Count frequency of K in given Binary Tree

Last Updated : 24 Mar, 2023

Given a binary tree of N nodes. Count the frequency of an integer K in the binary tree.

Examples:

Input: N = 7, K = 2
1
/ \
2 3
/ \ / \
4 2 2 5
Output: 3
Explanation: 2 occurs 3 times in the tree.

Input: N = 6, K = 5
1
/ \
4 5
/ \ / \
5 6 2 4
Output: 2
Explanation: 5 occurs 2 times in the tree.

Approach: The solution to the problem is based on the traversal of the given binary tree. Follow the steps as shown below:

Perform inorder traversal of the given Binary Tree
For each node in the tree, check whether it is equal to K or not
If it is equal to K, increment the required count by 1.
At the end, return the final count.

Below is the implementation of the above approach.

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
 
// Structure of a tree node
struct Node {
    int data;
    struct Node* left;
    struct Node* right;
    Node(int data)
    {
        this->data = data;
        left = right = NULL;
    }
};
 
// Function for inorder tree traversal
int countOccurrence(struct Node* root, int K)
{
    stack<Node*> s;
    Node* curr = root;
 
    // Variable for counting frequency of K
    int count = 0;
 
    while (curr != NULL || s.empty() == false) {
 
        // Reach the left most Node
        // of the curr Node
        while (curr != NULL) {
 
            // Place pointer to a tree node
            // on the stack before
            // traversing the node's
            // left subtree
            s.push(curr);
            curr = curr->left;
        }
 
        // Current must be NULL at this point
        curr = s.top();
        s.pop();
 
        // Increment count if element = K
        if (curr->data == K)
            count++;
 
        // Traverse the right subtree
        curr = curr->right;
    }
 
    return count;
}
 
// Driver code
int main()
{
 
    // Binary tree as shown in example
    struct Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(2);
    root->left->left = new Node(4);
    root->left->right = new Node(2);
 
    int K = 2;
 
    // Function call
    int ans = countOccurrence(root, K);
    cout << ans << endl;
    return 0;
}

Java

// JAVA code to implement the approach
import java.util.*;
 
// Structure of a tree node
class Node {
    int data;
    Node left;
    Node right;
    Node(int data)
    {
        this.data = data;
        left = right = null;
    }
}
class GFG {
    // Function for inorder tree traversal
    public static int countOccurrence(Node root, int K)
    {
        Stack<Node> s = new Stack<Node>();
        Node curr = root;
 
        // Variable for counting frequency of K
        int count = 0;
 
        while (curr != null || s.empty() == false) {
 
            // Reach the left most Node
            // of the curr Node
            while (curr != null) {
 
                // Place pointer to a tree node
                // on the stack before
                // traversing the node's
                // left subtree
                s.push(curr);
                curr = curr.left;
            }
 
            // Current must be NULL at this point
            curr = s.peek();
            s.pop();
 
            // Increment count if element = K
            if (curr.data == K)
                count++;
 
            // Traverse the right subtree
            curr = curr.right;
        }
 
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        // Binary tree as shown in example
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(2);
        root.left.left = new Node(4);
        root.left.right = new Node(2);
 
        int K = 2;
 
        // Function call
        int ans = countOccurrence(root, K);
        System.out.println(ans);
    }
}
 
// This code is contributed by Taranpreet

Python3

# Python code for the above approach
 
# Structure of a tree node
class Node:
    def __init__(self,d):
        self.data = d
        self.left = None
        self.right = None
 
# Function for inorder tree traversal
def countOccurrence(root, K):
    s = []
    curr = root
 
    # Variable for counting frequency of K
    count = 0
 
    while (curr != None or len(s) != 0):
 
        # Reach the left most Node
        # of the curr Node
        while (curr != None):
 
            # Place pointer to a tree node
            # on the stack before
            # traversing the node's
            # left subtree
            s.append(curr)
            curr = curr.left
 
        # Current must be None at this point
        curr = s[len(s) - 1]
        s.pop()
 
        # Increment count if element = K
        if (curr.data == K):
            count += 1
 
        # Traverse the right subtree
        curr = curr.right
 
    return count
 
# Driver code
 
# Binary tree as shown in example
root = Node(1)
root.left = Node(2)
root.right = Node(2)
root.left.left = Node(4)
root.left.right = Node(2)
 
