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Code Binary Tree

Find the sum of the node having child node x

Here given code implementation process.

/*
    C Program 
    Find the sum of the node having child node x
    Recursive solution
*/
#include <stdio.h>
#include <stdlib.h>

//Binary Tree node
struct Node
{
	int data;
	struct Node *left, *right;
};
//This is creating a binary tree node and return this
struct Node *get_node(int data)
{
	// Create dynamic node
	struct Node *new_node = (struct Node *) malloc(sizeof(struct Node));
	if (new_node != NULL)
	{
		//Set data and pointer values
		new_node->data = data;
		new_node->left = NULL;
		new_node->right = NULL;
	}
	else
	{
		//This is indicates, segmentation fault or memory overflow problem
		printf("Memory Overflow\n");
	}
	//return new node
	return new_node;
}
//Display pre order elements
void preorder(struct Node *node)
{
	if (node != NULL)
	{
		//Print node value
		printf("  %d", node->data);
		preorder(node->left);
		preorder(node->right);
	}
}
// Returns the sum of all parent that child  contains x values
int parent_sum_of_x(struct Node *node, int x)
{
	int sum = 0;
	if (node != NULL)
	{
		if (node->left != NULL && node->left->data == x)
		{
			sum = node->data;
		}
		else if (node->right != NULL && node->right->data == x)
		{
			sum = node->data;
		}
		// Find the x node parent in left and right subtree
		sum = sum + parent_sum_of_x(node->left, x) + parent_sum_of_x(node->right, x);
	}
	return sum;
}
int main()
{
	struct Node *root = NULL;
	/*
	constructor binary tree
	-----------------------

	      6                            
	   /    \    
	  7      3     
	 / \      \               
	1   10     1
	   /  \   / \
	  1    1 1   5

	-----------------------
	*/
	root = get_node(6);
	root->left = get_node(7);
	root->left->left = get_node(1);
	root->left->right = get_node(10);
	root->left->right->right = get_node(1);
	root->left->right->left = get_node(1);
	root->right = get_node(3);
	root->right->right = get_node(1);
	root->right->right->right = get_node(5);
	root->right->right->left = get_node(1);
	printf("\n Tree Nodes : ");
	preorder(root);
	int x = 1;
	//Display calculated result
	printf("\n Sum of parent node of [%d] is  : %d\n", x, parent_sum_of_x(root, x));
	return 0;
}

Output

 Tree Nodes :   6  7  1  10  1  1  3  1  1  5
 Sum of parent node of [1] is  : 21
/*
    Java Program 
    Find the sum of the node having child node x
    Recursive solution
*/

//Binary Tree node
class Node
{
    public int data;
    public Node left;
    public Node right;
    public Node(int data)
    {
        //set node value
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
public class BinaryTree
{
    public Node root;
    public BinaryTree()
    {
        //Set initial tree root to null
        this.root = null;
    }
    //Display pre order elements
    public void preorder(Node node)
    {
        if (node != null)
        {
            //Print node value
            System.out.print("  " + node.data);
            preorder(node.left);
            preorder(node.right);
        }
    }
    // Returns the sum of all parent that child  contains x values
    public int parent_sum_of_x(Node node, int x)
    {
        int sum = 0;
        if (node != null)
        {
            if (node.left != null && node.left.data == x)
            {
                sum = node.data;
            }
            else if (node.right != null && node.right.data == x)
            {
                sum = node.data;
            }
            // Find the x node parent in left and right subtree
            sum = sum + parent_sum_of_x(node.left, x) + parent_sum_of_x(node.right, x);
        }
        return sum;
    }
    public static void main(String[] args)
    {
        //Make object of binary tree
        BinaryTree tree = new BinaryTree();
        /*
            constructor binary tree
            -----------------------

                  6                            
               /    \    
              7      3     
             / \      \               
            1   10     1
               /  \   / \
              1    1 1   5

            -----------------------
            */
        tree.root = new Node(6);
        tree.root.left = new Node(7);
        tree.root.left.left = new Node(1);
        tree.root.left.right = new Node(10);
        tree.root.left.right.right = new Node(1);
        tree.root.left.right.left = new Node(1);
        tree.root.right = new Node(3);
        tree.root.right.right = new Node(1);
        tree.root.right.right.right = new Node(5);
        tree.root.right.right.left = new Node(1);
        System.out.print("\n Tree Nodes : ");
        tree.preorder(tree.root);
        int x = 1;
        //Display calculated result
        System.out.print("\n Sum of parent node of [" + x + "] is : " + tree.parent_sum_of_x(tree.root, x) + "\n");
    }
}

