
            res += 1
    return res
 
# Driver code 
if __name__ == '__main__':
 
    V = 5
    adj = [[] for i in range(V)] 
    addEdge(adj, 0, 1) 
    addEdge(adj, 0, 2) 
    addEdge(adj, 3, 4) 
    print(countTrees(adj, V))
 
# This code is contributed by PranchalK

C#

// C# program to count number of trees in a forest.
using System;
using System.Collections.Generic;
 
// This class represents a directed graph 
// using adjacency list representation 
class Graph 
{ 
    private int V; // No. of vertices 
 
    // Array of lists for 
    // Adjacency List Representation 
    private List<int> []adj; 
 
    // Constructor 
    Graph(int v) 
    { 
        V = v; 
        adj = new List<int>[v]; 
        for (int i = 0; i < v; ++i) 
            adj[i] = new List<int>(); 
    } 
 
    // Function to add an edge into the graph 
    void addEdge(int v, int w) 
    { 
        adj[v].Add(w); // Add w to v's list. 
    } 
 
    // A function used by DFS 
    void DFSUtil(int v, bool []visited) 
    { 
        // Mark the current node as 
        // visited and print it 
        visited[v] = true; 
         
        // Recur for all the vertices 
        // adjacent to this vertex 
        foreach(int i in adj[v]) 
        { 
            int n = i; 
            if (!visited[n])
            {
                DFSUtil(n, visited); 
            }
        }
    } 
 
    // The function to do DFS traversal.
    // It uses recursive DFSUtil() 
    int countTrees()
    { 
        // Mark all the vertices as not visited
        // (set as false by default in java) 
        bool []visited = new bool[V]; 
        int res = 0;
         
        // Call the recursive helper function 
        // to print DFS traversal starting from 
        // all vertices one by one 
        for (int i = 0; i < V; ++i) 
        {
            if (visited[i] == false)
            { 
                DFSUtil(i, visited); 
                res ++;
            }
        }
        return res;
    } 
 
    // Driver code
    public static void Main(String []args) 
    { 
        Graph g = new Graph(5); 
 
        g.addEdge(0, 1); 
        g.addEdge(0, 2); 
        g.addEdge(3, 4); 
 
        Console.WriteLine(g.countTrees()); 
    } 
} 
 
// This code is contributed by PrinciRaj1992

Javascript

<script>
// Javascript program to count number of trees in a forest.
 
// This class represents a directed graph 
// using adjacency list representation 
var V; // No. of vertices 
 
// Array of lists for 
// Adjacency List Representation 
var adj; 
// Constructor 
function Graph( v) 
{ 
    V = v; 
    adj = Array.from(Array(v), ()=>Array());
} 
// Function to add an edge into the graph 
function addEdge(v, w) 
{ 
    adj[v].push(w); // Add w to v's list. 
} 
// A function used by DFS 
function DFSUtil(v, visited) 
{ 
    // Mark the current node as 
    // visited and print it 
    visited[v] = true; 
     
    // Recur for all the vertices 
    // adjacent to this vertex 
    for(var i of adj[v]) 
    { 
        var n = i; 
        if (!visited[n])
        {
            DFSUtil(n, visited); 
        }
    }
} 
// The function to do DFS traversal.
// It uses recursive DFSUtil() 
function countTrees()
{ 
    // Mark all the vertices as not visited
    // (set as false by default in java) 
    var visited = Array(V).fill(false); 
    var res = 0;
     
    // Call the recursive helper function 
    // to print DFS traversal starting from 
    // all vertices one by one 
    for(var i = 0; i < V; ++i) 
    {
        if (visited[i] == false)
        { 
            DFSUtil(i, visited); 
            res ++;
        }
    }
    return res;
} 
 
// Driver code
Graph(5); 
addEdge(0, 1); 
addEdge(0, 2); 
addEdge(3, 4); 
document.write(countTrees()); 
 
// This code is contributed by rutvik_56.
</script>

Output

Time Complexity: O(V + E), where V is the number of vertices and E is the number of edges.

Space Complexity: O(V). We use an array of size V to store the visited nodes.

Approach:- Here’s an implementation of counting the number of trees in a forest using BFS in C++

Define a bfs function that takes the forest, a start node, and a visited array as inputs. The function performs BFS starting from the start node and marks all visited nodes in the visited array.
Inside the bfs function, create a queue q to store the nodes that are to be visited in the BFS. Initially, push the start node onto the queue and mark it as visited in the visited array.

