# Find maximum GCD value in binary tree path

Here given code implementation process.

``````import java.util.ArrayList;
/*
Java program
Find maximum GCD value from root to leaf in a binary tree
*/
// 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 result;
public BinaryTree()
{
// Set initial tree root to null
this.root = null;
this.result = 0;
}
// Finding gcd of two numbers using recursion
public int findGCD(int a, int b)
{
if (b == 0)
{
return a;
}
return findGCD(b, a % b);
}
// Find all path and detecting maximum gcd of path using recursion
public void maxGCD(TreeNode node, ArrayList < Integer > path)
{
if (node == null)
{
return;
}
if (node.left == null && node.right == null)
{
int i = 1;
// First node of path
int auxiliary = path.get(0);
while (i < path.size())
{
auxiliary = findGCD(auxiliary, path.get(i));
i++;
}
if (auxiliary > this.result)
{
// New largest gcd of current path
this.result = auxiliary;
}
}
else
{
maxGCD(node.left, path);
maxGCD(node.right, path);
}
// Remove last node in path
path.remove(path.size() - 1);
}
// Handles the request of finding maximum gcd in tree path
public void pathGCD()
{
if (this.root == null)
{
// Empty Tree
return;
}
else
{
// This is use to collect sort path
ArrayList < Integer > path = new ArrayList < Integer > ();
// Reset the resultant gcd
this.result = Integer.MIN_VALUE;
// Find max gcd in binary tree path
maxGCD(this.root, path);
// Display calculated result
System.out.println(" Result : "+this.result);
}
}
public static void main(String[] args)
{
// Create new binary tree
BinaryTree tree = new BinaryTree();
/*
4
/   \
20    2
/ \     \
3   5     12
/ \    / \
6   8  5   8
/   /    \
9   3     15

-----------------
Constructing binary tree
*/
tree.root = new TreeNode(4);
tree.root.left = new TreeNode(20);
tree.root.left.right = new TreeNode(5);
tree.root.left.right.left = new TreeNode(6);
tree.root.left.right.left.left = new TreeNode(9);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.left = new TreeNode(3);
tree.root.left.left = new TreeNode(3);
tree.root.right = new TreeNode(2);
tree.root.right.right = new TreeNode(12);
tree.root.right.right.right = new TreeNode(8);
tree.root.right.right.left = new TreeNode(5);
tree.root.right.right.left.right = new TreeNode(15);
// Test case
// [4 2 12 8] gcd is 2
tree.pathGCD();
}
}``````

#### input

`` Result : 2``
``````// Include header file
#include <iostream>
#include <limits.h>
#include <vector>
using namespace std;

/*
C++ program
Find maximum GCD value from root to leaf in a binary tree
*/

// 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 result;
BinaryTree()
{
this->root = NULL;
this->result = 0;
}
// Finding gcd of two numbers using recursion
int findGCD(int a, int b)
{
if (b == 0)
{
return a;
}
return this->findGCD(b, a % b);
}
// Find all path and detecting maximum gcd of path using recursion
void maxGCD(TreeNode *node, vector < int > path)
{
if (node == NULL)
{
return;
}
path.push_back(node->data);
if (node->left == NULL && node->right == NULL)
{
int i = 1;
// First node of path
int auxiliary = path.at(0);
while (i < path.size())
{
auxiliary = this->findGCD(auxiliary, path.at(i));
i++;
}
if (auxiliary > this->result)
{
// New largest gcd of current path
this->result = auxiliary;
}
}
else
{
this->maxGCD(node->left, path);
this->maxGCD(node->right, path);
}
// Remove last node in path
path.pop_back();
}
// Handles the request of finding maximum gcd in tree path
void pathGCD()
{
if (this->root == NULL)
{
// Empty Tree
return;
}
else
{
// This is use to collect sort path
vector < int > path ;
// Reset the resultant gcd
this->result = INT_MIN;
// Find max gcd in binary tree path
this->maxGCD(this->root, path);
// Display calculated result
cout << " Result : " << this->result << endl;
}
}
};
int main()
{
// Create new binary tree
BinaryTree *tree = new BinaryTree();
/*
4
/   \
20    2
/ \     \
3   5     12
/ \    / \
6   8  5   8
/   /    \
9   3     15