K = 2
 
# Function call
ans = countOccurrence(root, K)
print(ans)
 
 # This code is contributed by shinjanpatra

C#

// C# code to implement the approach
using System;
using System.Collections.Generic;
 
// Structure of a tree node
public  class Node {
  public int data;
  public Node left;
  public Node right;
  public Node(int data)
  {
    this.data = data;
    left = right = null;
  }
}
class GFG {
  // Function for inorder tree traversal
  public static int countOccurrence(Node root, int K)
  {
    Stack<Node> s = new Stack<Node> ();
    Node curr = root;
 
    // Variable for counting frequency of K
    int count = 0;
 
    while (curr != null || s.Count!=0) {
 
      // Reach the left most Node
      // of the curr Node
      while (curr != null) {
 
        // Place pointer to a tree node
        // on the stack before
        // traversing the node's
        // left subtree
        s.Push(curr);
        curr = curr.left;
      }
 
      // Current must be NULL at this point
      curr = s.Peek();
      s.Pop();
 
      // Increment count if element = K
      if (curr.data == K)
        count++;
 
      // Traverse the right subtree
      curr = curr.right;
    }
 
    return count;
  }
 
  // Driver Code
  public static void Main () {
    // Build a tree
    // Binary tree as shown in example
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(2);
    root.left.left = new Node(4);
    root.left.right = new Node(2);
 
    int K = 2;
 
    // Function call
    int ans = countOccurrence(root, K);
    Console.WriteLine(ans);
  }
}
 
// This code is contributed by jana_sayantan.

Javascript

<script>
       // JavaScript code for the above approach
 
 
       // Structure of a tree node
       class Node {
           constructor(d) {
               this.data = d;
               this.left = null;
               this.right = null;
           }
       };
 
       // Function for inorder tree traversal
       function countOccurrence(root, K) {
           let s = [];
           let curr = root;
 
           // Variable for counting frequency of K
           let count = 0;
 
           while (curr != null || s.length != 0) {
 
               // Reach the left most Node
               // of the curr Node
               while (curr != null) {
 
                   // Place pointer to a tree node
                   // on the stack before
                   // traversing the node's
                   // left subtree
                   s.push(curr);
                   curr = curr.left;
               }
 
               // Current must be null at this point
               curr = s[s.length - 1];
               s.pop();
 
               // Increment count if element = K
               if (curr.data == K)
                   count++;
 
               // Traverse the right subtree
               curr = curr.right;
           }
 
           return count;
       }
 
       // Driver code
 
 
       // Binary tree as shown in example
       let root = new Node(1);
       root.left = new Node(2);
       root.right = new Node(2);
       root.left.left = new Node(4);
       root.left.right = new Node(2);
 
       let K = 2;
 
       // Function call
       let ans = countOccurrence(root, K);
       document.write(ans + '<br>')
 
    // This code is contributed by Potta Lokesh
   </script>

Output

Time Complexity: O(N)
Auxiliary Space: O(N)

Another Approach(using Recursion):
follow the below steps to solve the given problem recursively:
1) traverse the given binary tree in preorder fashion and keep track to count at each node
2) if the current node value is equal to given value(K) then increment k
3) recursively call for left and right subtree.
4) print count answer.

Below is the implementation of above approach:

C++

// c++ program to count frequency of k
// in given binary tree
#include<bits/stdc++.h>
using namespace std;
 
// Structure of a tree node
struct Node {
    int data;
    struct Node* left;
    struct Node* right;
    Node(int data)
    {
        this->data = data;
        left = right = NULL;
    }
};
 
// Function for preorder tree traversal recursively
void countOccurrence(Node* root, int K, int &count){
    if(root == NULL) return;
    if(root->data == K) count++;
    countOccurrence(root->left, K, count);
    countOccurrence(root->right, K, count);
}
 
// Driver code
int main()
{
    // Binary tree as shown in example
    struct Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(2);
    root->left->left = new Node(4);
    root->left->right = new Node(2);
  
    int K = 2;
    int ans = 0;
    // Function call
    countOccurrence(root, K, ans);
    cout << ans << endl;
    return 0;
}
 
// this code is contributed by Yash Agarwal(yashagarwal2852002)

Java

/*package whatever //do not write package name here */
import java.io.*;
 
// Java program to count frequency of k
// in given binary tree
 
// structure of a tree node
class Node {
  int data;
  Node left;
  Node right;
  Node(int data)
  {
    this.data = data;
    this.left = null;
    this.right = null;
  }
}
class GFG {
  static int count = 0;
  public static void countOccurrence(Node root, int k)
  {
    if (root == null)
      return;
    if (root.data == k)
      count++;
    countOccurrence(root.left, k);
    countOccurrence(root.right, k);
  }
 
  // function topreorder tree traversal recursively
  public static void main(String[] args)
  {
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(2);
    root.left.left = new Node(4);
    root.left.right = new Node(2);
    int k = 2;
    int ans = 0;
    countOccurrence(root, k);
    System.out.println(count);
  }
}
 
// This code is contributed by anskalyan3.