Output

 Tree Nodes :   6  7  1  10  1  1  3  1  1  5
 Sum of parent node of [1] is : 21
//Include header file
#include <iostream>
using namespace std;

/*
    C++ Program 
    Find the sum of the node having child node x
    Recursive solution
*/

//Binary Tree node
class Node
{
    public: int data;
    Node *left;
    Node *right;
    Node(int data)
    {
        //set node value
        this->data = data;
        this->left = NULL;
        this->right = NULL;
    }
};
class BinaryTree
{
    public: Node *root;
    BinaryTree()
    {
        //Set initial tree root to null
        this->root = NULL;
    }
    //Display pre order elements
    void preorder(Node *node)
    {
        if (node != NULL)
        {
            //Print node value
            cout << "  " << node->data;
            this->preorder(node->left);
            this->preorder(node->right);
        }
    }
    // Returns the sum of all parent that child  contains x values
    int parent_sum_of_x(Node *node, int x)
    {
        int sum = 0;
        if (node != NULL)
        {
            if (node->left != NULL && node->left->data == x)
            {
                sum = node->data;
            }
            else if (node->right != NULL && node->right->data == x)
            {
                sum = node->data;
            }
            // Find the x node parent in left and right subtree
            sum = sum + this->parent_sum_of_x(node->left, x) + this->parent_sum_of_x(node->right, x);
        }
        return sum;
    }
};
int main()
{
    //Make object of binary tree
    BinaryTree tree = BinaryTree();
    /*
    constructor binary tree
    -----------------------

          6                            
       /    \    
      7      3     
     / \      \               
    1   10     1
       /  \   / \
      1    1 1   5

     -----------------------
    */
    tree.root = new Node(6);
    tree.root->left = new Node(7);
    tree.root->left->left = new Node(1);
    tree.root->left->right = new Node(10);
    tree.root->left->right->right = new Node(1);
    tree.root->left->right->left = new Node(1);
    tree.root->right = new Node(3);
    tree.root->right->right = new Node(1);
    tree.root->right->right->right = new Node(5);
    tree.root->right->right->left = new Node(1);
    cout << "\n Tree Nodes : ";
    tree.preorder(tree.root);
    int x = 1;
    //Display calculated result
    cout << "\n Sum of parent node of [" << x << "] is : " << tree.parent_sum_of_x(tree.root, x) << "\n";
    return 0;
}

Output

 Tree Nodes :   6  7  1  10  1  1  3  1  1  5
 Sum of parent node of [1] is : 21
//Include namespace system
using System;
/*
	C# Program 
	Find the sum of the node having child node x
	Recursive solution
*/
//Binary Tree node
public class Node
{
	public int data;
	public Node left;
	public Node right;
	public Node(int data)
	{
		//set node value
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
public class BinaryTree
{
	public Node root;
	public BinaryTree()
	{
		//Set initial tree root to null
		this.root = null;
	}
	//Display pre order elements
	public void preorder(Node node)
	{
		if (node != null)
		{
			//Print node value
			Console.Write("  " + node.data);
			preorder(node.left);
			preorder(node.right);
		}
	}
	// Returns the sum of all parent that child  contains x values
	public int parent_sum_of_x(Node node, int x)
	{
		int sum = 0;
		if (node != null)
		{
			if (node.left != null && node.left.data == x)
			{
				sum = node.data;
			}
			else if (node.right != null && node.right.data == x)
			{
				sum = node.data;
			}
			// Find the x node parent in left and right subtree
			sum = sum + parent_sum_of_x(node.left, x) + parent_sum_of_x(node.right, x);
		}
		return sum;
	}
	public static void Main(String[] args)
	{
		//Make object of binary tree
		BinaryTree tree = new BinaryTree();
		/*
			    constructor binary tree
			    -----------------------
			          6                            
			       /    \    
			      7      3     
			     / \      \               
			    1   10     1
			       /  \   / \
			      1    1 1   5
			  -----------------------
		*/
		tree.root = new Node(6);
		tree.root.left = new Node(7);
		tree.root.left.left = new Node(1);
		tree.root.left.right = new Node(10);
		tree.root.left.right.right = new Node(1);
		tree.root.left.right.left = new Node(1);
		tree.root.right = new Node(3);
		tree.root.right.right = new Node(1);
		tree.root.right.right.right = new Node(5);
		tree.root.right.right.left = new Node(1);
		Console.Write("\n Tree Nodes : ");
		tree.preorder(tree.root);
		int x = 1;
		//Display calculated result
		Console.Write("\n Sum of parent node of [" + x + "] is : " + tree.parent_sum_of_x(tree.root, x) + "\n");
	}
}

