Print all the paths in the binary tree whose xor is non-zero

Here given code implementation process.

import java.util.ArrayList;
/*
    Java Program
    Print all the paths in the binary tree whose xor is non-zero
*/
// Binary Tree node
class TreeNode
{
    public int data;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int data)
    {
        // Set node value
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
public class BinaryTree
{
    public TreeNode root;
    public int count;
    public BinaryTree()
    {
        // Set initial tree root to null
        this.root = null;
        this.count = 0;
    }
    // Display given path 
    public void printPath(ArrayList < Integer > path)
    {
        int i = 0;
        // print path
        while (i < path.size())
        {
            // print path node
            System.out.print(" " + path.get(i));
            i++;
        }
        System.out.print("\n");
    }
    // Find all root to given node path using recursion
    public void findNonZeoXorPath(TreeNode point, 
                                    ArrayList < Integer > path, int xOr)
    {
        if (point == null)
        {
            return;
        }
        // Add path element
        path.add(point.data);
        if (point.left == null && point.right == null && (xOr ^ point.data) != 0)
        {
            this.count++;
            printPath(path);
        }
        // Visit left and right subtree using recursion
        findNonZeoXorPath(point.left, path, xOr ^ point.data);
        findNonZeoXorPath(point.right, path,  xOr ^ point.data);
        // Remove last node in path
        path.remove(path.size() - 1);
    }
    // Handles the request of finding non zero xor path in tree
    public void nonZeoXorPath()
    {
        if (this.root == null)
        {
            // Empty Tree
            return;
        }
        else
        {
            // This is use to collect path
            ArrayList < Integer > path = new ArrayList < Integer > ();
            this.count = 0;
          	// print non zero xor path
            findNonZeoXorPath(this.root, path, 0);

            if (this.count == 0)
            {
                // When have no resultant path
                System.out.print("None");
            }
        }
    }
    public static void main(String[] args)
    {
        // Create new binary tree
        BinaryTree tree = new BinaryTree();
        /*
             4                            
           /   \    
          9     7    
         / \      \               
        2   4      4
           / \    / \
          6   9  5   7
         /        \
        9          15 
        -----------------
        Constructing binary tree    
        */
        tree.root = new TreeNode(4);
        tree.root.left = new TreeNode(9);
        tree.root.left.right = new TreeNode(4);
        tree.root.left.right.left = new TreeNode(6);
        tree.root.left.right.left.left = new TreeNode(9);
        tree.root.left.right.right = new TreeNode(9);
        tree.root.left.left = new TreeNode(2);
        tree.root.right = new TreeNode(7);
        tree.root.right.right = new TreeNode(4);
        tree.root.right.right.right = new TreeNode(7);
        tree.root.right.right.left = new TreeNode(5);
        tree.root.right.right.left.right = new TreeNode(15);

 		// Check
        tree.nonZeoXorPath();
       
    }
}

input

 4 9 2
 4 9 4 6 9
 4 7 4 5 15
// Include header file
#include <iostream>
#include <vector>
using namespace std;