Python

# Python program to count frequency of k
# in given binary tree
# structure of tree node
class Node:
    def __init__(self,key):
        self.data = key
        self.left = None
        self.right = None
     
 
# function to preorder tree traversal recursively
count = 0
def countOccurrence(root, K):
    if(root is None):
        return
    if(root.data == K):
        global count
        count = count + 1
    countOccurrence(root.left, K)
    countOccurrence(root.right, K)
 
 
# driver code
# binary tree as shown in example
root = Node(1)
root.left = Node(2)
root.right = Node(2)
root.left.left = Node(4)
root.left.right = Node(2)
 
K = 2
# function call
countOccurrence(root, K)
print(count)

C#

// C# program to count frequency of k
// in given binary tree
using System;
using System.Collections.Generic;
 
class Gfg
{
 
  static int count = 0;   
 
  // Structure of a tree node
  class Node {
    public int data;
    public Node left;
    public Node right;
    public Node(int data)
    {
      this.data = data;
      left = right = null;
    }
  }
 
  // Function for preorder tree traversal recursively
  static void countOccurrence(Node root, int K)
  {
    if(root == null) 
      return;
    if(root.data == K) 
      count++;
    countOccurrence(root.left, K);
    countOccurrence(root.right, K);
  }
 
  // Driver code
  public static void Main(string[] args)
  {
    // Binary tree as shown in example
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(2);
    root.left.left = new Node(4);
    root.left.right = new Node(2);
 
    int K = 2;
 
    // Function call
    countOccurrence(root, K);
    Console.Write(count);
  }
}
 
// This code is contributed by ratiagrawal.

Javascript

// Javascript program to count frequency of k
// in given binary tree
 
// structure of a tree node
class Node{
    constructor(data){
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
// function topreorder tree traversal recursively
let count = 0;
function countOccurrence(root, K){
    if(root == null) return;
    if(root.data == K) count++;
    countOccurrence(root.left, K);
    countOccurrence(root.right, K);
}
 
// driver code
// binary tree as shown in example
let root = new Node(1);
root.left = new Node(2);
root.right = new Node(2);
root.left.left = new Node(4);
root.left.right = new Node(2);
 
let K = 2;
// function call
countOccurrence(root, K);
console.log(count);
// THIS CODE IS CONTRIBUTED BY KIRTI AGARWAL(KIRITAGARWAL23121999)

Output

Time Complexity: O(N) where N is the number of nodes in given binary tree.
Auxiliary Space: O(h) where h is the height of the given tree due to recursion.

Another Iterative and Easiest Approach(Using Level Order Traversal with Queue):

Follow the below steps to solve the given problem:

Perform level order traversal using Queue data structure.
At each node in traversal check if it is equal to the given integer k then increment the count variable which is initialized by 0 in starting the level order traversal.
Simply return the value of count variable.

Below is the implementation of above approach:

C++

// c++ program to count frequency of k
// in given binary tree
#include<bits/stdc++.h>
using namespace std;
 
// Structure of a tree node
struct Node {
    int data;
    struct Node* left;
    struct Node* right;
    Node(int data)
    {
        this->data = data;
        left = right = NULL;
    }
};
 
// Function for preorder tree traversal recursively
void countOccurrence(Node* root, int K, int &count){
    // initialize queue for level order traversal
    queue<Node*> q;
    q.push(root);
    while(!q.empty()){
        Node* front_node = q.front();
        q.pop();
        if(front_node->data == K) count++;
        if(front_node->left) q.push(front_node->left);
        if(front_node->right) q.push(front_node->right);
    }
}
 
