Find maximum GCD value from root to leaf in a Binary tree

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Given a Binary Tree, the task is to find the maximum value of GCD from any path from the root node to the leaf node.

Examples:

Input: Below is the given tree:

Output: 3
Explanation:
Path 1: 15->3->5 = gcd(15, 3, 15) =3
Path 2: 15->3->1 =gcd(15, 3, 1) = 1
Path 3: 15->7->31=gcd(15, 7, 31)= 1
Path 4: 15->7->9 = gcd(15, 7, 9) =1, out of these 3 is the maximum.

Input: Below is the given tree:

Output: 1

Approach: The idea is to traverse all the paths from the root node to the leaf node and calculate the GCD of all the nodes that occurred in that path. Below are the steps:

Perform a preorder traversal on the given Binary Tree.
Iterate over all the paths and track all path values in an array.
Whenever encountered a leaf value then find the GCD of all the values in an array.
Update the GCD to the maximum value.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Initialise to update the maximum
// gcd value from all the path
int maxm = 0;
 
// Node structure
struct Node {
    int val;
 
    // Left & right child of the node
    Node *left, *right;
 
    // Initialize constructor
    Node(int x)
    {
        val = x;
        left = NULL;
        right = NULL;
    }
};
 
// Function to find gcd of a and b
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// function to find the gcd of a path
int find_gcd(vector<int> arr)
{
    if (arr.size() == 1)
        return arr[0];
 
    int g = arr[0];
 
    for (int i = 1; i < arr.size(); i++) {
        g = gcd(g, arr[i]);
    }
 
    return g;
}
 
// Function to find the maximum value
// of gcd from root to leaf
// in a Binary tree
void maxm_gcd(Node* root, vector<int> ans)
{
    // Check if root is not null
    if (!root)
        return;
 
    if (root->left == NULL
        and root->right == NULL) {
        ans.push_back(root->val);
 
        // Find the maximum gcd of
        // path value and store in
        // global maxm variable
        maxm = max(find_gcd(ans),
                   maxm);
 
        return;
    }
 
    // Traverse left of binary tree
    ans.push_back(root->val);
    maxm_gcd(root->left, ans);
 
    // Traverse right of the binary tree
    maxm_gcd(root->right, ans);
}
 
// Driver Code
int main()
{
    // Given Tree
    Node* root = new Node(15);
    root->left = new Node(3);
    root->right = new Node(7);
    root->left->left = new Node(15);
    root->left->right = new Node(1);
    root->right->left = new Node(31);
    root->right->right = new Node(9);
 
    // Function Call
    maxm_gcd(root, {});
 
    // Print the maximum AND value
    cout << maxm << endl;
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Initialise to update the maximum
// gcd value from all the path
static int maxm = 0;
 
// Node structure
static class Node 
{
    int val;
 
    // Left & right child of the node
    Node left, right;
 
    // Initialize constructor
    Node(int x)
    {
        val = x;
        left = null;
        right = null;
    }
};
 
// Function to find gcd of a and b
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function to find the gcd of a path
static int find_gcd(Vector<Integer> arr)
{
    if (arr.size() == 1)
        return arr.get(0);
 
    int g = arr.get(0);
 
    for(int i = 1; i < arr.size(); i++)
    {
        g = gcd(g, arr.get(1));
    }
    return g;
}
 
// Function to find the maximum value
// of gcd from root to leaf
// in a Binary tree
static void maxm_gcd(Node root, 
                     Vector<Integer> ans)
{
     
    // Check if root is not null
    if (root == null)
        return;
 
    if (root.left == null && 
        root.right == null) 
    {
        ans.add(root.val);
 
        // Find the maximum gcd of
        // path value and store in
        // global maxm variable
        maxm = Math.max(find_gcd(ans),
                        maxm);
                         
        return;
    }
 
    // Traverse left of binary tree
    ans.add(root.val);
    maxm_gcd(root.left, ans);
 
    // Traverse right of the binary tree
    maxm_gcd(root.right, ans);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Tree
    Node root = new Node(15);
    root.left = new Node(3);
    root.right = new Node(7);
    root.left.left = new Node(15);
    root.left.right = new Node(1);
    root.right.left = new Node(31);
    root.right.right = new Node(9);
 
    // Function call
    maxm_gcd(root, new Vector<>());
 
    // Print the maximum AND value
    System.out.print(maxm + "\n");
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of
# the above approach
 
# Initialise to update the maximum
# gcd value from all the path
global maxm
maxm = 0
 
# Node structure
class Node:
 
    # Initialize constructor
    def __init__(self, x):
 
        self.val = x
        self.left = None
        self.right = None
 
# Function to find gcd of a and b
def gcd(a, b):
 
    if(b == 0):
        return a
    return gcd(b, a % b)
 