/*
    C++ Program
    Print all the paths in the binary tree whose xor is non-zero
*/
// Binary Tree node
class TreeNode
{
	public: int data;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int data)
	{
		// Set node value
		this->data = data;
		this->left = NULL;
		this->right = NULL;
	}
};
class BinaryTree
{
	public: TreeNode *root;
	int count;
	BinaryTree()
	{
		this->root = NULL;
		this->count = 0;
	}
	// Display given path
	void printPath(vector < int > path)
	{
		int i = 0;
		// print path
		while (i < path.size())
		{
			// print path node
			cout << " " << path.at(i);
			i++;
		}
		cout << "\n";
	}
	// Find all root to given node path using recursion
	void findNonZeoXorPath(TreeNode *point, vector < int > path, int xOr)
	{
		if (point == NULL)
		{
			return;
		}
		// Add path element
		path.push_back(point->data);
		if (point->left == NULL 
            && point->right == NULL 
            && (xOr ^ point->data) != 0)
		{
			this->count++;
			this->printPath(path);
		}
		// Visit left and right subtree using recursion
		this->findNonZeoXorPath(point->left, path, xOr ^ point->data);
		this->findNonZeoXorPath(point->right, path, xOr ^ point->data);
		// Remove last node in path
		path.pop_back();
	}
	// Handles the request of finding non zero xor path in tree
	void nonZeoXorPath()
	{
		if (this->root == NULL)
		{
			// Empty Tree
			return;
		}
		else
		{
			// This is use to collect path
			vector < int > path ;
			this->count = 0;
			// print non zero xor path
			this->findNonZeoXorPath(this->root, path, 0);
			if (this->count == 0)
			{
				// When have no resultant path
				cout << "None";
			}
		}
	}
};
int main()
{
	// Create new binary tree
	BinaryTree *tree = new BinaryTree();
	/*
	     4                            
	   /   \    
	  9     7    
	 / \      \               
	2   4      4
	   / \    / \
	  6   9  5   7
	 /        \
	9          15 
	-----------------
	Constructing binary tree    
	*/
	tree->root = new TreeNode(4);
	tree->root->left = new TreeNode(9);
	tree->root->left->right = new TreeNode(4);
	tree->root->left->right->left = new TreeNode(6);
	tree->root->left->right->left->left = new TreeNode(9);
	tree->root->left->right->right = new TreeNode(9);
	tree->root->left->left = new TreeNode(2);
	tree->root->right = new TreeNode(7);
	tree->root->right->right = new TreeNode(4);
	tree->root->right->right->right = new TreeNode(7);
	tree->root->right->right->left = new TreeNode(5);
	tree->root->right->right->left->right = new TreeNode(15);
	// Check
	tree->nonZeoXorPath();
	return 0;
}

input

 4 9 2
 4 9 4 6 9
 4 7 4 5 15
// Include namespace system
using System;
using System.Collections.Generic;
/*
    Csharp Program
    Print all the paths in the binary tree whose xor is non-zero
*/
// Binary Tree node
public class TreeNode
{
	public int data;
	public TreeNode left;
	public TreeNode right;
	public TreeNode(int data)
	{
		// Set node value
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
public class BinaryTree
{
	public TreeNode root;
	public int count;
	public BinaryTree()
	{
		// Set initial tree root to null
		this.root = null;
		this.count = 0;
	}
	// Display given path
	public void printPath(List < int > path)
	{
		int i = 0;
		// print path
		while (i < path.Count)
		{
			// print path node
			Console.Write(" " + path[i]);
			i++;
		}
		Console.Write("\n");
	}
	// Find all root to given node path using recursion
	public void findNonZeoXorPath(TreeNode point, List < int > path, int xOr)
	{
		if (point == null)
		{
			return;
		}
		// Add path element
		path.Add(point.data);
		if (point.left == null && point.right == null 
            && (xOr ^ point.data) != 0)
		{
			this.count++;
			this.printPath(path);
		}
		// Visit left and right subtree using recursion
		this.findNonZeoXorPath(point.left, path, xOr ^ point.data);
		this.findNonZeoXorPath(point.right, path, xOr ^ point.data);
		// Remove last node in path
		path.RemoveAt(path.Count - 1);
	}
	// Handles the request of finding non zero xor path in tree
	public void nonZeoXorPath()
	{
		if (this.root == null)
		{
			// Empty Tree
			return;
		}
		else
		{
			// This is use to collect path
			List < int > path = new List < int > ();
			this.count = 0;
			// print non zero xor path
			this.findNonZeoXorPath(this.root, path, 0);
			if (this.count == 0)
			{
				// When have no resultant path
				Console.Write("None");
			}
		}
	}
	public static void Main(String[] args)
	{
		// Create new binary tree
		BinaryTree tree = new BinaryTree();
		/*
		     4                            
		   /   \    
		  9     7    
		 / \      \               
		2   4      4
		   / \    / \
		  6   9  5   7
		 /        \
		9          15 
		-----------------
		Constructing binary tree    
		*/
		tree.root = new TreeNode(4);
		tree.root.left = new TreeNode(9);
		tree.root.left.right = new TreeNode(4);
		tree.root.left.right.left = new TreeNode(6);
		tree.root.left.right.left.left = new TreeNode(9);
		tree.root.left.right.right = new TreeNode(9);
		tree.root.left.left = new TreeNode(2);
		tree.root.right = new TreeNode(7);
		tree.root.right.right = new TreeNode(4);
		tree.root.right.right.right = new TreeNode(7);
		tree.root.right.right.left = new TreeNode(5);
		tree.root.right.right.left.right = new TreeNode(15);
		// Check
		tree.nonZeoXorPath();
	}
}