-----------------
Constructing binary tree
*/
tree->root = new TreeNode(4);
tree->root->left = new TreeNode(20);
tree->root->left->right = new TreeNode(5);
tree->root->left->right->left = new TreeNode(6);
tree->root->left->right->left->left = new TreeNode(9);
tree->root->left->right->right = new TreeNode(8);
tree->root->left->right->right->left = new TreeNode(3);
tree->root->left->left = new TreeNode(3);
tree->root->right = new TreeNode(2);
tree->root->right->right = new TreeNode(12);
tree->root->right->right->right = new TreeNode(8);
tree->root->right->right->left = new TreeNode(5);
tree->root->right->right->left->right = new TreeNode(15);
// Test case
// [4 2 12 8] gcd is 2
tree->pathGCD();
return 0;
}``````

#### input

`` Result : 2``
``````// Include namespace system
using System;
using System.Collections.Generic;
/*
Csharp program
Find maximum GCD value from root to leaf in a binary tree
*/
// 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 result;
public BinaryTree()
{
// Set initial tree root to null
this.root = null;
this.result = 0;
}
// Finding gcd of two numbers using recursion
public int findGCD(int a, int b)
{
if (b == 0)
{
return a;
}
return this.findGCD(b, a % b);
}
// Find all path and detecting maximum gcd of path using recursion
public void maxGCD(TreeNode node, List < int > path)
{
if (node == null)
{
return;
}
if (node.left == null && node.right == null)
{
int i = 1;
// First node of path
int auxiliary = path[0];
while (i < path.Count)
{
auxiliary = this.findGCD(auxiliary, path[i]);
i++;
}
if (auxiliary > this.result)
{
// New largest gcd of current path
this.result = auxiliary;
}
}
else
{
this.maxGCD(node.left, path);
this.maxGCD(node.right, path);
}
// Remove last node in path
path.RemoveAt(path.Count - 1);
}
// Handles the request of finding maximum gcd in tree path
public void pathGCD()
{
if (this.root == null)
{
// Empty Tree
return;
}
else
{
// This is use to collect sort path
List < int > path = new List < int > ();
// Reset the resultant gcd
this.result = int.MinValue;
// Find max gcd in binary tree path
this.maxGCD(this.root, path);
// Display calculated result
Console.WriteLine(" Result : " + this.result);
}
}
public static void Main(String[] args)
{
// Create new binary tree
BinaryTree tree = new BinaryTree();
/*
4
/   \
20    2
/ \     \
3   5     12
/ \    / \
6   8  5   8
/   /    \
9   3     15

-----------------
Constructing binary tree
*/
tree.root = new TreeNode(4);
tree.root.left = new TreeNode(20);
tree.root.left.right = new TreeNode(5);
tree.root.left.right.left = new TreeNode(6);
tree.root.left.right.left.left = new TreeNode(9);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.left = new TreeNode(3);
tree.root.left.left = new TreeNode(3);
tree.root.right = new TreeNode(2);
tree.root.right.right = new TreeNode(12);
tree.root.right.right.right = new TreeNode(8);
tree.root.right.right.left = new TreeNode(5);
tree.root.right.right.left.right = new TreeNode(15);
// Test case
// [4 2 12 8] gcd is 2
tree.pathGCD();
}
}``````

#### input

`` Result : 2``
``````<?php
/*
Php program
Find maximum GCD value from root to leaf in a binary tree
*/
// 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 \$result;
public	function __construct()
{
\$this->root = NULL;
\$this->result = 0;
}
// Finding gcd of two numbers using recursion
public	function findGCD(\$a, \$b)
{
if (\$b == 0)
{
return \$a;
}
return \$this->findGCD(\$b, \$a % \$b);
}
// Find all path and detecting maximum gcd of path using recursion
public	function maxGCD(\$node, \$path)
{
if (\$node == NULL)
{
return;
}
\$path[] = \$node->data;
if (\$node->left == NULL && \$node->right == NULL)
{
\$i = 1;
// First node of path
\$auxiliary = \$path[0];
while (\$i < count(\$path))
{
\$auxiliary = \$this->findGCD(\$auxiliary, \$path[\$i]);
\$i++;
}
if (\$auxiliary > \$this->result)
{
// New largest gcd of current path
\$this->result = \$auxiliary;
}
}
else
{
\$this->maxGCD(\$node->left, \$path);
\$this->maxGCD(\$node->right, \$path);
}
// Remove last node in path
array_pop(\$path);
}
// Handles the request of finding maximum gcd in tree path
public	function pathGCD()
{
if (\$this->root == NULL)
{
// Empty Tree
return;
}
else
{
// This is use to collect sort path
\$path = array();
// Reset the resultant gcd
\$this->result = -PHP_INT_MAX;
// Find max gcd in binary tree path
\$this->maxGCD(\$this->root, \$path);
// Display calculated result
echo(" Result : ".\$this->result.
"\n");
}
}
}

