Find occurrence of most repeated nodes in a binary tree

Here given code implementation process.

import java.util.HashMap;
/*
  Java program
  Find occurrence of most repeated nodes in a binary tree
*/
class TreeNode
{
    public int data;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
public class BinarySearchTree
{
    public TreeNode root;
    public int result;
    public BinarySearchTree()
    {
        this.root = null;
        this.result = 0;
    }
    public void addNode(int data)
    {
        // Create a new node 
        TreeNode node = new TreeNode(data);
        if (this.root == null)
        {
            // When add a first node
            this.root = node;
        }
        else
        {
            // Get root of tree
            TreeNode find = this.root;

            // Find position to insert new node
            while (find != null)
            {
                if (find.data >= data)
                {
                    if (find.left == null)
                    {
                        find.left = node;
                        return;
                    }
                    else
                    {
                        // Visit left sub-tree
                        find = find.left;
                    }
                }
                else
                {
                    if (find.right == null)
                    {
                        find.right = node;
                        return;
                    }
                    else
                    {
                        // Visit right sub-tree
                        find = find.right;
                    }
                }
            }
        }
    }
    public void nodeCount(TreeNode node, HashMap < Integer, Integer > record)
    {
        if (node != null)
        {
            int count = 1;

            if (record.containsKey(node.data))
            {
                // Update frequency
                count = record.get(node.data) + 1;
            }
            if (this.result < count)
            {
                // When node occurs most time
                // Collect frequency
                this.result = count;
            }
            record.put(node.data, count);
            // Visit left and right sub-tree
            nodeCount(node.left, record);
            nodeCount(node.right, record);
        }
    }
    public void findMostRepeatedOccurrence()
    {
        if (this.root == null)
        {
            return;
        }
        this.result = 0;

        HashMap < Integer, Integer > record = new HashMap < Integer, Integer > ();
        
        this.nodeCount(this.root, record);
        
        // Display calculated result
        System.out.println(this.result);
    }
    public static void main(String[] args)
    {
        BinarySearchTree tree = new BinarySearchTree();

        // Input tree nodes
        int[] inputs = {
            6 , 2 , 1 , 0 , 3 , 2 , 7 , 3 , 3 , 4 , 0 , 1 , 7 , 3 , 9 , 8 , 1
        };
        // Get the number of elements
        int n = inputs.length;

        // Add tree elements
        for (int i = n - 1; i >= 0; --i)
        {
            tree.addNode(inputs[i]);
        }
        /*
            
            6 occurs = 1 time
            2 occurs = 2 time
            1 occurs = 3 time
            0 occurs = 2 time
            3 occurs = 4 time
            7 occurs = 2 time
            4 occurs = 1 time
            9 occurs = 1 time
            8 occurs = 1 time
            ------------------
            Most occurrence
            Result 4 [because 3 appears (4) times]
        */
        tree.findMostRepeatedOccurrence();
    }
}