input

 4 9 2
 4 9 4 6 9
 4 7 4 5 15
<?php
/*
    Php Program
    Print all the paths in the binary tree whose xor is non-zero
*/
// Binary Tree node
class TreeNode
{
	public $data;
	public $left;
	public $right;
	public	function __construct($data)
	{
		// Set node value
		$this->data = $data;
		$this->left = NULL;
		$this->right = NULL;
	}
}
class BinaryTree
{
	public $root;
	public $count;
	public	function __construct()
	{
		$this->root = NULL;
		$this->count = 0;
	}
	// Display given path
	public	function printPath($path)
	{
		$i = 0;
		// print path
		while ($i < count($path))
		{
			// print path node
			echo(" ".$path[$i]);
			$i++;
		}
		echo("\n");
	}
	// Find all root to given node path using recursion
	public	function findNonZeoXorPath($point, $path, $xOr)
	{
		if ($point == NULL)
		{
			return;
		}
		// Add path element
		$path[] = $point->data;
		if ($point->left == NULL 
            && $point->right == NULL 
            && ($xOr ^ $point->data) != 0)
		{
			$this->count++;
			$this->printPath($path);
		}
		// Visit left and right subtree using recursion
		$this->findNonZeoXorPath($point->left, 
                                 $path, $xOr ^ $point->data);
		$this->findNonZeoXorPath($point->right, 
                                 $path, $xOr ^ $point->data);
		// Remove last node in path
		array_pop($path);
	}
	// Handles the request of finding non zero xor path in tree
	public	function nonZeoXorPath()
	{
		if ($this->root == NULL)
		{
			// Empty Tree
			return;
		}
		else
		{
			// This is use to collect path
			$path = array();
			$this->count = 0;
			// print non zero xor path
			$this->findNonZeoXorPath($this->root, $path, 0);
			if ($this->count == 0)
			{
				// When have no resultant path
				echo("None");
			}
		}
	}
}

function main()
{
	// Create new binary tree
	$tree = new BinaryTree();
	/*
	     4                            
	   /   \    
	  9     7    
	 / \      \               
	2   4      4
	   / \    / \
	  6   9  5   7
	 /        \
	9          15 
	-----------------
	Constructing binary tree    
	*/
	$tree->root = new TreeNode(4);
	$tree->root->left = new TreeNode(9);
	$tree->root->left->right = new TreeNode(4);
	$tree->root->left->right->left = new TreeNode(6);
	$tree->root->left->right->left->left = new TreeNode(9);
	$tree->root->left->right->right = new TreeNode(9);
	$tree->root->left->left = new TreeNode(2);
	$tree->root->right = new TreeNode(7);
	$tree->root->right->right = new TreeNode(4);
	$tree->root->right->right->right = new TreeNode(7);
	$tree->root->right->right->left = new TreeNode(5);
	$tree->root->right->right->left->right = new TreeNode(15);
	// Check
	$tree->nonZeoXorPath();
}
main();

