Sum of all left leaves nodes in a binary tree

Here given code implementation process.

/*
    C Program 
    Sum of all left leaves nodes in a binary tree
    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 preorder 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 left leaves nodes in binary tree
int left_leaves_sum(struct Node *node)
{
	int sum = 0;
	if (node != NULL)
	{
		if (node->left != NULL && node->left->left == NULL && node->left->right == NULL)
		{
			// When get left leaf node
			sum = node->left->data;
		}
		// Find the left leaf nodes in left and right subtree
		sum = sum + left_leaves_sum(node->left) + left_leaves_sum(node->right);
	}
	return sum;
}
int main()
{
	struct Node *root = NULL;
	/*
	constructor binary tree
	-----------------
	     6                            
	   /   \    
	  2     3     
	 / \      \               
	8   10     1
	   /      /  \
	  6      4    5
	.................
	*/
	root = get_node(6);
	root->left = get_node(2);
	root->left->left = get_node(8);
	root->left->right = get_node(10);
	root->left->right->left = get_node(6);
	root->right = get_node(3);
	root->right->right = get_node(1);
	root->right->right->left = get_node(4);
	root->right->right->right = get_node(5);
	printf("\n Tree Nodes : ");
	preorder(root);
	printf("\n Left leaves nodes sum  : %d\n", left_leaves_sum(root));
	return 0;
}

Output

 Tree Nodes :   6  2  8  10  6  3  1  4  5
 Left leaves nodes sum  : 18
/*
    Java Program 
    Sum of all left leaves nodes in a binary tree
    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;
    }
}
class BinaryTree
{
    public Node root;
    public BinaryTree()
    {
        //Set initial tree root to null
        this.root = null;
    }
    //Display preorder 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 left leaves nodes in binary tree
    public int left_leaves_sum(Node node)
    {
        int sum = 0;
        if (node != null)
        {
            if (node.left != null && node.left.left == null && node.left.right == null)
            {
                // When get left leaf node
                sum = node.left.data;
            }
            // Find the left leaf nodes in left and right subtree
            sum = sum + left_leaves_sum(node.left) + left_leaves_sum(node.right);
        }
        return sum;
    }
    public static void main(String[] args)
    {
        //Make object of binary tree
        BinaryTree tree = new BinaryTree();
        /*
            constructor binary tree
            -----------------
                 6                            
               /   \    
              2     3     
             / \      \               
            8   10     1
               /      /  \
              6      4    5
            .................
        */
        tree.root = new Node(6);
        tree.root.left = new Node(2);
        tree.root.left.left = new Node(8);
        tree.root.left.right = new Node(10);
        tree.root.left.right.left = new Node(6);
        tree.root.right = new Node(3);
        tree.root.right.right = new Node(1);
        tree.root.right.right.left = new Node(4);
        tree.root.right.right.right = new Node(5);
        System.out.print("\n Tree Nodes : ");
        tree.preorder(tree.root);
        System.out.print("\n Left leaves nodes sum : " + tree.left_leaves_sum(tree.root) + "\n");
    }
}

Output

 Tree Nodes :   6  2  8  10  6  3  1  4  5
 Left leaves nodes sum : 18
//Include header file
#include <iostream>
using namespace std;

/*
    C++ Program 
    Sum of all left leaves nodes in a binary tree
    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 preorder 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 left leaves nodes in binary tree
	int left_leaves_sum(Node *node)
	{
		int sum = 0;
		if (node != NULL)
		{
			if (node->left != NULL && node->left->left == NULL && node->left->right == NULL)
			{
				// When get left leaf node
				sum = node->left->data;
			}
			// Find the left leaf nodes in left and right subtree
			sum = sum + this->left_leaves_sum(node->left) + this->left_leaves_sum(node->right);
		}
		return sum;
	}
};
int main()
{
	//Make object of binary tree
	BinaryTree tree = BinaryTree();
	/*
	            constructor binary tree
	            -----------------
	                 6                            
	               /   \    
	              2     3     
	             / \      \               
	            8   10     1
	               /      /  \
	              6      4    5
	            .................
	        */
	tree.root = new Node(6);
	tree.root->left = new Node(2);
	tree.root->left->left = new Node(8);
	tree.root->left->right = new Node(10);
	tree.root->left->right->left = new Node(6);
	tree.root->right = new Node(3);
	tree.root->right->right = new Node(1);
	tree.root->right->right->left = new Node(4);
	tree.root->right->right->right = new Node(5);
	cout << "\n Tree Nodes : ";
	tree.preorder(tree.root);
	cout << "\n Left leaves nodes sum : " << tree.left_leaves_sum(tree.root) << "\n";
	return 0;
}