Output

 Tree Nodes :   6  7  1  10  1  1  3  1  1  5
 Sum of parent node of [1] is : 21
<?php
/*
	Php Program 
	Find the sum of the node having child node x
	Recursive solution
*/
//Binary Tree node
class Node
{
	public $data;
	public $left;
	public $right;

	function __construct($data)
	{
		//set node value
		$this->data = $data;
		$this->left = null;
		$this->right = null;
	}
}
class BinaryTree
{
	public $root;

	function __construct()
	{
		//Set initial tree root to null
		$this->root = null;
	}
	//Display pre order elements
	public	function preorder($node)
	{
		if ($node != null)
		{
			//Print node value
			echo "  ". $node->data;
			$this->preorder($node->left);
			$this->preorder($node->right);
		}
	}
	// Returns the sum of all parent that child  contains x values
	public	function parent_sum_of_x($node, $x)
	{
		$sum = 0;
		if ($node != null)
		{
			if ($node->left != null && $node->left->data == $x)
			{
				$sum = $node->data;
			}
			else if ($node->right != null && $node->right->data == $x)
			{
				$sum = $node->data;
			}
			// Find the x node parent in left and right subtree
			$sum = $sum + $this->parent_sum_of_x($node->left, $x) + $this->parent_sum_of_x($node->right, $x);
		}
		return $sum;
	}
}

function main()
{
	//Make object of binary tree
	$tree = new BinaryTree();
	/*
		    constructor binary tree
		    -----------------------
		          6                            
		       /    \    
		      7      3     
		     / \      \               
		    1   10     1
		       /  \   / \
		      1    1 1   5
		  -----------------------
	*/
	$tree->root = new Node(6);
	$tree->root->left = new Node(7);
	$tree->root->left->left = new Node(1);
	$tree->root->left->right = new Node(10);
	$tree->root->left->right->right = new Node(1);
	$tree->root->left->right->left = new Node(1);
	$tree->root->right = new Node(3);
	$tree->root->right->right = new Node(1);
	$tree->root->right->right->right = new Node(5);
	$tree->root->right->right->left = new Node(1);
	echo "\n Tree Nodes : ";
	$tree->preorder($tree->root);
	$x = 1;
	//Display calculated result
	echo "\n Sum of parent node of [". $x ."] is : ". $tree->parent_sum_of_x($tree->root, $x) ."\n";
}
main();

Output

 Tree Nodes :   6  7  1  10  1  1  3  1  1  5
 Sum of parent node of [1] is : 21
/*
	Node Js Program 
	Find the sum of the node having child node x
	Recursive solution
*/
//Binary Tree node
class Node
{
	constructor(data)
	{
		//set node value
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
class BinaryTree
{
	constructor()
	{
		//Set initial tree root to null
		this.root = null;
	}
	//Display pre order elements
	preorder(node)
	{
		if (node != null)
		{
			//Print node value
			process.stdout.write("  " + node.data);
			this.preorder(node.left);
			this.preorder(node.right);
		}
	}
	// Returns the sum of all parent that child  contains x values
	parent_sum_of_x(node, x)
	{
		var sum = 0;
		if (node != null)
		{
			if (node.left != null && node.left.data == x)
			{
				sum = node.data;
			}
			else if (node.right != null && node.right.data == x)
			{
				sum = node.data;
			}
			// Find the x node parent in left and right subtree
			sum = sum + this.parent_sum_of_x(node.left, x) + this.parent_sum_of_x(node.right, x);
		}
		return sum;
	}
}

function main()
{
	//Make object of binary tree
	var tree = new BinaryTree();
	/*
		    constructor binary tree
		    -----------------------
		          6                            
		       /    \    
		      7      3     
		     / \      \               
		    1   10     1
		       /  \   / \
		      1    1 1   5
		  -----------------------
	*/
	tree.root = new Node(6);
	tree.root.left = new Node(7);
	tree.root.left.left = new Node(1);
	tree.root.left.right = new Node(10);
	tree.root.left.right.right = new Node(1);
	tree.root.left.right.left = new Node(1);
	tree.root.right = new Node(3);
	tree.root.right.right = new Node(1);
	tree.root.right.right.right = new Node(5);
	tree.root.right.right.left = new Node(1);
	process.stdout.write("\n Tree Nodes : ");
	tree.preorder(tree.root);
	var x = 1;
	//Display calculated result
	process.stdout.write("\n Sum of parent node of [" + x + "] is : " + tree.parent_sum_of_x(tree.root, x) + "\n");
}
main();

Output

 Tree Nodes :   6  7  1  10  1  1  3  1  1  5
 Sum of parent node of [1] is : 21
#   Python 3 Program 
#   Find the sum of the node having child node x
#   Recursive solution