input

 4 9 2
 4 9 4 6 9
 4 7 4 5 15
/*
    Node JS Program
    Print all the paths in the binary tree whose xor is non-zero
*/
// Binary Tree node
class TreeNode
{
	constructor(data)
	{
		// Set node value
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
class BinaryTree
{
	constructor()
	{
		this.root = null;
		this.count = 0;
	}
	// Display given path
	printPath(path)
	{
		var i = 0;
		// print path
		while (i < path.length)
		{
			// print path node
			process.stdout.write(" " + path[i]);
			i++;
		}
		process.stdout.write("\n");
	}
	// Find all root to given node path using recursion
	findNonZeoXorPath(point, path, xOr)
	{
		if (point == null)
		{
			return;
		}
		// Add path element
		path.push(point.data);
		if (point.left == null 
            && point.right == null 
            && (xOr ^ point.data) != 0)
		{
			this.count++;
			this.printPath(path);
		}
		// Visit left and right subtree using recursion
		this.findNonZeoXorPath(point.left, path, xOr ^ point.data);
		this.findNonZeoXorPath(point.right, path, xOr ^ point.data);
		// Remove last node in path
		path.pop();
	}
	// Handles the request of finding non zero xor path in tree
	nonZeoXorPath()
	{
		if (this.root == null)
		{
			// Empty Tree
			return;
		}
		else
		{
			// This is use to collect path
			var path = [];
			this.count = 0;
			// print non zero xor path
			this.findNonZeoXorPath(this.root, path, 0);
			if (this.count == 0)
			{
				// When have no resultant path
				process.stdout.write("None");
			}
		}
	}
}

function main()
{
	// Create new binary tree
	var tree = new BinaryTree();
	/*
	     4                            
	   /   \    
	  9     7    
	 / \      \               
	2   4      4
	   / \    / \
	  6   9  5   7
	 /        \
	9          15 
	-----------------
	Constructing binary tree    
	*/
	tree.root = new TreeNode(4);
	tree.root.left = new TreeNode(9);
	tree.root.left.right = new TreeNode(4);
	tree.root.left.right.left = new TreeNode(6);
	tree.root.left.right.left.left = new TreeNode(9);
	tree.root.left.right.right = new TreeNode(9);
	tree.root.left.left = new TreeNode(2);
	tree.root.right = new TreeNode(7);
	tree.root.right.right = new TreeNode(4);
	tree.root.right.right.right = new TreeNode(7);
	tree.root.right.right.left = new TreeNode(5);
	tree.root.right.right.left.right = new TreeNode(15);
	// Check
	tree.nonZeoXorPath();
}
main();

input

 4 9 2
 4 9 4 6 9
 4 7 4 5 15
#    Python 3 Program
#    Print all the paths in the binary tree whose xor is non-zero

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

class BinaryTree :
	def __init__(self) :
		self.root = None
		self.count = 0
	
	#  Display given path
	def printPath(self, path) :
		i = 0
		#  print path
		while (i < len(path)) :
			#  print path node
			print(" ", path[i], end = "")
			i += 1
		
		print(end = "\n")
	
	#  Find all root to given node path using recursion
	def findNonZeoXorPath(self, point, path, xOr) :
		if (point == None) :
			return
		
		#  Add path element
		path.append(point.data)
		if (point.left == None and point.right == None 
            and (xOr ^ point.data) != 0) :
			self.count += 1
			self.printPath(path)
		
		#  Visit left and right subtree using recursion
		self.findNonZeoXorPath(point.left, path, xOr ^ point.data)
		self.findNonZeoXorPath(point.right, path, xOr ^ point.data)
		#  Remove last node in path
		del path[len(path) - 1]
	
	#  Handles the request of finding non zero xor path in tree
	def nonZeoXorPath(self) :
		if (self.root == None) :
			#  Empty Tree
			return
		else :
			#  This is use to collect path
			path = []
			self.count = 0
			#  print non zero xor path
			self.findNonZeoXorPath(self.root, path, 0)
			if (self.count == 0) :
				#  When have no resultant path
				print("None", end = "")
			