function main()
{
// Create new binary tree
\$tree = new BinaryTree();
/*
4
/   \
20    2
/ \     \
3   5     12
/ \    / \
6   8  5   8
/   /    \
9   3     15

-----------------
Constructing binary tree
*/
\$tree->root = new TreeNode(4);
\$tree->root->left = new TreeNode(20);
\$tree->root->left->right = new TreeNode(5);
\$tree->root->left->right->left = new TreeNode(6);
\$tree->root->left->right->left->left = new TreeNode(9);
\$tree->root->left->right->right = new TreeNode(8);
\$tree->root->left->right->right->left = new TreeNode(3);
\$tree->root->left->left = new TreeNode(3);
\$tree->root->right = new TreeNode(2);
\$tree->root->right->right = new TreeNode(12);
\$tree->root->right->right->right = new TreeNode(8);
\$tree->root->right->right->left = new TreeNode(5);
\$tree->root->right->right->left->right = new TreeNode(15);
// Test case
// [4 2 12 8] gcd is 2
\$tree->pathGCD();
}
main();``````

#### input

`` Result : 2``
``````/*
Node JS program
Find maximum GCD value from root to leaf in a binary tree
*/
// 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.result = 0;
}
// Finding gcd of two numbers using recursion
findGCD(a, b)
{
if (b == 0)
{
return a;
}
return this.findGCD(b, a % b);
}
// Find all path and detecting maximum gcd of path using recursion
maxGCD(node, path)
{
if (node == null)
{
return;
}
path.push(node.data);
if (node.left == null && node.right == null)
{
var i = 1;
// First node of path
var auxiliary = path[0];
while (i < path.length)
{
auxiliary = this.findGCD(auxiliary, path[i]);
i++;
}
if (auxiliary > this.result)
{
// New largest gcd of current path
this.result = auxiliary;
}
}
else
{
this.maxGCD(node.left, path);
this.maxGCD(node.right, path);
}
// Remove last node in path
path.pop();
}
// Handles the request of finding maximum gcd in tree path
pathGCD()
{
if (this.root == null)
{
// Empty Tree
return;
}
else
{
// This is use to collect sort path
var path = [];
// Reset the resultant gcd
this.result = -Number.MAX_VALUE;
// Find max gcd in binary tree path
this.maxGCD(this.root, path);
// Display calculated result
console.log(" Result : " + this.result);
}
}
}

function main()
{
// Create new binary tree
var tree = new BinaryTree();
/*
4
/   \
20    2
/ \     \
3   5     12
/ \    / \
6   8  5   8
/   /    \
9   3     15

-----------------
Constructing binary tree
*/
tree.root = new TreeNode(4);
tree.root.left = new TreeNode(20);
tree.root.left.right = new TreeNode(5);
tree.root.left.right.left = new TreeNode(6);
tree.root.left.right.left.left = new TreeNode(9);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.left = new TreeNode(3);
tree.root.left.left = new TreeNode(3);
tree.root.right = new TreeNode(2);
tree.root.right.right = new TreeNode(12);
tree.root.right.right.right = new TreeNode(8);
tree.root.right.right.left = new TreeNode(5);
tree.root.right.right.left.right = new TreeNode(15);
// Test case
// [4 2 12 8] gcd is 2
tree.pathGCD();
}
main();``````

#### input

`` Result : 2``
``````import sys
#  Python 3 program
#  Find maximum GCD value from root to leaf in a binary tree