Output

4
// Include header file
#include <iostream>
#include <unordered_map>

using namespace std;
/*
  C++ program
  Find occurrence of most repeated nodes in a binary tree
*/
class TreeNode
{
	public: int data;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int data)
	{
		this->data = data;
		this->left = NULL;
		this->right = NULL;
	}
};
class BinarySearchTree
{
	public: TreeNode *root;
	int result;
	BinarySearchTree()
	{
		this->root = NULL;
		this->result = 0;
	}
	void addNode(int data)
	{
		// Create a new node
		TreeNode *node = new TreeNode(data);
		if (this->root == NULL)
		{
			// When add a first node
			this->root = node;
		}
		else
		{
			// Get root of tree
			TreeNode *find = this->root;
			// Find position to insert new node
			while (find != NULL)
			{
				if (find->data >= data)
				{
					if (find->left == NULL)
					{
						find->left = node;
						return;
					}
					else
					{
						// Visit left sub-tree
						find = find->left;
					}
				}
				else
				{
					if (find->right == NULL)
					{
						find->right = node;
						return;
					}
					else
					{
						// Visit right sub-tree
						find = find->right;
					}
				}
			}
		}
	}
	void nodeCount(TreeNode *node, unordered_map < int, int > record)
	{
		if (node != NULL)
		{
			int count = 1;
			if (record.find(node->data) != record.end())
			{
				// Update frequency
				count = record[node->data] + 1;
			}
			if (this->result < count)
			{
				// When node occurs most time
				// Collect frequency
				this->result = count;
			}
			record[node->data] = count;
			// Visit left and right sub-tree
			this->nodeCount(node->left, record);
			this->nodeCount(node->right, record);
		}
	}
	void findMostRepeatedOccurrence()
	{
		if (this->root == NULL)
		{
			return;
		}
		this->result = 0;
		unordered_map < int, int > record;
		this->nodeCount(this->root, record);
		// Display calculated result
		cout << this->result << endl;
	}
};
int main()
{
	BinarySearchTree *tree = new BinarySearchTree();
	// Input tree nodes
	int inputs[] = {
		6 , 2 , 1 , 0 , 3 , 2 , 7 , 3 , 3 , 4 , 0 , 1 , 7 , 3 , 9 , 8 , 1
	};
	// Get the number of elements
	int n = sizeof(inputs) / sizeof(inputs[0]);
	// Add tree elements
	for (int i = n - 1; i >= 0; --i)
	{
		tree->addNode(inputs[i]);
	}
	/*
	    6 occurs = 1 time
	    2 occurs = 2 time
	    1 occurs = 3 time
	    0 occurs = 2 time
	    3 occurs = 4 time
	    7 occurs = 2 time
	    4 occurs = 1 time
	    9 occurs = 1 time
	    8 occurs = 1 time
	    ------------------
	    Most occurrence
	    Result 4 [because 3 appears (4) times]
	*/
	tree->findMostRepeatedOccurrence();
	return 0;
}

Output

4
package main
import "fmt"
/*
  Go program
  Find occurrence of most repeated nodes in a binary tree
*/
type TreeNode struct {
	data int
	left * TreeNode
	right * TreeNode
}
func getTreeNode(data int) * TreeNode {
	var me *TreeNode = &TreeNode {}
	me.data = data
	me.left = nil
	me.right = nil
	return me
}
type BinarySearchTree struct {
	root * TreeNode
	result int
}
func getBinarySearchTree() * BinarySearchTree {
	var me *BinarySearchTree = &BinarySearchTree {}
	me.root = nil
	me.result = 0
	return me
}
func(this *BinarySearchTree) addNode(data int) {
	// Create a new node
	var node * TreeNode = getTreeNode(data)
	if this.root == nil {
		// When add a first node
		this.root = node
	} else {
		// Get root of tree
		var find * TreeNode = this.root
		// Find position to insert new node
		for (find != nil) {
			if find.data >= data {
				if find.left == nil {
					find.left = node
					return
				} else {
					// Visit left sub-tree
					find = find.left
				}
			} else {
				if find.right == nil {
					find.right = node
					return
				} else {
					// Visit right sub-tree
					find = find.right
				}
			}
		}
	}
}
func(this *BinarySearchTree) nodeCount(node * TreeNode, record map[int]int) {
	if node != nil {
		var count int = 1
		if _, found := record[node.data] ; found {
			// Update frequency
			count = record[node.data] + 1
		}
		if this.result < count {
			// When node occurs most time
			// Collect frequency
			this.result = count
		}
		record[node.data] = count
		// Visit left and right sub-tree
		this.nodeCount(node.left, record)
		this.nodeCount(node.right, record)
	}
}
func(this BinarySearchTree) findMostRepeatedOccurrence() {
	if this.root == nil {
		return
	}
	this.result = 0
	var record = make(map[int] int)
	this.nodeCount(this.root, record)
	// Display calculated result
	fmt.Println(this.result)
}
func main() {
	var tree * BinarySearchTree = getBinarySearchTree()
	// Input tree nodes
	var inputs = [] int {
		6,
		2,
		1,
		0,
		3,
		2,
		7,
		3,
		3,
		4,
		0,
		1,
		7,
		3,
		9,
		8,
		1,
	}
	// Get the number of elements
	var n int = len(inputs)
	// Add tree elements
	for i := n - 1 ; i >= 0 ; i-- {
		tree.addNode(inputs[i])
	}
	/*
	    6 occurs = 1 time
	    2 occurs = 2 time
	    1 occurs = 3 time
	    0 occurs = 2 time
	    3 occurs = 4 time
	    7 occurs = 2 time
	    4 occurs = 1 time
	    9 occurs = 1 time
	    8 occurs = 1 time
	    ------------------
	    Most occurrence
	    Result 4 [because 3 appears (4) times]
	*/
	tree.findMostRepeatedOccurrence()
}