# Function to find the gcd of a path
def find_gcd(arr):
 
    if(len(arr) == 1):
        return arr[0]
 
    g = arr[0]
 
    for i in range(1, len(arr)):
        g = gcd(g, arr[i])
 
    return g
 
# Function to find the maximum value
# of gcd from root to leaf
# in a Binary tree
def maxm_gcd(root, ans):
 
    global maxm
 
    # Check if root is not null
    if(not root):
        return
 
    if(root.left == None and
       root.right == None):
        ans.append(root.val)
 
        # Find the maximum gcd of
        # path value and store in
        # global maxm variable
        maxm = max(find_gcd(ans), maxm)
 
        return
 
    # Traverse left of binary tree
    ans.append(root.val)
    maxm_gcd(root.left, ans)
 
    # Traverse right of the binary tree
    maxm_gcd(root.right, ans)
 
# Driver Code
if __name__ == '__main__':
 
    # Given Tree
    root = Node(15)
    root.left = Node(3)
    root.right = Node(7)
    root.left.left = Node(15)
    root.left.right = Node(1)
    root.right.left = Node(31)
    root.right.right = Node(9)
 
    # Function call
    maxm_gcd(root, [])
     
    # Print the maximum AND value
    print(maxm)
 
# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Initialise to update the maximum
// gcd value from all the path
static int maxm = 0;
 
// Node structure
class Node 
{
    public int val;
 
    // Left & right child of the node
    public Node left, right;
 
    // Initialize constructor
    public Node(int x)
    {
        val = x;
        left = null;
        right = null;
    }
};
 
// Function to find gcd of a and b
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function to find the gcd of a path
static int find_gcd(List<int> arr)
{
    if (arr.Count == 1)
        return arr[0];
 
    int g = arr[0];
 
    for(int i = 1; i < arr.Count; i++)
    {
        g = gcd(g, arr[1]);
    }
    return g;
}
 
// Function to find the maximum value
// of gcd from root to leaf
// in a Binary tree
static void maxm_gcd(Node root, 
                     List<int> ans)
{
     
    // Check if root is not null
    if (root == null)
        return;
 
    if (root.left == null && 
        root.right == null) 
    {
        ans.Add(root.val);
 
        // Find the maximum gcd of
        // path value and store in
        // global maxm variable
        maxm = Math.Max(find_gcd(ans),
                                 maxm);
                         
        return;
    }
 
    // Traverse left of binary tree
    ans.Add(root.val);
    maxm_gcd(root.left, ans);
 
    // Traverse right of the binary tree
    maxm_gcd(root.right, ans);
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given Tree
    Node root = new Node(15);
    root.left = new Node(3);
    root.right = new Node(7);
    root.left.left = new Node(15);
    root.left.right = new Node(1);
    root.right.left = new Node(31);
    root.right.right = new Node(9);
 
    // Function call
    maxm_gcd(root, new List<int>());
 
    // Print the maximum AND value
    Console.Write(maxm + "\n");
}
}
 
// This code is contributed by sapnasingh4991

Javascript

<script>
 
// JavaScript program for the above approach
 
// Initialise to update the maximum
// gcd value from all the path
let maxm = 0;
 
// Node structure
class Node 
{
 
    // Initialize constructor
    constructor(x)
    {
        this.val = x;
        this.left = null;
        this.right = null;
    }
}
 
var root;
 
// Function to find gcd of a and b
function gcd(a, b)
{
    if (b == 0)
        return a;
         
    return gcd(b, a % b);
}
 
// Function to find the gcd of a path
function find_gcd(arr)
{
    if (arr.length == 1)
        return arr[0];
 
    let g = arr[0];
 
    for(let i = 1; i < arr.length; i++)
    {
        g = gcd(g, arr[1]);
    }
    return g;
}
 
// Function to find the maximum value
// of gcd from root to leaf
// in a Binary tree
function maxm_gcd(root, ans)
{
     
    // Check if root is not null
    if (root == null)
        return;
 
    if (root.left == null && 
        root.right == null) 
    {
        ans.push(root.val);
 
        // Find the maximum gcd of
        // path value and store in
        // global maxm variable
        maxm = Math.max(find_gcd(ans),
                        maxm);
                         
        return;
    }
 
    // Traverse left of binary tree
    ans.push(root.val);
    maxm_gcd(root.left, ans);
 
    // Traverse right of the binary tree
    maxm_gcd(root.right, ans);
}
 
// Driver Code
 
// Given Tree
root = new Node(15);
root.left = new Node(3);
root.right = new Node(7);
root.left.left = new Node(15);
root.left.right = new Node(1);
root.right.left = new Node(31);
root.right.right = new Node(9);
 
// Function call
let arr = [];
maxm_gcd(root, arr);
 
// Print the maximum AND value
document.write(maxm);
 
// This code is contributed by Dharanendra L V.
 
</script>

Output:

Time Complexity: O(N²)
Auxiliary Space: O(N)

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