#  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.result = 0

#  Finding gcd of two numbers using recursion
def findGCD(self, a, b) :
if (b == 0) :
return a

return self.findGCD(b, a % b)

#  Find all path and detecting maximum gcd of path using recursion
def maxGCD(self, node, path) :
if (node == None) :
return

path.append(node.data)
if (node.left == None and node.right == None) :
i = 1
#  First node of path
auxiliary = path[0]
while (i < len(path)) :
auxiliary = self.findGCD(auxiliary, path[i])
i += 1

if (auxiliary > self.result) :
#  New largest gcd of current path
self.result = auxiliary

else :
self.maxGCD(node.left, path)
self.maxGCD(node.right, path)

#  Remove last node in path
del path[len(path) - 1]

#  Handles the request of finding maximum gcd in tree path
def pathGCD(self) :
if (self.root == None) :
#  Empty Tree
return
else :
#  This is use to collect sort path
path = []
#  Reset the resultant gcd
self.result = -sys.maxsize
#  Find max gcd in binary tree path
self.maxGCD(self.root, path)
#  Display calculated result
print(" Result : ", self.result)

def main() :
#  Create new binary tree
tree = BinaryTree()
#         4
#       /   \
#      20    2
#     / \     \
#    3   5     12
#       / \    / \
#      6   8  5   8
#     /   /    \
#    9   3     15
# -----------------
# Constructing binary tree
tree.root = TreeNode(4)
tree.root.left = TreeNode(20)
tree.root.left.right = TreeNode(5)
tree.root.left.right.left = TreeNode(6)
tree.root.left.right.left.left = TreeNode(9)
tree.root.left.right.right = TreeNode(8)
tree.root.left.right.right.left = TreeNode(3)
tree.root.left.left = TreeNode(3)
tree.root.right = TreeNode(2)
tree.root.right.right = TreeNode(12)
tree.root.right.right.right = TreeNode(8)
tree.root.right.right.left = TreeNode(5)
tree.root.right.right.left.right = TreeNode(15)
#  Test case
#  [4 2 12 8] gcd is 2
tree.pathGCD()

if __name__ == "__main__": main()``````

#### input

`` Result :  2``
``````#  Ruby program
#  Find maximum GCD value from root to leaf in a binary tree
#  Binary Tree node
class TreeNode
# Define the accessor and reader of class TreeNode
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_accessor :root, :result
def initialize()
self.root = nil
self.result = 0
end

#  Finding gcd of two numbers using recursion
def findGCD(a, b)
if (b == 0)
return a
end

return self.findGCD(b, a % b)
end

#  Find all path and detecting maximum gcd of path using recursion
def maxGCD(node, path)
if (node == nil)
return
end

path.push(node.data)
if (node.left == nil && node.right == nil)
i = 1
#  First node of path
auxiliary = path[0]
while (i < path.length)
auxiliary = self.findGCD(auxiliary, path[i])
i += 1
end

if (auxiliary > self.result)
#  New largest gcd of current path
self.result = auxiliary
end

else
self.maxGCD(node.left, path)
self.maxGCD(node.right, path)
end

#  Remove last node in path
path.delete_at(path.length - 1)
end

#  Handles the request of finding maximum gcd in tree path
def pathGCD()
if (self.root == nil)
#  Empty Tree
return
else

#  This is use to collect sort path
path = []
#  Reset the resultant gcd
self.result = -(2 ** (0. size * 8 - 2))
#  Find max gcd in binary tree path
self.maxGCD(self.root, path)
#  Display calculated result
print(" Result : ", self.result, "\n")
end

end

end

def main()
#  Create new binary tree
tree = BinaryTree.new()
#         4
#       /   \
#      20    2
#     / \     \
#    3   5     12
#       / \    / \
#      6   8  5   8
#     /   /    \
#    9   3     15
# -----------------
# Constructing binary tree
tree.root = TreeNode.new(4)
tree.root.left = TreeNode.new(20)
tree.root.left.right = TreeNode.new(5)
tree.root.left.right.left = TreeNode.new(6)
tree.root.left.right.left.left = TreeNode.new(9)
tree.root.left.right.right = TreeNode.new(8)
tree.root.left.right.right.left = TreeNode.new(3)
tree.root.left.left = TreeNode.new(3)
tree.root.right = TreeNode.new(2)
tree.root.right.right = TreeNode.new(12)
tree.root.right.right.right = TreeNode.new(8)
tree.root.right.right.left = TreeNode.new(5)
tree.root.right.right.left.right = TreeNode.new(15)
#  Test case
#  [4 2 12 8] gcd is 2
tree.pathGCD()
end

main()``````

#### input