		
	

def main() :
	#  Create new binary tree
	tree = BinaryTree()
	#     4                            
	#    /   \    
	#   9     7    
	#  / \      \               
	# 2   4      4
	#    / \    / \
	#   6   9  5   7
	#  /        \
	# 9          15 
	# -----------------
	# Constructing binary tree    
	tree.root = TreeNode(4)
	tree.root.left = TreeNode(9)
	tree.root.left.right = TreeNode(4)
	tree.root.left.right.left = TreeNode(6)
	tree.root.left.right.left.left = TreeNode(9)
	tree.root.left.right.right = TreeNode(9)
	tree.root.left.left = TreeNode(2)
	tree.root.right = TreeNode(7)
	tree.root.right.right = TreeNode(4)
	tree.root.right.right.right = TreeNode(7)
	tree.root.right.right.left = TreeNode(5)
	tree.root.right.right.left.right = TreeNode(15)
	#  Check
	tree.nonZeoXorPath()

if __name__ == "__main__": main()

input

  4  9  2
  4  9  4  6  9
  4  7  4  5  15
#    Ruby Program
#    Print all the paths in the binary tree whose xor is non-zero

#  Binary Tree node
class TreeNode 
	# Define the accessor and reader of class TreeNode
	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, :count
	attr_accessor :root, :count
	def initialize() 
		self.root = nil
		self.count = 0
	end

	#  Display given path
	def printPath(path) 
		i = 0
		#  print path
		while (i < path.length) 
			#  print path node
			print(" ", path[i])
			i += 1
		end

		print("\n")
	end

	#  Find all root to given node path using recursion
	def findNonZeoXorPath(point, path, xOr) 
		if (point == nil) 
			return
		end

		#  Add path element
		path.push(point.data)
		if (point.left == nil && 
            point.right == nil && 
            (xOr ^ point.data) != 0) 
			self.count += 1
			self.printPath(path)
		end

		#  Visit left and right subtree using recursion
		self.findNonZeoXorPath(point.left, path, xOr ^ point.data)
		self.findNonZeoXorPath(point.right, path, xOr ^ point.data)
		#  Remove last node in path
		path.delete_at(path.length - 1)
	end

	#  Handles the request of finding non zero xor path in tree
	def nonZeoXorPath() 
		if (self.root == nil) 
			#  Empty Tree
			return
		else
 
			#  This is use to collect path
			path = []
			self.count = 0
			#  print non zero xor path
			self.findNonZeoXorPath(self.root, path, 0)
			if (self.count == 0) 
				#  When have no resultant path
				print("None")
			end

		end

	end

end

def main() 
	#  Create new binary tree
	tree = BinaryTree.new()
	#     4                            
	#    /   \    
	#   9     7    
	#  / \      \               
	# 2   4      4
	#    / \    / \
	#   6   9  5   7
	#  /        \
	# 9          15 
	# -----------------
	# Constructing binary tree    
	tree.root = TreeNode.new(4)
	tree.root.left = TreeNode.new(9)
	tree.root.left.right = TreeNode.new(4)
	tree.root.left.right.left = TreeNode.new(6)
	tree.root.left.right.left.left = TreeNode.new(9)
	tree.root.left.right.right = TreeNode.new(9)
	tree.root.left.left = TreeNode.new(2)
	tree.root.right = TreeNode.new(7)
	tree.root.right.right = TreeNode.new(4)
	tree.root.right.right.right = TreeNode.new(7)
	tree.root.right.right.left = TreeNode.new(5)
	tree.root.right.right.left.right = TreeNode.new(15)
	#  Check
	tree.nonZeoXorPath()
end

main()