C++

#include <iostream>
#include <queue>
#include <vector>
using namespace std;
 
// define a pair to represent a node in the forest
typedef pair<int, int> Node;
 
// function to perform BFS from a given node and mark all visited nodes
void bfs(vector<vector<int>>& forest, Node start, vector<vector<bool>>& visited) {
    // create a queue for BFS
    queue<Node> q;
    q.push(start);
    visited[start.first][start.second] = true;
 
    // BFS loop
    while (!q.empty()) {
        Node curr = q.front();
        q.pop();
 
        // add unvisited neighboring nodes to the queue
        int dx[] = {-1, 0, 1, 0};
        int dy[] = {0, 1, 0, -1};
        for (int i = 0; i < 4; i++) {
            int nx = curr.first + dx[i];
            int ny = curr.second + dy[i];
            if (nx >= 0 && nx < forest.size() && ny >= 0 && ny < forest[0].size() && forest[nx][ny] == 1 && !visited[nx][ny]) {
                q.push(make_pair(nx, ny));
                visited[nx][ny] = true;
            }
        }
    }
}
 
// function to count the number of trees in a forest using BFS
int count_trees_in_forest(vector<vector<int>>& forest) {
    int count = 0;
    int n = forest.size();
    int m = forest[0].size();
 
    // create a 2D boolean array to keep track of visited nodes
    vector<vector<bool>> visited(n, vector<bool>(m, false));
 
    // iterate over all nodes in the forest and perform BFS from each unvisited tree
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (forest[i][j] == 1 && !visited[i][j]) {
                bfs(forest, make_pair(i, j), visited);
                count++;
            }
        }
    }
 
    return count;
}
 
int main() {
    // example usage
    vector<vector<int>> forest = {
        {0, 1, 1, 0, 0},
        {0, 0, 0, 0, 0},
        {0, 0, 0, 0, 0},
        {0, 0, 0, 0, 1},
        {0, 0, 0, 0, 0}
    };
    int num_trees = count_trees_in_forest(forest);
    cout << "The forest has " << num_trees << " trees." << endl;
    return 0;
}

Java

import java.util.*;
 
public class Forest {
    // define a pair to represent a node in the forest
    static class Node {
        int x;
        int y;
        Node(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }
 
    // function to perform BFS from a given node and mark all visited nodes
    static void bfs(int[][] forest, Node start, boolean[][] visited) {
        // create a queue for BFS
        Queue<Node> q = new LinkedList<>();
        q.add(start);
        visited[start.x][start.y] = true;
 
        // BFS loop
        while (!q.isEmpty()) {
            Node curr = q.poll();
 
            // add unvisited neighboring nodes to the queue
            int[] dx = {-1, 0, 1, 0};
            int[] dy = {0, 1, 0, -1};
            for (int i = 0; i < 4; i++) {
                int nx = curr.x + dx[i];
                int ny = curr.y + dy[i];
                if (nx >= 0 && nx < forest.length && ny >= 0 && ny < forest[0].length && forest[nx][ny] == 1 && !visited[nx][ny]) {
                    q.add(new Node(nx, ny));
                    visited[nx][ny] = true;
                }
            }
        }
    }
 
    // function to count the number of trees in a forest using BFS
    static int count_trees_in_forest(int[][] forest) {
        int count = 0;
        int n = forest.length;
        int m = forest[0].length;
 
        // create a 2D boolean array to keep track of visited nodes
        boolean[][] visited = new boolean[n][m];
 
        // iterate over all nodes in the forest and perform BFS from each unvisited tree
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (forest[i][j] == 1 && !visited[i][j]) {
                    bfs(forest, new Node(i, j), visited);
                    count++;
                }
            }
        }
 
        return count;
    }
 
    public static void main(String[] args) {
        // example usage
        int[][] forest = {
            {0, 1, 1, 0, 0},
            {0, 0, 0, 0, 0},
            {0, 0, 0, 0, 0},
            {0, 0, 0, 0, 1},
            {0, 0, 0, 0, 0}
        };
        int num_trees = count_trees_in_forest(forest);
        System.out.println("The forest has " + num_trees + " trees.");
    }
}

Python3

from typing import List, Tuple
from queue import Queue
 
# define a tuple to represent a node in the forest
Node = Tuple[int, int]
 
# function to perform BFS from a given node and mark all visited nodes
 
 
def bfs(forest: List[List[int]], start: Node, visited: List[List[bool]]) -> None:
    # create a queue for BFS
    q = Queue()
    q.put(start)
    visited[start[0]][start[1]] = True
 
    # BFS loop
    while not q.empty():
        curr = q.get()
 
        # add unvisited neighboring nodes to the queue
        dx = [-1, 0, 1, 0]
        dy = [0, 1, 0, -1]
        for i in range(4):
            nx = curr[0] + dx[i]
            ny = curr[1] + dy[i]
            if 0 <= nx < len(forest) and 0 <= ny < len(forest[0]) and forest[nx][ny] == 1 and not visited[nx][ny]:
                q.put((nx, ny))
                visited[nx][ny] = True
 
# function to count the number of trees in a forest using BFS
 
 
def count_trees_in_forest(forest: List[List[int]]) -> int:
    count = 0
    n, m = len(forest), len(forest[0])
 
    # create a 2D boolean array to keep track of visited nodes
    visited = [[False for _ in range(m)] for _ in range(n)]
 
    # iterate over all nodes in the forest and perform BFS from each unvisited tree
    for i in range(n):
        for j in range(m):
            if forest[i][j] == 1 and not visited[i][j]:
                bfs(forest, (i, j), visited)
                count += 1
 
    return count
 
 
# example usage
forest = [
    [0, 1, 1, 0, 0],
    [0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0],
    [0, 0, 0, 0, 1],
    [0, 0, 0, 0, 0]
]
num_trees = count_trees_in_forest(forest)
print(f"The forest has {num_trees} trees.")