Output

4
// Include namespace system
using System;
using System.Collections.Generic;
/*
  Csharp program
  Find occurrence of most repeated nodes in a binary tree
*/
public class TreeNode
{
	public int data;
	public TreeNode left;
	public TreeNode right;
	public TreeNode(int data)
	{
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
public class BinarySearchTree
{
	public TreeNode root;
	public int result;
	public BinarySearchTree()
	{
		this.root = null;
		this.result = 0;
	}
	public void addNode(int data)
	{
		// Create a new node
		TreeNode node = new TreeNode(data);
		if (this.root == null)
		{
			// When add a first node
			this.root = node;
		}
		else
		{
			// Get root of tree
			TreeNode find = this.root;
			// Find position to insert new node
			while (find != null)
			{
				if (find.data >= data)
				{
					if (find.left == null)
					{
						find.left = node;
						return;
					}
					else
					{
						// Visit left sub-tree
						find = find.left;
					}
				}
				else
				{
					if (find.right == null)
					{
						find.right = node;
						return;
					}
					else
					{
						// Visit right sub-tree
						find = find.right;
					}
				}
			}
		}
	}
	public void nodeCount(TreeNode node, Dictionary < int, int > record)
	{
		if (node != null)
		{
			int count = 1;
			if (record.ContainsKey(node.data))
			{
				// Update frequency
				count = record[node.data] + 1;
				record[node.data] = count;
			}
			else
			{
				record.Add(node.data, count);
			}
			if (this.result < count)
			{
				// When node occurs most time
				// Collect frequency
				this.result = count;
			}
			// Visit left and right sub-tree
			this.nodeCount(node.left, record);
			this.nodeCount(node.right, record);
		}
	}
	public void findMostRepeatedOccurrence()
	{
		if (this.root == null)
		{
			return;
		}
		this.result = 0;
		Dictionary < int, int > record = new Dictionary < int, int > ();
		this.nodeCount(this.root, record);
		// Display calculated result
		Console.WriteLine(this.result);
	}
	public static void Main(String[] args)
	{
		BinarySearchTree tree = new BinarySearchTree();
		// Input tree nodes
		int[] inputs = {
			6 , 2 , 1 , 0 , 3 , 2 , 7 , 3 , 3 , 4 , 0 , 1 , 7 , 3 , 9 , 8 , 1
		};
		// Get the number of elements
		int n = inputs.Length;
		// Add tree elements
		for (int i = n - 1; i >= 0; --i)
		{
			tree.addNode(inputs[i]);
		}
		/*
		    6 occurs = 1 time
		    2 occurs = 2 time
		    1 occurs = 3 time
		    0 occurs = 2 time
		    3 occurs = 4 time
		    7 occurs = 2 time
		    4 occurs = 1 time
		    9 occurs = 1 time
		    8 occurs = 1 time
		    ------------------
		    Most occurrence
		    Result 4 [because 3 appears (4) times]
		*/
		tree.findMostRepeatedOccurrence();
	}
}