`````` Result : 2
``````
``````import scala.collection.mutable._;
/*
Scala program
Find maximum GCD value from root to leaf in a binary tree
*/
// 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 result: Int)
{
def this()
{
this(null, 0);
}
// Finding gcd of two numbers using recursion
def findGCD(a: Int, b: Int): Int = {
if (b == 0)
{
return a;
}
return findGCD(b, a % b);
}
// Find all path and detecting maximum gcd of path using recursion
def maxGCD(node: TreeNode, path: ArrayBuffer[Int]): Unit = {
if (node == null)
{
return;
}
path += node.data;
if (node.left == null && node.right == null)
{
var i: Int = 1;
// First node of path
var auxiliary: Int = path(0);
while (i < path.size)
{
auxiliary = findGCD(auxiliary, path(i));
i += 1;
}
if (auxiliary > this.result)
{
// New largest gcd of current path
this.result = auxiliary;
}
}
else
{
maxGCD(node.left, path);
maxGCD(node.right, path);
}
// Remove last node in path
path.remove(path.size - 1);
}
// Handles the request of finding maximum gcd in tree path
def pathGCD(): Unit = {
if (this.root == null)
{
// Empty Tree
return;
}
else
{
// This is use to collect sort path
var path: ArrayBuffer[Int] = new ArrayBuffer[Int]();
// Reset the resultant gcd
this.result = Int.MinValue;
// Find max gcd in binary tree path
maxGCD(this.root, path);
// Display calculated result
println(" Result : " + this.result);
}
}
}
object Main
{
def main(args: Array[String]): Unit = {
// Create new binary tree
var tree: BinaryTree = new BinaryTree();
/*
4
/   \
20    2
/ \     \
3   5     12
/ \    / \
6   8  5   8
/   /    \
9   3     15
-----------------
Constructing binary tree
*/
tree.root = new TreeNode(4);
tree.root.left = new TreeNode(20);
tree.root.left.right = new TreeNode(5);
tree.root.left.right.left = new TreeNode(6);
tree.root.left.right.left.left = new TreeNode(9);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.left = new TreeNode(3);
tree.root.left.left = new TreeNode(3);
tree.root.right = new TreeNode(2);
tree.root.right.right = new TreeNode(12);
tree.root.right.right.right = new TreeNode(8);
tree.root.right.right.left = new TreeNode(5);
tree.root.right.right.left.right = new TreeNode(15);
// Test case
// [4 2 12 8] gcd is 2
tree.pathGCD();
}
}``````