// Driver code
int main()
{
    // Binary tree as shown in example
    struct Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(2);
    root->left->left = new Node(4);
    root->left->right = new Node(2);
  
    int K = 2;
    int ans = 0;
    // Function call
    countOccurrence(root, K, ans);
    cout << ans << endl;
    return 0;
}
 
// this code is contributed by Kirti Agarwal(kirtiagarwal23121999)

Java

import java.util.LinkedList;
import java.util.Queue;
 
// Structure of a tree node
class Node {
    int data;
    Node left;
    Node right;
 
    Node(int data) {
        this.data = data;
        left = right = null;
    }
}
 
public class Main {
    // Function for preorder tree traversal recursively
    static void countOccurrence(Node root, int K, int[] count) {
        // Initialize queue for level order traversal
        Queue<Node> q = new LinkedList<Node>();
        q.add(root);
 
        while (!q.isEmpty()) {
            Node front_node = q.poll();
            if (front_node.data == K) {
                count[0]++;
            }
            if (front_node.left != null) {
                q.add(front_node.left);
            }
            if (front_node.right != null) {
                q.add(front_node.right);
            }
        }
    }
 
    // Driver code
    public static void main(String[] args) {
        // Binary tree as shown in example
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(2);
        root.left.left = new Node(4);
        root.left.right = new Node(2);
 
        int K = 2;
        int[] ans = {0};
        // Function call
        countOccurrence(root, K, ans);
        System.out.println(ans[0]);
    }
}
// This code is contributed by divyansh2212

Python3

# Python3 program to count frequency of k
# in given binary tree
 
# Structure of a tree node
class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# Function for preorder tree traversal recursively
def countOccurrence(root, k):
    if root is None:
        return 0
 
    count = 0
    # initialize queue for level order traversal
    queue = []
    queue.append(root)
  
    while(len(queue) > 0):
        node = queue.pop(0)
        if (node.data == k):
            count += 1
 
        if node.left is not None:
            queue.append(node.left)
  
        if node.right is not None:
            queue.append(node.right)
              
    return count
  
# Driver code
if __name__ == '__main__':
      # Binary tree as shown in example
    root = Node(1)
    root.left = Node(2)
    root.right = Node(2)
    root.left.left = Node(4)
    root.left.right = Node(2)
 
    k = 2
     
    # Function Call
    print(countOccurrence(root, k))

C#

// C# program to count frequency of k
// in given binary tree
 
using System;
using System.Collections.Generic;
 
// Structure of a tree node
public class Node {
    public int data;
    public Node left, right;
    public Node(int item) {
        data = item;
        left = right = null;
    }
}
 
public class BinaryTree {
    Node root;
 
      // Function for preorder tree traversal recursively
    public void CountOccurrence(int k, ref int count) {
        if (root == null)
            return;
         
        //initialize queue for level order traversal
        Queue<Node> queue = new Queue<Node>();
        queue.Enqueue(root);
 
        while (queue.Count > 0) {
            Node frontNode = queue.Dequeue();
 
            if (frontNode.data == k)
                count++;
 
            if (frontNode.left != null)
                queue.Enqueue(frontNode.left);
 
            if (frontNode.right != null)
                queue.Enqueue(frontNode.right);
        }
    }
     
      // Driver code
    public static void Main(string[] args) {
        // Binary tree as shown in example
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(2);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(2);
 
        int k = 2;
        int count = 0;
           
          // Function Call
        tree.CountOccurrence(k, ref count);
        Console.WriteLine(count);
    }
}

Javascript

// Structure of a tree node
class Node {
    constructor(data) {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
// Function for preorder tree traversal recursively
function countOccurrence(root, K) {
    let count = 0;
    // initialize queue for level order traversal
    let q = [];
    q.push(root);
    while(q.length > 0){
        let front_node = q.shift();
        if(front_node.data == K) count++;
        if(front_node.left) q.push(front_node.left);
        if(front_node.right) q.push(front_node.right);
    }
    return count;
}
 
// Driver code
// Binary tree as shown in example
let root = new Node(1);
root.left = new Node(2);
root.right = new Node(2);
root.left.left = new Node(4);
root.left.right = new Node(2);
  
let K = 2;
let ans = countOccurrence(root, K);
console.log(ans);

Output

Time Complexity: O(N) where N is the number of nodes in given Binary tree because we simply traverse the each node only once.
Auxiliary space: O(N) due to queue data structure for storing the node level-wise.

Suggest improvement

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Count frequency of K in given Binary Tree

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