Output

4
<?php
/*
  Php program
  Find occurrence of most repeated nodes in a binary tree
*/
class TreeNode
{
	public $data;
	public $left;
	public $right;
	public	function __construct($data)
	{
		$this->data = $data;
		$this->left = NULL;
		$this->right = NULL;
	}
}
class BinarySearchTree
{
	public $root;
	public $result;
	public	function __construct()
	{
		$this->root = NULL;
		$this->result = 0;
	}
	public	function addNode($data)
	{
		// Create a new node
		$node = new TreeNode($data);
		if ($this->root == NULL)
		{
			// When add a first node
			$this->root = $node;
		}
		else
		{
			// Get root of tree
			$find = $this->root;
			// Find position to insert new node
			while ($find != NULL)
			{
				if ($find->data >= $data)
				{
					if ($find->left == NULL)
					{
						$find->left = $node;
						return;
					}
					else
					{
						// Visit left sub-tree
						$find = $find->left;
					}
				}
				else
				{
					if ($find->right == NULL)
					{
						$find->right = $node;
						return;
					}
					else
					{
						// Visit right sub-tree
						$find = $find->right;
					}
				}
			}
		}
	}
	public	function nodeCount($node, $record)
	{
		if ($node != NULL)
		{
			$count = 1;
			if (array_key_exists($node->data, $record))
			{
				// Update frequency
				$count = $record[$node->data] + 1;
			}
			if ($this->result < $count)
			{
				// When node occurs most time
				// Collect frequency
				$this->result = $count;
			}
			$record[$node->data] = $count;
			// Visit left and right sub-tree
			$this->nodeCount($node->left, $record);
			$this->nodeCount($node->right, $record);
		}
	}
	public	function findMostRepeatedOccurrence()
	{
		if ($this->root == NULL)
		{
			return;
		}
		$this->result = 0;
		$record = array();
		$this->nodeCount($this->root, $record);
		// Display calculated result
		echo($this->result.
			"\n");
	}
}

function main()
{
	$tree = new BinarySearchTree();
	// Input tree nodes
	$inputs = array(6, 2, 1, 0, 3, 2, 7, 3, 3, 4, 0, 1, 7, 3, 9, 8, 1);
	// Get the number of elements
	$n = count($inputs);
	// Add tree elements
	for ($i = $n - 1; $i >= 0; --$i)
	{
		$tree->addNode($inputs[$i]);
	}
	/*
	    6 occurs = 1 time
	    2 occurs = 2 time
	    1 occurs = 3 time
	    0 occurs = 2 time
	    3 occurs = 4 time
	    7 occurs = 2 time
	    4 occurs = 1 time
	    9 occurs = 1 time
	    8 occurs = 1 time
	    ------------------
	    Most occurrence
	    Result 4 [because 3 appears (4) times]
	*/
	$tree->findMostRepeatedOccurrence();
}
main();

Output

4
/*
  Node JS program
  Find occurrence of most repeated nodes in a binary tree
*/
class TreeNode
{
	constructor(data)
	{
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
class BinarySearchTree
{
	constructor()
	{
		this.root = null;
		this.result = 0;
	}
	addNode(data)
	{
		// Create a new node
		var node = new TreeNode(data);
		if (this.root == null)
		{
			// When add a first node
			this.root = node;
		}
		else
		{
			// Get root of tree
			var find = this.root;
			// Find position to insert new node
			while (find != null)
			{
				if (find.data >= data)
				{
					if (find.left == null)
					{
						find.left = node;
						return;
					}
					else
					{
						// Visit left sub-tree
						find = find.left;
					}
				}
				else
				{
					if (find.right == null)
					{
						find.right = node;
						return;
					}
					else
					{
						// Visit right sub-tree
						find = find.right;
					}
				}
			}
		}
	}
	nodeCount(node, record)
	{
		if (node != null)
		{
			var count = 1;
			if (record.has(node.data))
			{
				// Update frequency
				count = record.get(node.data) + 1;
			}
			if (this.result < count)
			{
				// When node occurs most time
				// Collect frequency
				this.result = count;
			}
			record.set(node.data, count);
			// Visit left and right sub-tree
			this.nodeCount(node.left, record);
			this.nodeCount(node.right, record);
		}
	}
	findMostRepeatedOccurrence()
	{
		if (this.root == null)
		{
			return;
		}
		this.result = 0;
		var record = new Map();
		this.nodeCount(this.root, record);
		// Display calculated result
		console.log(this.result);
	}
}