# Binary Tree node
class Node :
	
	def __init__(self, data) :
		# set node value
		self.data = data
		self.left = None
		self.right = None
	

class BinaryTree :
	
	def __init__(self) :
		# Set initial tree root to null
		self.root = None
	
	# Display pre order elements
	def preorder(self, node) :
		if (node != None) :
			# Print node value
			print("  ", node.data, end = "")
			self.preorder(node.left)
			self.preorder(node.right)
		
	
	#  Returns the sum of all parent that child  contains x values
	def parent_sum_of_x(self, node, x) :
		sum = 0
		if (node != None) :
			if (node.left != None and node.left.data == x) :
				sum = node.data
			
			elif(node.right != None and node.right.data == x) :
				sum = node.data
			
			#  Find the x node parent in left and right subtree
			sum = sum + self.parent_sum_of_x(node.left, x) + self.parent_sum_of_x(node.right, x)
		
		return sum
	

def main() :
	# Make object of binary tree
	tree = BinaryTree()
	# 
	#             constructor binary tree
	#             -----------------------
	#                   6                            
	#                /    \    
	#               7      3     
	#              / \      \               
	#             1   10     1
	#                /  \   / \
	#               1    1 1   5
	#           -----------------------
	#         
	
	tree.root = Node(6)
	tree.root.left = Node(7)
	tree.root.left.left = Node(1)
	tree.root.left.right = Node(10)
	tree.root.left.right.right = Node(1)
	tree.root.left.right.left = Node(1)
	tree.root.right = Node(3)
	tree.root.right.right = Node(1)
	tree.root.right.right.right = Node(5)
	tree.root.right.right.left = Node(1)
	print("\n Tree Nodes : ", end = "")
	tree.preorder(tree.root)
	x = 1
	# Display calculated result
	print("\n Sum of parent node of [", x ,"] is : ", tree.parent_sum_of_x(tree.root, x) ,"\n", end = "")

if __name__ == "__main__": main()

Output

 Tree Nodes :    6   7   1   10   1   1   3   1   1   5
 Sum of parent node of [ 1 ] is :  21
#   Ruby Program 
#   Find the sum of the node having child node x
#   Recursive solution

# Binary Tree node
class Node  
	# Define the accessor and reader of class Node  
	attr_reader :data, :left, :right
	attr_accessor :data, :left, :right
 
	
	def initialize(data) 
		# set node value
		self.data = data
		self.left = nil
		self.right = nil
	end

end

class BinaryTree  
	# Define the accessor and reader of class BinaryTree  
	attr_reader :root
	attr_accessor :root
 
	
	def initialize() 
		# Set initial tree root to null
		self.root = nil
	end

	# Display pre order elements
	def preorder(node) 
		if (node != nil) 
			# Print node value
			print("  ", node.data)
			self.preorder(node.left)
			self.preorder(node.right)
		end

	end

	#  Returns the sum of all parent that child  contains x values
	def parent_sum_of_x(node, x) 
		sum = 0
		if (node != nil) 
			if (node.left != nil && node.left.data == x) 
				sum = node.data
			elsif(node.right != nil && node.right.data == x) 
				sum = node.data
			end

			#  Find the x node parent in left and right subtree
			sum = sum + self.parent_sum_of_x(node.left, x) + self.parent_sum_of_x(node.right, x)
		end

		return sum
	end

end

def main() 
	# Make object of binary tree
	tree = BinaryTree.new()
	# 
	#             constructor binary tree
	#             -----------------------
	#                   6                            
	#                /    \    
	#               7      3     
	#              / \      \               
	#             1   10     1
	#                /  \   / \
	#               1    1 1   5
	#           -----------------------
	#         
	
	tree.root = Node.new(6)
	tree.root.left = Node.new(7)
	tree.root.left.left = Node.new(1)
	tree.root.left.right = Node.new(10)
	tree.root.left.right.right = Node.new(1)
	tree.root.left.right.left = Node.new(1)
	tree.root.right = Node.new(3)
	tree.root.right.right = Node.new(1)
	tree.root.right.right.right = Node.new(5)
	tree.root.right.right.left = Node.new(1)
	print("\n Tree Nodes : ")
	tree.preorder(tree.root)
	x = 1
	# Display calculated result
	print("\n Sum of parent node of [", x ,"] is : ", tree.parent_sum_of_x(tree.root, x) ,"\n")
end

main()

Output

 Tree Nodes :   6  7  1  10  1  1  3  1  1  5
 Sum of parent node of [1] is : 21

/*
	Scala Program 
	Find the sum of the node having child node x
	Recursive solution
*/

