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
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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
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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
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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
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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
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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
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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
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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
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4 9 4 6 9
4 7 4 5 15import 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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