function main()
{
	var tree = new BinarySearchTree();
	// Input tree nodes
	var inputs = [6, 2, 1, 0, 3, 2, 7, 3, 3, 4, 0, 1, 7, 3, 9, 8, 1];
	// Get the number of elements
	var n = inputs.length;
	// Add tree elements
	for (var i = n - 1; i >= 0; --i)
	{
		tree.addNode(inputs[i]);
	}
	/*
	    6 occurs = 1 time
	    2 occurs = 2 time
	    1 occurs = 3 time
	    0 occurs = 2 time
	    3 occurs = 4 time
	    7 occurs = 2 time
	    4 occurs = 1 time
	    9 occurs = 1 time
	    8 occurs = 1 time
	    ------------------
	    Most occurrence
	    Result 4 [because 3 appears (4) times]
	*/
	tree.findMostRepeatedOccurrence();
}
main();

Output

4
#  Python 3 program
#  Find occurrence of most repeated nodes in a binary tree
class TreeNode :
	def __init__(self, data) :
		self.data = data
		self.left = None
		self.right = None
	

class BinarySearchTree :
	def __init__(self) :
		self.root = None
		self.result = 0
	
	def addNode(self, data) :
		#  Create a new node
		node = TreeNode(data)
		if (self.root == None) :
			#  When add a first node
			self.root = node
		else :
			#  Get root of tree
			find = self.root
			#  Find position to insert new node
			while (find != None) :
				if (find.data >= data) :
					if (find.left == None) :
						find.left = node
						return
					else :
						#  Visit left sub-tree
						find = find.left
					
				else :
					if (find.right == None) :
						find.right = node
						return
					else :
						#  Visit right sub-tree
						find = find.right
					
				
			
		
	
	def nodeCount(self, node, record) :
		if (node != None) :
			count = 1
			if ((node.data in record.keys())) :
				#  Update frequency
				count = record.get(node.data) + 1
			
			if (self.result < count) :
				#  When node occurs most time
				#  Collect frequency
				self.result = count
			
			record[node.data] = count
			#  Visit left and right sub-tree
			self.nodeCount(node.left, record)
			self.nodeCount(node.right, record)
		
	
	def findMostRepeatedOccurrence(self) :
		if (self.root == None) :
			return
		
		self.result = 0
		record = dict()
		self.nodeCount(self.root, record)
		#  Display calculated result
		print(self.result)
	

def main() :
	tree = BinarySearchTree()
	#  Input tree nodes
	inputs = [6, 2, 1, 0, 3, 2, 7, 3, 3, 4, 0, 1, 7, 3, 9, 8, 1]
	#  Get the number of elements
	n = len(inputs)
	i = n - 1
	#  Add tree elements
	while (i >= 0) :
		tree.addNode(inputs[i])
		i -= 1
	
	#    6 occurs = 1 time
	#    2 occurs = 2 time
	#    1 occurs = 3 time
	#    0 occurs = 2 time
	#    3 occurs = 4 time
	#    7 occurs = 2 time
	#    4 occurs = 1 time
	#    9 occurs = 1 time
	#    8 occurs = 1 time
	#    ------------------
	#    Most occurrence
	#    Result 4 [because 3 appears (4) times]
	tree.findMostRepeatedOccurrence()

if __name__ == "__main__": main()

