Find the maximum GCD of the siblings of a Binary Tree
Last Updated :
12 Jul, 2025
Given a 2d-array arr[][] which represents the nodes of a Binary tree, the task is to find the maximum GCD of the siblings of this tree without actually constructing it.
Example:
Input: arr[][] = {{4, 5}, {4, 2}, {2, 3}, {2, 1}, {3, 6}, {3, 12}}
Output: 6
Explanation:
For the above tree, the maximum GCD for the siblings is formed for the nodes 6 and 12 for the children of node 3.
Input: arr[][] = {{5, 4}, {5, 8}, {4, 6}, {4, 9}, {8, 10}, {10, 20}, {10, 30}}
Output: 10
Approach: The idea is to form a vector and store the tree in the form of the vector. After storing the tree in the form of a vector, the following cases occur:
- If the vector size is 0 or 1, then print 0 as GCD could not be found.
- For all other cases, since we store the tree in the form of a pair, we consider the first values of two pairs and compare them. But first, you need to Sort the pair of edges.
For example, let's assume there are two pairs in the vector A and B. We check if:
A.first == B.first
- If both of them match, then both of them belongs to the same parent. Therefore, we compute the GCD of the second values in the pairs and finally print the maximum of all such GCD's.
Below is the implementation of the above approach:
C++
// C++ program to find the maximum
// GCD of the siblings of a binary tree
#include <bits/stdc++.h>
using namespace std;
// Function to find maximum GCD
int max_gcd(vector<pair<int, int> >& v)
{
// No child or Single child
if (v.size() == 1 || v.size() == 0)
return 0;
sort(v.begin(), v.end());
// To get the first pair
pair<int, int> a = v[0];
pair<int, int> b;
int ans = INT_MIN;
for (int i = 1; i < v.size(); i++) {
b = v[i];
// If both the pairs belongs to
// the same parent
if (b.first == a.first)
// Update ans with the max
// of current gcd and
// gcd of both children
ans
= max(ans,
__gcd(a.second,
b.second));
// Update previous
// for next iteration
a = b;
}
return ans;
}
// Driver function
int main()
{
vector<pair<int, int> > v;
v.push_back(make_pair(5, 4));
v.push_back(make_pair(5, 8));
v.push_back(make_pair(4, 6));
v.push_back(make_pair(4, 9));
v.push_back(make_pair(8, 10));
v.push_back(make_pair(10, 20));
v.push_back(make_pair(10, 30));
cout << max_gcd(v);
return 0;
}
// Java program to find the maximum
// GCD of the siblings of a binary tree
import java.util.*;
import java.lang.*;
class GFG{
// Function to find maximum GCD
static int max_gcd(ArrayList<int[]> v)
{
// No child or Single child
if (v.size() == 1 || v.size() == 0)
return 0;
Collections.sort(v, new Comparator<int[]>() {
public int compare(int[] a, int[] b) {
return a[0]-b[0];
}
});
// To get the first pair
int[] a = v.get(0);
int[] b = new int[2];
int ans = Integer.MIN_VALUE;
for(int i = 1; i < v.size(); i++)
{
b = v.get(i);
// If both the pairs belongs to
// the same parent
if (b[0] == a[0])
// Update ans with the max
// of current gcd and
// gcd of both children
ans = Math.max(ans,
gcd(a[1],
b[1]));
// Update previous
// for next iteration
a = b;
}
return ans;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Driver code
public static void main(String[] args)
{
ArrayList<int[]> v = new ArrayList<>();
v.add(new int[]{5, 4});
v.add(new int[]{5, 8});
v.add(new int[]{4, 6});
v.add(new int[]{4, 9});
v.add(new int[]{8, 10});
v.add(new int[]{10, 20});
v.add(new int[]{10, 30});
System.out.println(max_gcd(v));
}
}
// This code is contributed by offbeat
# Python3 program to find the maximum
# GCD of the siblings of a binary tree
from math import gcd
# Function to find maximum GCD
def max_gcd(v):
# No child or Single child
if (len(v) == 1 or len(v) == 0):
return 0
v.sort()
# To get the first pair
a = v[0]
ans = -10**9
for i in range(1, len(v)):
b = v[i]
# If both the pairs belongs to
# the same parent
if (b[0] == a[0]):
# Update ans with the max
# of current gcd and
# gcd of both children
ans = max(ans, gcd(a[1], b[1]))
# Update previous
# for next iteration
a = b
return ans