#### input

`` Result : 2``
``````import Foundation;
/*
Swift 4 program
Find maximum GCD value from root to leaf in a binary tree
*/
// 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 result: Int;
init()
{
self.root = nil;
self.result = 0;
}
// Finding gcd of two numbers using recursion
func findGCD(_ a: Int, _ b: Int) -> Int
{
if (b == 0)
{
return a;
}
return self.findGCD(b, a % b);
}
// Find all path and detecting maximum gcd of path using recursion
func maxGCD(_ node: TreeNode? , _ path : inout[Int])
{
if (node == nil)
{
return;
}
path.append(node!.data);
if (node!.left == nil && node!.right == nil)
{
var i = 1;
// First node of path
var auxiliary = path[0];
while (i < path.count)
{
auxiliary = self.findGCD(auxiliary, path[i]);
i += 1;
}
if (auxiliary > self.result)
{
// New largest gcd of current path
self.result = auxiliary;
}
}
else
{
self.maxGCD(node!.left, &path);
self.maxGCD(node!.right, &path);
}
// Remove last node in path
path.removeLast();
}
// Handles the request of finding maximum gcd in tree path
func pathGCD()
{
if (self.root == nil)
{
// Empty Tree
return;
}
else
{
// This is use to collect sort path
var path = [Int]();
// Reset the resultant gcd
self.result = Int.min;
// Find max gcd in binary tree path
self.maxGCD(self.root, &path);
// Display calculated result
print(" Result : ", self.result);
}
}
}
func main()
{
// Create new binary tree
let tree = BinaryTree();
/*
4
/   \
20    2
/ \     \
3   5     12
/ \    / \
6   8  5   8
/   /    \
9   3     15
-----------------
Constructing binary tree
*/
tree.root = TreeNode(4);
tree.root!.left = TreeNode(20);
tree.root!.left!.right = TreeNode(5);
tree.root!.left!.right!.left = TreeNode(6);
tree.root!.left!.right!.left!.left = TreeNode(9);
tree.root!.left!.right!.right = TreeNode(8);
tree.root!.left!.right!.right!.left = TreeNode(3);
tree.root!.left!.left = TreeNode(3);
tree.root!.right = TreeNode(2);
tree.root!.right!.right = TreeNode(12);
tree.root!.right!.right!.right = TreeNode(8);
tree.root!.right!.right!.left = TreeNode(5);
tree.root!.right!.right!.left!.right = TreeNode(15);
// Test case
// [4 2 12 8] gcd is 2
tree.pathGCD();
}
main();``````

#### input

`` Result :  2``
``````/*
Kotlin program
Find maximum GCD value from root to leaf in a binary tree
*/
// 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 result: Int;
constructor()
{
this.root = null;
this.result = 0;
}
// Finding gcd of two numbers using recursion
fun findGCD(a: Int, b: Int): Int
{
if (b == 0)
{
return a;
}
return this.findGCD(b, a % b);
}
// Find all path and detecting maximum gcd of path using recursion
fun maxGCD(node: TreeNode ? , path : MutableList <Int>  ): Unit
{
if (node == null)
{
return;
}
if (node.left == null && node.right == null)
{
var i: Int = 1;
// First node of path
var auxiliary: Int = path[0];
while (i < path.size)
{
auxiliary = this.findGCD(auxiliary, path[i]);
i += 1;
}
if (auxiliary > this.result)
{
// New largest gcd of current path
this.result = auxiliary;
}
}
else
{
this.maxGCD(node.left, path);
this.maxGCD(node.right, path);
}
// Remove last node in path
path.removeAt(path.size - 1);
}
// Handles the request of finding maximum gcd in tree path
fun pathGCD(): Unit
{
if (this.root == null)
{
// Empty Tree
return;
}
else
{
// This is use to collect sort path
val path: MutableList < Int > = mutableListOf < Int > ();
// Reset the resultant gcd
this.result = Int.MIN_VALUE;
// Find max gcd in binary tree path
this.maxGCD(this.root, path);
// Display calculated result
println(" Result : " + this.result);
}
}
}
fun main(args: Array < String > ): Unit
{
// Create new binary tree
val tree: BinaryTree = BinaryTree();
/*
4
/   \
20    2
/ \     \
3   5     12
/ \    / \
6   8  5   8
/   /    \
9   3     15
-----------------
Constructing binary tree
*/
tree.root = TreeNode(4);
tree.root?.left = TreeNode(20);
tree.root?.left?.right = TreeNode(5);
tree.root?.left?.right?.left = TreeNode(6);
tree.root?.left?.right?.left?.left = TreeNode(9);
tree.root?.left?.right?.right = TreeNode(8);
tree.root?.left?.right?.right?.left = TreeNode(3);
tree.root?.left?.left = TreeNode(3);
tree.root?.right = TreeNode(2);
tree.root?.right?.right = TreeNode(12);
tree.root?.right?.right?.right = TreeNode(8);
tree.root?.right?.right?.left = TreeNode(5);
tree.root?.right?.right?.left?.right = TreeNode(15);
// Test case
// [4 2 12 8] gcd is 2
tree.pathGCD();
}``````

#### input

`` Result : 2``

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