Output

4
#  Ruby program
#  Find occurrence of most repeated nodes in a binary tree
class TreeNode 
	# Define the accessor and reader of class TreeNode
	attr_reader :data, :left, :right
	attr_accessor :data, :left, :right
	def initialize(data) 
		self.data = data
		self.left = nil
		self.right = nil
	end

end

class BinarySearchTree 
	# Define the accessor and reader of class BinarySearchTree
	attr_reader :root, :result
	attr_accessor :root, :result
	def initialize() 
		self.root = nil
		self.result = 0
	end

	def addNode(data) 
		#  Create a new node
		node = TreeNode.new(data)
		if (self.root == nil) 
			#  When add a first node
			self.root = node
		else
 
			#  Get root of tree
			find = self.root
			#  Find position to insert new node
			while (find != nil) 
				if (find.data >= data) 
					if (find.left == nil) 
						find.left = node
						return
					else
 
						#  Visit left sub-tree
						find = find.left
					end

				else
 
					if (find.right == nil) 
						find.right = node
						return
					else
 
						#  Visit right sub-tree
						find = find.right
					end

				end

			end

		end

	end

	def nodeCount(node, record) 
		if (node != nil) 
			count = 1
			if (record.key?(node.data)) 
				#  Update frequency
				count = record[node.data] + 1
			end

			if (self.result < count) 
				#  When node occurs most time
				#  Collect frequency
				self.result = count
			end

			record[node.data] = count
			#  Visit left and right sub-tree
			self.nodeCount(node.left, record)
			self.nodeCount(node.right, record)
		end

	end

	def findMostRepeatedOccurrence() 
		if (self.root == nil) 
			return
		end

		self.result = 0
		record = Hash.new()
		self.nodeCount(self.root, record)
		#  Display calculated result
		print(self.result, "\n")
	end

end

def main() 
	tree = BinarySearchTree.new()
	#  Input tree nodes
	inputs = [6, 2, 1, 0, 3, 2, 7, 3, 3, 4, 0, 1, 7, 3, 9, 8, 1]
	#  Get the number of elements
	n = inputs.length
	i = n - 1
	#  Add tree elements
	while (i >= 0) 
		tree.addNode(inputs[i])
		i -= 1
	end

	#    6 occurs = 1 time
	#    2 occurs = 2 time
	#    1 occurs = 3 time
	#    0 occurs = 2 time
	#    3 occurs = 4 time
	#    7 occurs = 2 time
	#    4 occurs = 1 time
	#    9 occurs = 1 time
	#    8 occurs = 1 time
	#    ------------------
	#    Most occurrence
	#    Result 4 [because 3 appears (4) times]
	tree.findMostRepeatedOccurrence()
end

main()