Output

 Tree Nodes :   6  2  8  10  6  3  1  4  5
 Left leaves nodes sum : 18
//Include namespace system
using System;

/*
    C# Program 
    Sum of all left leaves nodes in a binary tree
    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;
	}
}
class BinaryTree
{
	public Node root;
	public BinaryTree()
	{
		//Set initial tree root to null
		this.root = null;
	}
	//Display preorder 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 left leaves nodes in binary tree
	public int left_leaves_sum(Node node)
	{
		int sum = 0;
		if (node != null)
		{
			if (node.left != null && node.left.left == null && node.left.right == null)
			{
				// When get left leaf node
				sum = node.left.data;
			}
			// Find the left leaf nodes in left and right subtree
			sum = sum + left_leaves_sum(node.left) + left_leaves_sum(node.right);
		}
		return sum;
	}
	public static void Main(String[] args)
	{
		//Make object of binary tree
		BinaryTree tree = new BinaryTree();
		/*
		            constructor binary tree
		            -----------------
		                 6                            
		               /   \    
		              2     3     
		             / \      \               
		            8   10     1
		               /      /  \
		              6      4    5
		            .................
		        */
		tree.root = new Node(6);
		tree.root.left = new Node(2);
		tree.root.left.left = new Node(8);
		tree.root.left.right = new Node(10);
		tree.root.left.right.left = new Node(6);
		tree.root.right = new Node(3);
		tree.root.right.right = new Node(1);
		tree.root.right.right.left = new Node(4);
		tree.root.right.right.right = new Node(5);
		Console.Write("\n Tree Nodes : ");
		tree.preorder(tree.root);
		Console.Write("\n Left leaves nodes sum : " + tree.left_leaves_sum(tree.root) + "\n");
	}
}

Output

 Tree Nodes :   6  2  8  10  6  3  1  4  5
 Left leaves nodes sum : 18
<?php
/*
    Php Program 
    Sum of all left leaves nodes in a binary tree
    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
		$this->root = null;
	}
	//Display preorder elements
	public	function preorder($node)
	{
		if ($node != null)
		{
			//Print node value
			echo "  ". $node->data;
			$this->preorder($node->left);
			$this->preorder($node->right);
		}
	}
	//Returns the sum of left leaves nodes in binary tree
	public	function left_leaves_sum($node)
	{
		$sum = 0;
		if ($node != null)
		{
			if ($node->left != null && $node->left->left == null && $node->left->right == null)
			{
				// When get left leaf node
				$sum = $node->left->data;
			}
			// Find the left leaf nodes in left and right subtree
			$sum = $sum + $this->left_leaves_sum($node->left) + $this->left_leaves_sum($node->right);
		}
		return $sum;
	}
}

function main()
{
	//Make object of binary tree
	$tree = new BinaryTree();
	/*
	            constructor binary tree
	            -----------------
	                 6                            
	               /   \    
	              2     3     
	             / \      \               
	            8   10     1
	               /      /  \
	              6      4    5
	            .................
	        */
	$tree->root = new Node(6);
	$tree->root->left = new Node(2);
	$tree->root->left->left = new Node(8);
	$tree->root->left->right = new Node(10);
	$tree->root->left->right->left = new Node(6);
	$tree->root->right = new Node(3);
	$tree->root->right->right = new Node(1);
	$tree->root->right->right->left = new Node(4);
	$tree->root->right->right->right = new Node(5);
	echo "\n Tree Nodes : ";
	$tree->preorder($tree->root);
	echo "\n Left leaves nodes sum : ". $tree->left_leaves_sum($tree->root) ."\n";
}
main();