# Driver function
if __name__ == '__main__':
v=[]
v.append([5, 4])
v.append([5, 8])
v.append([4, 6])
v.append([4, 9])
v.append([8, 10])
v.append([10, 20])
v.append([10, 30])
print(max_gcd(v))
# This code is contributed by mohit kumar 29
// C# program to find the maximum
// GCD of the siblings of a binary tree
using System.Collections;
using System;
class GFG{
// Function to find maximum GCD
static int max_gcd(ArrayList v)
{
// No child or Single child
if (v.Count == 1 || v.Count == 0)
return 0;
v.Sort();
// To get the first pair
int[] a = (int [])v[0];
int[] b = new int[2];
int ans = -10000000;
for(int i = 1; i < v.Count; i++)
{
b = (int[])v[i];
// If both the pairs belongs to
// the same parent
if (b[0] == a[0])
// Update ans with the max
// of current gcd and
// gcd of both children
ans = Math.Max(ans, gcd(a[1], b[1]));
// Update previous
// for next iteration
a = b;
}
return ans;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Driver code
public static void Main(string[] args)
{
ArrayList v = new ArrayList();
v.Add(new int[]{5, 4});
v.Add(new int[]{5, 8});
v.Add(new int[]{4, 6});
v.Add(new int[]{4, 9});
v.Add(new int[]{8, 10});
v.Add(new int[]{10, 20});
v.Add(new int[]{10, 30});
Console.Write(max_gcd(v));
}
}
// This code is contributed by rutvik_56
// JavaScript program to find the maximum
// GCD of the siblings of a binary tree
// Function to find maximum GCD
function max_gcd(v)
{
// No child or Single child
if (v.length == 1 || v.length == 0)
return 0;
v.sort((a, b) => a - b);
// To get the first pair
let a = v[0];
let b;
let ans = Number.MIN_SAFE_INTEGER;
for (let i = 1; i < v.length; i++) {
b = v[i];
// If both the pairs belongs to
// the same parent
if (b[0] == a[0])
// Update ans with the max
// of current gcd and
// gcd of both children
ans
= Math.max(ans, gcd(a[1], b[1]));
// Update previous
// for next iteration
a = b;
}
return ans;
}
function gcd(a, b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Driver function
let v = new Array();
v.push([5, 4]);
v.push([5, 8]);
v.push([4, 6]);
v.push([4, 9]);
v.push([8, 10]);
v.push([10, 20]);
v.push([10, 30]);
console.log(max_gcd(v));
// This code is contributed by gfgking
Time Complexity: O(nlogn), as sort() takes O(nlogn) time.
Auxiliary Space: O(1)
Another Approach:
The approach is to traverse the binary tree in a postorder manner and for each node, compute the greatest common divisor (GCD) of the values of its left and right children. If both children exist and their GCD is greater than the current maximum GCD, update the maximum GCD. Then return the GCD of the current node value and the maximum of its children's GCDs.
Here is an implementation of above approach:
C++
// C++ Implementation
#include <bits/stdc++.h>
using namespace std;
// Structure of Tree Node
struct TreeNode {
int val;
TreeNode* left;
TreeNode* right;
TreeNode(int x)
: val(x)
, left(NULL)
, right(NULL)
{
}
};
// Find out gcd
int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
int max_gcd_helper(TreeNode* root, int& ans)
{
if (!root)
return 0;
int left_gcd = max_gcd_helper(root->left, ans);
int right_gcd = max_gcd_helper(root->right, ans);
if (left_gcd != 0 && right_gcd != 0) {
int siblings_gcd = gcd(left_gcd, right_gcd);
ans = max(ans, siblings_gcd);
}
return (root->left) ? gcd(root->val, left_gcd)
: root->val;
}
int max_gcd(TreeNode* root)
{
int ans = 0;
max_gcd_helper(root, ans);
return ans;
}
// Driver code
int main()
{
TreeNode* root = new TreeNode(10);
root->left = new TreeNode(5);
root->right = new TreeNode(15);
root->left->left = new TreeNode(4);
root->left->right = new TreeNode(8);
root->right->left = new TreeNode(10);
root->right->right = new TreeNode(20);
// Function call
cout << max_gcd(root); // Output: 10
return 0;
}
import java.io.*;
import java.util.Objects;
// Definition of TreeNode
class TreeNode {
int val;
TreeNode left;
TreeNode right;
public TreeNode(int x) {
val = x;
left = null;
right = null;
}
}
public class GFG {
// Find out gcd