Output

4
import scala.collection.mutable._;
/*
  Scala program
  Find occurrence of most repeated nodes in a binary tree
*/
class TreeNode(var data: Int,
	var left: TreeNode,
		var right: TreeNode)
{
	def this(data: Int)
	{
		this(data, null, null);
	}
}
class BinarySearchTree(var root: TreeNode,
	var result: Int)
{
	def this()
	{
		this(null, 0);
	}
	def addNode(data: Int): Unit = {
		// Create a new node
		var node: TreeNode = new TreeNode(data);
		if (this.root == null)
		{
			// When add a first node
			this.root = node;
		}
		else
		{
			// Get root of tree
			var find: TreeNode = this.root;
			// Find position to insert new node
			while (find != null)
			{
				if (find.data >= data)
				{
					if (find.left == null)
					{
						find.left = node;
						return;
					}
					else
					{
						// Visit left sub-tree
						find = find.left;
					}
				}
				else
				{
					if (find.right == null)
					{
						find.right = node;
						return;
					}
					else
					{
						// Visit right sub-tree
						find = find.right;
					}
				}
			}
		}
	}
	def nodeCount(node: TreeNode, record: HashMap[Int, Int]): Unit = {
		if (node != null)
		{
			var count: Int = 1;
			if (record.contains(node.data))
			{
				// Update frequency
				count = record.get(node.data).get + 1;
			}
			if (this.result < count)
			{
				// When node occurs most time
				// Collect frequency
				this.result = count;
			}
			record.addOne(node.data, count);
			// Visit left and right sub-tree
			nodeCount(node.left, record);
			nodeCount(node.right, record);
		}
	}
	def findMostRepeatedOccurrence(): Unit = {
		if (this.root == null)
		{
			return;
		}
		this.result = 0;
		var record: HashMap[Int, Int] = new HashMap[Int, Int]();
		this.nodeCount(this.root, record);
		// Display calculated result
		println(this.result);
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var tree: BinarySearchTree = new BinarySearchTree();
		// Input tree nodes
		var inputs: Array[Int] = Array(6, 2, 1, 0, 3, 2, 7, 3, 3, 4, 0, 1, 7, 3, 9, 8, 1);
		// Get the number of elements
		var n: Int = inputs.length;
		var i: Int = n - 1;
		// Add tree elements
		while (i >= 0)
		{
			tree.addNode(inputs(i));
			i -= 1;
		}
		/*
		    6 occurs = 1 time
		    2 occurs = 2 time
		    1 occurs = 3 time
		    0 occurs = 2 time
		    3 occurs = 4 time
		    7 occurs = 2 time
		    4 occurs = 1 time
		    9 occurs = 1 time
		    8 occurs = 1 time
		    ------------------
		    Most occurrence
		    Result 4 [because 3 appears (4) times]
		*/
		tree.findMostRepeatedOccurrence();
	}
}

Output

4
import Foundation;
/*
  Swift 4 program
  Find occurrence of most repeated nodes in a binary tree
*/
class TreeNode
{
	var data: Int;
	var left: TreeNode? ;
	var right: TreeNode? ;
	init(_ data: Int)
	{
		self.data = data;
		self.left = nil;
		self.right = nil;
	}
}
class BinarySearchTree
{
	var root: TreeNode? ;
	var result: Int;
	init()
	{
		self.root = nil;
		self.result = 0;
	}
	func addNode(_ data: Int)
	{
		// Create a new node
		let node: TreeNode = TreeNode(data);
		if (self.root == nil)
		{
			// When add a first node
			self.root = node;
		}
		else
		{
			// Get root of tree
			var find: TreeNode? = self.root;
			// Find position to insert new node
			while (find  != nil)
			{
				if (find!.data >= data)
				{
					if (find!.left == nil)
					{
						find!.left = node;
						return;
					}
					else
					{
						// Visit left sub-tree
						find = find!.left;
					}
				}
				else
				{
					if (find!.right == nil)
					{
						find!.right = node;
						return;
					}
					else
					{
						// Visit right sub-tree
						find = find!.right;
					}
				}
			}
		}
	}
	func nodeCount(_ node: TreeNode? , _ record : inout[Int : Int])
	{
		if (node  != nil)
		{
			var count: Int = 1;
			if (record.keys.contains(node!.data))
			{
				// Update frequency
				count = record[node!.data]! + 1;
			}
			if (self.result < count)
			{
				// When node occurs most time
				// Collect frequency
				self.result = count;
			}
			record[node!.data] = count;
			// Visit left and right sub-tree
			self.nodeCount(node!.left, &record);
			self.nodeCount(node!.right, &record);
		}
	}
	func findMostRepeatedOccurrence()
	{
		if (self.root == nil)
		{
			return;
		}
		self.result = 0;
		var record: [Int : Int] = [Int : Int]();
		self.nodeCount(self.root, &record);
		// Display calculated result
		print(self.result);
	}
}
func main()
{
	let tree: BinarySearchTree = BinarySearchTree();
	// Input tree nodes
	let inputs: [Int] = [6, 2, 1, 0, 3, 2, 7, 3, 3, 4, 0, 1, 7, 3, 9, 8, 1];
	// Get the number of elements
	let n: Int = inputs.count;
	var i: Int = n - 1;
	// Add tree elements
	while (i >= 0)
	{
		tree.addNode(inputs[i]);
		i -= 1;
	}
	/*
	    6 occurs = 1 time
	    2 occurs = 2 time
	    1 occurs = 3 time
	    0 occurs = 2 time
	    3 occurs = 4 time
	    7 occurs = 2 time
	    4 occurs = 1 time
	    9 occurs = 1 time
	    8 occurs = 1 time
	    ------------------
	    Most occurrence
	    Result 4 [because 3 appears (4) times]
	*/
	tree.findMostRepeatedOccurrence();
}
main();