Output

 Tree Nodes :   6  2  8  10  6  3  1  4  5
 Left leaves nodes sum : 18
/*
    Node Js Program 
    Sum of all left leaves nodes in a binary tree
    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 preorder 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 left leaves nodes in binary tree
	left_leaves_sum(node)
	{
		var sum = 0;
		if (node != null)
		{
			if (node.left != null && node.left.left == null && node.left.right == null)
			{
				// When get left leaf node
				sum = node.left.data;
			}
			// Find the left leaf nodes in left and right subtree
			sum = sum + this.left_leaves_sum(node.left) + this.left_leaves_sum(node.right);
		}
		return sum;
	}
}

function main()
{
	//Make object of binary tree
	var tree = new BinaryTree();
	/*
	            constructor binary tree
	            -----------------
	                 6                            
	               /   \    
	              2     3     
	             / \      \               
	            8   10     1
	               /      /  \
	              6      4    5
	            .................
	        */
	tree.root = new Node(6);
	tree.root.left = new Node(2);
	tree.root.left.left = new Node(8);
	tree.root.left.right = new Node(10);
	tree.root.left.right.left = new Node(6);
	tree.root.right = new Node(3);
	tree.root.right.right = new Node(1);
	tree.root.right.right.left = new Node(4);
	tree.root.right.right.right = new Node(5);
	process.stdout.write("\n Tree Nodes : ");
	tree.preorder(tree.root);
	process.stdout.write("\n Left leaves nodes sum : " + tree.left_leaves_sum(tree.root) + "\n");
}
main();

Output

 Tree Nodes :   6  2  8  10  6  3  1  4  5
 Left leaves nodes sum : 18
#     Python 3 Program 
#     Sum of all left leaves nodes in a binary tree
#     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 preorder 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 left leaves nodes in binary tree
	def left_leaves_sum(self, node) :
		sum = 0
		if (node != None) :
			if (node.left != None and node.left.left == None and node.left.right == None) :
				#  When get left leaf node
				sum = node.left.data
			
			#  Find the left leaf nodes in left and right subtree
			sum = sum + self.left_leaves_sum(node.left) + self.left_leaves_sum(node.right)
		
		return sum
	

def main() :
	# Make object of binary tree
	tree = BinaryTree()
	# 
	#             constructor binary tree
	#             -----------------
	#                  6                            
	#                /   \    
	#               2     3     
	#              / \      \               
	#             8   10     1
	#                /      /  \
	#               6      4    5
	#             .................
	#         
	
	tree.root = Node(6)
	tree.root.left = Node(2)
	tree.root.left.left = Node(8)
	tree.root.left.right = Node(10)
	tree.root.left.right.left = Node(6)
	tree.root.right = Node(3)
	tree.root.right.right = Node(1)
	tree.root.right.right.left = Node(4)
	tree.root.right.right.right = Node(5)
	print("\n Tree Nodes : ", end = "")
	tree.preorder(tree.root)
	print("\n Left leaves nodes sum : ", tree.left_leaves_sum(tree.root) ,"\n", end = "")

if __name__ == "__main__": main()

Output

 Tree Nodes :    6   2   8   10   6   3   1   4   5
 Left leaves nodes sum :  18
#     Ruby Program 
#     Sum of all left leaves nodes in a binary tree
#     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 preorder 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 left leaves nodes in binary tree
	def left_leaves_sum(node) 
		sum = 0
		if (node != nil) 
			if (node.left != nil && node.left.left == nil && node.left.right == nil) 
				#  When get left leaf node
				sum = node.left.data
			end