//Binary Tree node
class Node(var data: Int , var left: Node , var right: Node)
{
	def this(data: Int)
	{
		this(data, null, null);
	}
}
class BinaryTree(var root: Node)
{
	def this()
	{
		this(null);
	}
	//Display pre order elements
	def preorder(node: Node): Unit = {
		if (node != null)
		{
			//Print node value
			print("  " + node.data);
			preorder(node.left);
			preorder(node.right);
		}
	}
	// Returns the sum of all parent that child  contains x values
	def parent_sum_of_x(node: Node, x: Int): Int = {
		var sum: Int = 0;
		if (node != null)
		{
			if (node.left != null && node.left.data == x)
			{
				sum = node.data;
			}
			else if (node.right != null && node.right.data == x)
			{
				sum = node.data;
			}
			// Find the x node parent in left and right subtree
			sum = sum + parent_sum_of_x(node.left, x) + parent_sum_of_x(node.right, x);
		}
		return sum;
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		//Make object of binary tree
		var tree: BinaryTree = new BinaryTree();
		/*
			    constructor binary tree
			    -----------------------
			          6                            
			       /    \    
			      7      3     
			     / \      \               
			    1   10     1
			       /  \   / \
			      1    1 1   5
			  -----------------------
		*/
		tree.root = new Node(6);
		tree.root.left = new Node(7);
		tree.root.left.left = new Node(1);
		tree.root.left.right = new Node(10);
		tree.root.left.right.right = new Node(1);
		tree.root.left.right.left = new Node(1);
		tree.root.right = new Node(3);
		tree.root.right.right = new Node(1);
		tree.root.right.right.right = new Node(5);
		tree.root.right.right.left = new Node(1);
		print("\n Tree Nodes : ");
		tree.preorder(tree.root);
		var x: Int = 1;
		//Display calculated result
		print("\n Sum of parent node of [" + x + "] is : " + tree.parent_sum_of_x(tree.root, x) + "\n");
	}
}

Output

 Tree Nodes :   6  7  1  10  1  1  3  1  1  5
 Sum of parent node of [1] is : 21
/*
	Swift 4 Program 
	Find the sum of the node having child node x
	Recursive solution
*/

// Binary Tree node
class Node
{
	var data: Int;
	var left: Node? ;
	var right: Node? ;
	init(_ data: Int)
	{
		//set node value
		self.data = data;
		self.left = nil;
		self.right = nil;
	}
}
class BinaryTree
{
	var root: Node? ;
	init()
	{
		// Set initial tree root to null
		self.root = nil;
	}
	// Display pre order elements
	func preorder(_ node: Node? )
	{
		if (node != nil)
		{
			//Print node value
			print("  ", node!.data, terminator: "");
			self.preorder(node!.left);
			self.preorder(node!.right);
		}
	}
	// Returns the sum of all parent that child  contains x values
	func parent_sum_of_x(_ node: Node? , _ x : Int)->Int
	{
		var sum: Int = 0;
		if (node != nil)
		{
			if (node!.left != nil && node!.left!.data == x)
			{
				sum = node!.data;
			}
			else if (node!.right != nil && node!.right!.data == x)
			{
				sum = node!.data;
			}
			// Find the x node parent in left and right subtree
			sum = sum + self.parent_sum_of_x(node!.left, x) + self.parent_sum_of_x(node!.right, x);
		}
		return sum;
	}
}
func main()
{
	//Make object of binary tree
	let tree: BinaryTree = BinaryTree();
	/*
	    constructor binary tree
	    -----------------------
	          6                            
	       /    \    
	      7      3     
	     / \      \               
	    1   10     1
	       /  \   / \
	      1    1 1   5
	  -----------------------
*/
	tree.root = Node(6);
	tree.root!.left = Node(7);
	tree.root!.left!.left = Node(1);
	tree.root!.left!.right = Node(10);
	tree.root!.left!.right!.right = Node(1);
	tree.root!.left!.right!.left = Node(1);
	tree.root!.right = Node(3);
	tree.root!.right!.right = Node(1);
	tree.root!.right!.right!.right = Node(5);
	tree.root!.right!.right!.left = Node(1);
	print("\n Tree Nodes : ", terminator: "");
	tree.preorder(tree.root);
	let x: Int = 1;
	//Display calculated result
	print("\n Sum of parent node of [", x ,"] is : ", tree.parent_sum_of_x(tree.root, x) ,"\n", terminator: "");
}
main();

Output

 Tree Nodes :    6   7   1   10   1   1   3   1   1   5
 Sum of parent node of [ 1 ] is :  21

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