Output

4
/*
  Kotlin program
  Find occurrence of most repeated nodes in a binary tree
*/
class TreeNode
{
	var data: Int;
	var left: TreeNode ? ;
	var right: TreeNode ? ;
	constructor(data: Int)
	{
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
class BinarySearchTree
{
	var root: TreeNode ? ;
	var result: Int;
	constructor()
	{
		this.root = null;
		this.result = 0;
	}
	fun addNode(data: Int): Unit
	{
		// Create a new node
		val node: TreeNode = TreeNode(data);
		if (this.root == null)
		{
			// When add a first node
			this.root = node;
		}
		else
		{
			// Get root of tree
			var find: TreeNode ? = this.root;
			// Find position to insert new node
			while (find != null)
			{
				if (find.data >= data)
				{
					if (find.left == null)
					{
						find.left = node;
						return;
					}
					else
					{
						// Visit left sub-tree
						find = find.left;
					}
				}
				else
				{
					if (find.right == null)
					{
						find.right = node;
						return;
					}
					else
					{
						// Visit right sub-tree
						find = find.right;
					}
				}
			}
		}
	}
	fun nodeCount(node: TreeNode ? , record : HashMap < Int, Int > ): Unit
	{
		if (node != null)
		{
			var count: Int = 1;
			if (record.containsKey(node.data))
			{
				// Update frequency
				count = record.getValue(node.data) + 1;
			}
			if (this.result < count)
			{
				// When node occurs most time
				// Collect frequency
				this.result = count;
			}
			record.put(node.data, count);
			// Visit left and right sub-tree
			this.nodeCount(node.left, record);
			this.nodeCount(node.right, record);
		}
	}
	fun findMostRepeatedOccurrence(): Unit
	{
		if (this.root == null)
		{
			return;
		}
		this.result = 0;
		val record: HashMap < Int, Int > = HashMap < Int, Int > ();
		this.nodeCount(this.root, record);
		// Display calculated result
		println(this.result);
	}
}
fun main(args: Array < String > ): Unit
{
	val tree: BinarySearchTree = BinarySearchTree();
	// Input tree nodes
	val inputs: Array < Int > = arrayOf(6, 2, 1, 0, 3, 2, 7, 3, 3, 4, 0, 1, 7, 3, 9, 8, 1);
	// Get the number of elements
	val n: Int = inputs.count();
	var i: Int = n - 1;
	// Add tree elements
	while (i >= 0)
	{
		tree.addNode(inputs[i]);
		i -= 1;
	}
	/*
	    6 occurs = 1 time
	    2 occurs = 2 time
	    1 occurs = 3 time
	    0 occurs = 2 time
	    3 occurs = 4 time
	    7 occurs = 2 time
	    4 occurs = 1 time
	    9 occurs = 1 time
	    8 occurs = 1 time
	    ------------------
	    Most occurrence
	    Result 4 [because 3 appears (4) times]
	*/
	tree.findMostRepeatedOccurrence();
}

Output

4


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