			#  Find the left leaf nodes in left and right subtree
			sum = sum + self.left_leaves_sum(node.left) + self.left_leaves_sum(node.right)
		end

		return sum
	end

end

def main() 
	# Make object of binary tree
	tree = BinaryTree.new()
	# 
	#             constructor binary tree
	#             -----------------
	#                  6                            
	#                /   \    
	#               2     3     
	#              / \      \               
	#             8   10     1
	#                /      /  \
	#               6      4    5
	#             .................
	#         
	
	tree.root = Node.new(6)
	tree.root.left = Node.new(2)
	tree.root.left.left = Node.new(8)
	tree.root.left.right = Node.new(10)
	tree.root.left.right.left = Node.new(6)
	tree.root.right = Node.new(3)
	tree.root.right.right = Node.new(1)
	tree.root.right.right.left = Node.new(4)
	tree.root.right.right.right = Node.new(5)
	print("\n Tree Nodes : ")
	tree.preorder(tree.root)
	print("\n Left leaves nodes sum : ", tree.left_leaves_sum(tree.root) ,"\n")
end

main()

Output

 Tree Nodes :   6  2  8  10  6  3  1  4  5
 Left leaves nodes sum : 18
/*
    Scala Program 
    Sum of all left leaves nodes in a binary tree
    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 preorder 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 left leaves nodes in binary tree
	def left_leaves_sum(node: Node): Int = {
		var sum: Int = 0;
		if (node != null)
		{
			if (node.left != null && node.left.left == null && node.left.right == null)
			{
				// When get left leaf node
				sum = node.left.data;
			}
			// Find the left leaf nodes in left and right subtree
			sum = sum + left_leaves_sum(node.left) + left_leaves_sum(node.right);
		}
		return sum;
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		//Make object of binary tree
		var tree: BinaryTree = new BinaryTree();
		/*
		            constructor binary tree
		            -----------------
		                 6                            
		               /   \    
		              2     3     
		             / \      \               
		            8   10     1
		               /      /  \
		              6      4    5
		            .................
		        */
		tree.root = new Node(6);
		tree.root.left = new Node(2);
		tree.root.left.left = new Node(8);
		tree.root.left.right = new Node(10);
		tree.root.left.right.left = new Node(6);
		tree.root.right = new Node(3);
		tree.root.right.right = new Node(1);
		tree.root.right.right.left = new Node(4);
		tree.root.right.right.right = new Node(5);
		print("\n Tree Nodes : ");
		tree.preorder(tree.root);
		print("\n Left leaves nodes sum : " + tree.left_leaves_sum(tree.root) + "\n");
	}
}

Output

 Tree Nodes :   6  2  8  10  6  3  1  4  5
 Left leaves nodes sum : 18
/*
    Swift 4 Program 
    Sum of all left leaves nodes in a binary tree
    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 preorder 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 left leaves nodes in binary tree
	func left_leaves_sum(_ node: Node? )->Int
	{
		var sum: Int = 0;
		if (node != nil)
		{
			if (node!.left != nil && node!.left!.left == nil && node!.left!.right == nil)
			{
				// When get left leaf node
				sum = node!.left!.data;
			}
			// Find the left leaf nodes in left and right subtree
			sum = sum + self.left_leaves_sum(node!.left) + self.left_leaves_sum(node!.right);
		}
		return sum;
	}
}
func main()
{
	//Make object of binary tree
	let tree: BinaryTree = BinaryTree();
	/*
	            constructor binary tree
	            -----------------
	                 6                            
	               /   \    
	              2     3     
	             / \      \               
	            8   10     1
	               /      /  \
	              6      4    5
	            .................
	        */
	tree.root = Node(6);
	tree.root!.left = Node(2);
	tree.root!.left!.left = Node(8);
	tree.root!.left!.right = Node(10);
	tree.root!.left!.right!.left = Node(6);
	tree.root!.right = Node(3);
	tree.root!.right!.right = Node(1);
	tree.root!.right!.right!.left = Node(4);
	tree.root!.right!.right!.right = Node(5);
	print("\n Tree Nodes : ", terminator: "");
	tree.preorder(tree.root);
	print("\n Left leaves nodes sum : ", tree.left_leaves_sum(tree.root) ,"\n", terminator: "");
}
main();

Output

 Tree Nodes :    6   2   8   10   6   3   1   4   5
 Left leaves nodes sum :  18


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