private static int gcd(int a, int b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
// Helper function to find the
// maximum GCD in the tree
private static int maxGCDHelper(TreeNode root, int[] ans) {
if (root == null)
return 0;
int leftGCD = maxGCDHelper(root.left, ans);
int rightGCD = maxGCDHelper(root.right, ans);
if (leftGCD != 0 && rightGCD != 0) {
int siblingsGCD = gcd(leftGCD, rightGCD);
ans[0] = Math.max(ans[0], siblingsGCD);
}
return (root.left != null) ? gcd(root.val, leftGCD) : root.val;
}
// Function to find the maximum GCD in the tree
public static int maxGCD(TreeNode root) {
int[] ans = { 0 };
// Store the result in an array to
// pass by reference
maxGCDHelper(root, ans);
return ans[0];
}
// Driver code
public static void main(String[] args) {
TreeNode root = new TreeNode(10);
root.left = new TreeNode(5);
root.right = new TreeNode(15);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(8);
root.right.left = new TreeNode(10);
root.right.right = new TreeNode(20);
// Function call
System.out.println(maxGCD(root));
}
}
# Python3 Implementation
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
def max_gcd_helper(root, ans):
if not root:
return 0
left_gcd = max_gcd_helper(root.left, ans)
right_gcd = max_gcd_helper(root.right, ans)
if left_gcd != 0 and right_gcd != 0:
siblings_gcd = gcd(left_gcd, right_gcd)
ans[0] = max(ans[0], siblings_gcd)
return root.val if not root.left else gcd(root.val, left_gcd)
def max_gcd(root):
ans = [0]
max_gcd_helper(root, ans)
return ans[0]
# Driver code
if __name__ == "__main__":
root = TreeNode(10)
root.left = TreeNode(5)
root.right = TreeNode(15)
root.left.left = TreeNode(4)
root.left.right = TreeNode(8)
root.right.left = TreeNode(10)
root.right.right = TreeNode(20)
# Function call
print(max_gcd(root)) # Output: 10
// C# implementation
using System;
public class TreeNode
{
public int val;
public TreeNode left;
public TreeNode right;
public TreeNode(int x)
{
val = x;
left = null;
right = null;
}
}
public class Program
{
// Find out gcd
public static int GCD(int a, int b)
{
if (b == 0)
return a;
return GCD(b, a % b);
}
public static int max_gcd_helper(TreeNode root, ref int ans)
{
if (root == null)
return 0;
int leftGCD = max_gcd_helper(root.left, ref ans);
int rightGCD = max_gcd_helper(root.right, ref ans);
if (leftGCD != 0 && rightGCD != 0)
{
int siblingsGCD = GCD(leftGCD, rightGCD);
ans = Math.Max(ans, siblingsGCD);
}
return (root.left != null) ? GCD(root.val, leftGCD) : root.val;
}
public static int MaxGCD(TreeNode root)
{
int ans = 0;
max_gcd_helper(root, ref ans);
return ans;
}
// Driver code
public static void Main(string[] args)
{
TreeNode root = new TreeNode(10);
root.left = new TreeNode(5);
root.right = new TreeNode(15);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(8);
root.right.left = new TreeNode(10);
root.right.right = new TreeNode(20);
Console.WriteLine(MaxGCD(root)); // Output: 10
}
}
// Code is Added by Avinash Wani
// Structure of Tree Node
class TreeNode {
constructor(val) {
this.val = val;
this.left = null;
this.right = null;
}
}
// Find out gcd
function gcd(a, b) {
if (b === 0)
return a;
return gcd(b, a % b);
}
function max_gcd_helper(root, ans) {
if (!root)
return 0;
const left_gcd = max_gcd_helper(root.left, ans);
const right_gcd = max_gcd_helper(root.right, ans);
if (left_gcd !== 0 && right_gcd !== 0) {
const siblings_gcd = gcd(left_gcd, right_gcd);
ans[0] = Math.max(ans[0], siblings_gcd);
}
return (root.left) ? gcd(root.val, left_gcd) : root.val;
}
function max_gcd(root) {
const ans = [0];
max_gcd_helper(root, ans);
return ans[0];
}
// Driver code
function main() {
const root = new TreeNode(10);
root.left = new TreeNode(5);
root.right = new TreeNode(15);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(8);
root.right.left = new TreeNode(10);
root.right.right = new TreeNode(20);
// Function call
console.log(max_gcd(root)); // Output: 10
}
main();
Time Complexity: O(nlogn)
Auxiliary Space: O(1)
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