input

 4 9 2
 4 9 4 6 9
 4 7 4 5 15
import scala.collection.mutable._;
/*
    Scala Program
    Print all the paths in the binary tree whose xor is non-zero
*/
// Binary Tree node
class TreeNode(var data: Int,
	var left: TreeNode,
		var right: TreeNode)
{
	def this(data: Int)
	{
		// Set node value
		this(data,null,null);
	}
}
class BinaryTree(var root: TreeNode,
	var count: Int)
{
	def this()
	{
		this(null,0);
	}
	// Display given path
	def printPath(path: ArrayBuffer[Int]): Unit = {
		var i: Int = 0;
		// print path
		while (i < path.size)
		{
			// print path node
			print(" " + path(i));
			i += 1;
		}
		print("\n");
	}
	// Find all root to given node path using recursion
	def findNonZeoXorPath(point: TreeNode, 
                          path: ArrayBuffer[Int], xOr: Int): Unit = {
		if (point == null)
		{
			return;
		}
		// Add path element
		path += point.data;
		if (point.left == null && point.right == null 
      && (xOr ^ point.data) != 0)
		{
			this.count += 1;
			printPath(path);
		}
		// Visit left and right subtree using recursion
		findNonZeoXorPath(point.left, path, xOr ^ point.data);
		findNonZeoXorPath(point.right, path, xOr ^ point.data);
		// Remove last node in path
		path.remove(path.size - 1);
	}
	// Handles the request of finding non zero xor path in tree
	def nonZeoXorPath(): Unit = {
		if (this.root == null)
		{
			// Empty Tree
			return;
		}
		else
		{
			// This is use to collect path
			var path: ArrayBuffer[Int] = new ArrayBuffer[Int]();
			this.count = 0;
			// print non zero xor path
			findNonZeoXorPath(this.root, path, 0);
			if (this.count == 0)
			{
				// When have no resultant path
				print("None");
			}
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		// Create new binary tree
		var tree: BinaryTree = new BinaryTree();
		/*
		     4                            
		   /   \    
		  9     7    
		 / \      \               
		2   4      4
		   / \    / \
		  6   9  5   7
		 /        \
		9          15 
		-----------------
		Constructing binary tree    
		*/
		tree.root = new TreeNode(4);
		tree.root.left = new TreeNode(9);
		tree.root.left.right = new TreeNode(4);
		tree.root.left.right.left = new TreeNode(6);
		tree.root.left.right.left.left = new TreeNode(9);
		tree.root.left.right.right = new TreeNode(9);
		tree.root.left.left = new TreeNode(2);
		tree.root.right = new TreeNode(7);
		tree.root.right.right = new TreeNode(4);
		tree.root.right.right.right = new TreeNode(7);
		tree.root.right.right.left = new TreeNode(5);
		tree.root.right.right.left.right = new TreeNode(15);
		// Check
		tree.nonZeoXorPath();
	}
}

input

 4 9 2
 4 9 4 6 9
 4 7 4 5 15
import Foundation;
/*
    Swift 4 Program
    Print all the paths in the binary tree whose xor is non-zero
*/
// Binary Tree node
class TreeNode
{
	var data: Int;
	var left: TreeNode? ;
	var right: TreeNode? ;
	init(_ data: Int)
	{
		// Set node value
		self.data = data;
		self.left = nil;
		self.right = nil;
	}
}
class BinaryTree
{
	var root: TreeNode? ;
	var count: Int;
	init()
	{
		self.root = nil;
		self.count = 0;
	}
	// Display given path
	func printPath(_ path: [Int])
	{
		var i = 0;
		// print path
		while (i < path.count)
		{
			// print path node
			print(" ", path[i], terminator: "");
			i += 1;
		}
		print(terminator: "\n");
	}
	// Find all root to given node path using recursion
	func findNonZeoXorPath(_ point: TreeNode? , 
                           _ path : inout[Int], _ xOr: Int)
	{
		if (point == nil)
		{
			return;
		}
		// Add path element
		path.append(point!.data);
		if (point!.left == nil && point!.right == nil 
            && (xOr ^ point!.data)  != 0)
		{
			self.count += 1;
			self.printPath(path);
		}
		// Visit left and right subtree using recursion
		self.findNonZeoXorPath(point!.left, &path, xOr ^ point!.data);
		self.findNonZeoXorPath(point!.right, &path, xOr ^ point!.data);
		// Remove last node in path
		path.removeLast();
	}
	// Handles the request of finding non zero xor path in tree
	func nonZeoXorPath()
	{
		if (self.root == nil)
		{
			// Empty Tree
			return;
		}
		else
		{
			// This is use to collect path
			var path = [Int]();
			self.count = 0;
			// print non zero xor path
			self.findNonZeoXorPath(self.root, &path, 0);
			if (self.count == 0)
			{
				// When have no resultant path
				print("None", terminator: "");
			}
		}
	}
}
func main()
{
	// Create new binary tree
	let tree = BinaryTree();
	/*
	     4                            
	   /   \    
	  9     7    
	 / \      \               
	2   4      4
	   / \    / \
	  6   9  5   7
	 /        \
	9          15 
	-----------------
	Constructing binary tree    
	*/
	tree.root = TreeNode(4);
	tree.root!.left = TreeNode(9);
	tree.root!.left!.right = TreeNode(4);
	tree.root!.left!.right!.left = TreeNode(6);
	tree.root!.left!.right!.left!.left = TreeNode(9);
	tree.root!.left!.right!.right = TreeNode(9);
	tree.root!.left!.left = TreeNode(2);
	tree.root!.right = TreeNode(7);
	tree.root!.right!.right = TreeNode(4);
	tree.root!.right!.right!.right = TreeNode(7);
	tree.root!.right!.right!.left = TreeNode(5);
	tree.root!.right!.right!.left!.right = TreeNode(15);
	// Check
	tree.nonZeoXorPath();
}
main();