C#

// C# code for above mentioned approach
using System;
using System.Collections.Generic;
 
class Program {
    // define a pair to represent a node in the forest
    class Node {
        public int x;
        public int y;
        public Node(int x, int y)
        {
            this.x = x;
            this.y = y;
        }
    }
 
    // function to perform BFS from a given node and mark
    // all visited nodes
    static void bfs(int[][] forest, Node start,
                    bool[][] visited)
    {
        // create a queue for BFS
        Queue<Node> q = new Queue<Node>();
        q.Enqueue(start);
        visited[start.x][start.y] = true;
 
        // BFS loop
        while (q.Count != 0) {
            Node curr = q.Dequeue();
 
            // add unvisited neighboring nodes to the queue
            int[] dx = { -1, 0, 1, 0 };
            int[] dy = { 0, 1, 0, -1 };
            for (int i = 0; i < 4; i++) {
                int nx = curr.x + dx[i];
                int ny = curr.y + dy[i];
                if (nx >= 0 && nx < forest.Length && ny >= 0
                    && ny < forest[0].Length
                    && forest[nx][ny] == 1
                    && !visited[nx][ny]) {
                    q.Enqueue(new Node(nx, ny));
                    visited[nx][ny] = true;
                }
            }
        }
    }
 
    // function to count the number of trees in a forest
    // using BFS
    static int count_trees_in_forest(int[][] forest)
    {
        int count = 0;
        int n = forest.Length;
        int m = forest[0].Length;
 
        // create a 2D boolean array to keep track of
        // visited nodes
        bool[][] visited = new bool[n][];
        for (int i = 0; i < n; i++) {
            visited[i] = new bool[m];
        }
 
        // iterate over all nodes in the forest and perform
        // BFS from each unvisited tree
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (forest[i][j] == 1 && !visited[i][j]) {
                    bfs(forest, new Node(i, j), visited);
                    count++;
                }
            }
        }
 
        return count;
    }
 
    static void Main(string[] args)
    {
        // example usage
        int[][] forest = { new int[] { 0, 1, 1, 0, 0 },
                           new int[] { 0, 0, 0, 0, 0 },
                           new int[] { 0, 0, 0, 0, 0 },
                           new int[] { 0, 0, 0, 0, 1 },
                           new int[] { 0, 0, 0, 0, 0 } };
        int num_trees = count_trees_in_forest(forest);
        Console.WriteLine("The forest has " + num_trees
                          + " trees.");
    }
}
 
// This code is contributed by Tapesh(tapeshdua420)

Javascript

// define a pair to represent a node in the forest
class Node {
    constructor(x, y) {
        this.x = x;
        this.y = y;
    }
}
 
// function to perform BFS from a given node and mark all visited nodes
function bfs(forest, start, visited) {
    // create a queue for BFS
    let q = [];
    q.push(start);
    visited[start.x][start.y] = true;
 
    // BFS loop
    while (q.length > 0) {
        let curr = q.shift();
 
        // add unvisited neighboring nodes to the queue
        let dx = [-1, 0, 1, 0];
        let dy = [0, 1, 0, -1];
        for (let i = 0; i < 4; i++) {
            let nx = curr.x + dx[i];
            let ny = curr.y + dy[i];
            if (nx >= 0 && nx < forest.length && ny >= 0 && ny < forest[0].length && forest[nx][ny] == 1 && !visited[nx][ny]) {
                q.push(new Node(nx, ny));
                visited[nx][ny] = true;
            }
        }
    }
}
 
// function to count the number of trees in a forest using BFS
function count_trees_in_forest(forest) {
    let count = 0;
    let n = forest.length;
    let m = forest[0].length;
 
    // create a 2D boolean array to keep track of visited nodes
    let visited = new Array(n);
    for (let i=0; i<n; i++) {
      visited[i] = new Array(m).fill(false);
    }
 
    // iterate over all nodes in the forest and perform BFS from each unvisited tree
    for (let i=0; i<n; i++) {
      for (let j=0; j<m; j++) {
          if (forest[i][j] == 1 && !visited[i][j]) {
              bfs(forest,new Node(i,j),visited);
              count++;
          }
      }
   }
 
   return count;
}
 
let forest=[
 [0 ,1 ,1 ,0 ,0],
 [0 ,0 ,0 ,0 ,0],
 [0, 0, 0, 0, 0],
 [0, 0, 0, 0, 1],
 [0, 0, 0, 0, 0]];
let num_trees = count_trees_in_forest(forest);
console.log("The forest has " + num_trees + " trees.");

Output

The forest has 2 trees.

Time complexity : – O(NM)
Auxiliary Space :- O(MN)

Bibhu Pala

Improve

Find size of the largest region in Boolean Matrix

A Peterson Graph Problem

BFS and DFS on Graph

Cycle in a Graph

Shortest Paths in Graph

Minimum Spanning Tree in Graph

Topological Sorting in Graph

Connectivity of Graph

Maximum flow in a Graph

Some must do problems on Graph

Count number of trees in a forest

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

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