input

  4  9  2
  4  9  4  6  9
  4  7  4  5  15
/*
    Kotlin Program
    Print all the paths in the binary tree whose xor is non-zero
*/
// Binary Tree node
class TreeNode
{
	var data: Int;
	var left: TreeNode ? ;
	var right: TreeNode ? ;
	constructor(data: Int)
	{
		// Set node value
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
class BinaryTree
{
	var root: TreeNode ? ;
	var count: Int;
	constructor()
	{
		this.root = null;
		this.count = 0;
	}
	// Display given path
	fun printPath(path: MutableList < Int > ): Unit
	{
		var i: Int = 0;
		// print path
		while (i < path.size)
		{
			// print path node
			print(" " + path[i]);
			i += 1;
		}
		print("\n");
	}
	// Find all root to given node path using recursion
	fun findNonZeoXorPath(point: TreeNode ? , 
                          path : MutableList < Int >  , xOr : Int): Unit
	{
		if (point == null)
		{
			return;
		}
		// Add path element
		path.add(point.data);
		if (point.left == null 
            && point.right == null 
            && (xOr xor point.data) != 0)
		{
			this.count += 1;
			this.printPath(path);
		}
		// Visit left and right subtree using recursion
		this.findNonZeoXorPath(point.left, path, xOr xor point.data);
		this.findNonZeoXorPath(point.right, path, xOr xor point.data);
		// Remove last node in path
		path.removeAt(path.size - 1);
	}
	// Handles the request of finding non zero xor path in tree
	fun nonZeoXorPath(): Unit
	{
		if (this.root == null)
		{
			// Empty Tree
			return;
		}
		else
		{
			// This is use to collect path
			val path: MutableList < Int > = mutableListOf < Int > ();
			this.count = 0;
			// print non zero xor path
			this.findNonZeoXorPath(this.root, path, 0);
			if (this.count == 0)
			{
				// When have no resultant path
				print("None");
			}
		}
	}
}
fun main(args: Array < String > ): Unit
{
	// Create new binary tree
	val tree: BinaryTree = BinaryTree();
	/*
	     4                            
	   /   \    
	  9     7    
	 / \      \               
	2   4      4
	   / \    / \
	  6   9  5   7
	 /        \
	9          15 
	-----------------
	Constructing binary tree    
	*/
	tree.root = TreeNode(4);
	tree.root?.left = TreeNode(9);
	tree.root?.left?.right = TreeNode(4);
	tree.root?.left?.right?.left = TreeNode(6);
	tree.root?.left?.right?.left?.left = TreeNode(9);
	tree.root?.left?.right?.right = TreeNode(9);
	tree.root?.left?.left = TreeNode(2);
	tree.root?.right = TreeNode(7);
	tree.root?.right?.right = TreeNode(4);
	tree.root?.right?.right?.right = TreeNode(7);
	tree.root?.right?.right?.left = TreeNode(5);
	tree.root?.right?.right?.left?.right = TreeNode(15);
	// Check
	tree.nonZeoXorPath();
}

input

 4 9 2
 4 9 4 6 9
 4 7 4 5 15


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