Find the parent node of maximum product Siblings in given Binary Tree
Last Updated :
23 Jul, 2025
Given a binary tree, the task is to find the node whose children have maximum Sibling product in the given Binary Tree. If there are multiple such nodes, return the node which has the maximum value.
Examples:
Input: Tree:
4
/ \
5 2
/ \
3 1
/ \
6 12
Output: 3
Explanation: For the above tree, the maximum product for the siblings is formed for nodes 6 and 12 which are the children of node 3.
Input: Tree:
1
/ \
3 5
/ \ / \
6 9 4 8
Output: 3
Explanation: For the above tree, the maximum product for the siblings is formed for nodes 6 and 9 which are the children of node 3.
Approach using Level Order Traversal:
To solve this problem, level order traversal of the Binary Tree can be used to find the maximum sum of siblings. Follow the following steps:
- Start level order traversal of the tree from root of the tree.
- For each node, check if it has both the child.
- If yes, then find the node with maximum product of children and store this node value in a reference variable.
- Update the node value in reference variable if any node is found with greater product of children.
- If the current node don't have both children, then skip that node
- Return the node value in reference variable, as it contains the node with maximum product of children, or the parent of maximum product siblings.
Below is the implementation of the above approach:
C++
// C++ code to find the Parent Node
// of maximum product Siblings
// in given Binary Tree
#include <bits/stdc++.h>
using namespace std;
// Structure for Node
struct Node {
int data;
Node *left, *right;
};
// Function to get a new node
Node* getNode(int data)
{
// Allocate space
Node* newNode
= (Node*)malloc(sizeof(Node));
// Put in the data
newNode->data = data;
newNode->left = newNode->right = NULL;
return newNode;
}
// Function to get the parent
// of siblings with maximum product
int maxproduct(Node* root)
{
int mproduct = INT_MIN;
int ans = 0;
// Checking base case
if (root == NULL
|| (root->left == NULL
&& root->right == NULL))
return 0;
// Declaration of queue to run
// level order traversal
queue<Node*> q;
q.push(root);
// Loop to implement level order traversal
while (!q.empty()) {
Node* temp = q.front();
q.pop();
// If both the siblings are present
// then take their product
if (temp->right && temp->left) {
int curr_max
= temp->right->data
* temp->left->data;
if (mproduct < curr_max) {
mproduct = curr_max;
ans = temp->data;
}
else if (mproduct == curr_max) {
// if max product is equal to
// curr_max then consider node
// which has maximum value
ans = max(ans, temp->data);
}
}
// pushing childs in the queue
if (temp->right) {
q.push(temp->right);
}
if (temp->left) {
q.push(temp->left);
}
}
return ans;
}
// Driver Code
int main()
{
/* Binary tree creation
1
/ \
3 5
/ \ / \
6 9 4 8
*/
Node* root = getNode(1);
root->left = getNode(3);
root->right = getNode(5);
root->left->left = getNode(6);
root->left->right = getNode(9);
root->right->left = getNode(4);
root->right->right = getNode(8);
cout << maxproduct(root) << endl;
return 0;
}
// Java code to find the Parent Node
// of maximum product Siblings
// in given Binary Tree
import java.util.LinkedList;
import java.util.Queue;
class GFG {
// Structure for Node
static class Node {
int data;
Node left;
Node right;
public Node(int data) {
this.data = data;
this.left = null;
this.right = null;
}
};
// Function to get a new node
public static Node getNode(int data) {
// Allocate space
Node newNode = new Node(data);
// Put in the data
newNode.data = data;
newNode.left = newNode.right = null;
return newNode;
}
// Function to get the parent
// of siblings with maximum product
public static int maxproduct(Node root) {
int mproduct = Integer.MIN_VALUE;
int ans = 0;
// Checking base case
if (root == null
|| (root.left == null
&& root.right == null))
return 0;
// Declaration of queue to run
// level order traversal
Queue<Node> q = new LinkedList<Node>();
q.add(root);
// Loop to implement level order traversal
while (!q.isEmpty()) {
Node temp = q.peek();
q.remove();
// If both the siblings are present
// then take their product
if (temp.right != null && temp.left != null) {
int curr_max = temp.right.data
* temp.left.data;
if (mproduct < curr_max) {
mproduct = curr_max;
ans = temp.data;
} else if (mproduct == curr_max) {
// if max product is equal to
// curr_max then consider node
// which has maximum value
ans = Math.max(ans, temp.data);
}
}
// pushing childs in the queue
if (temp.right != null) {
q.add(temp.right);
}
if (temp.left != null) {
q.add(temp.left);
}
}
return ans;
}
// Driver Code
public static void main(String args[]) {
/*
* Binary tree creation
* 1
* / \
* 3 5
* / \ / \
* 6 9 4 8
*/
Node root = getNode(1);
root.left = getNode(3);
root.right = getNode(5);
root.left.left = getNode(6);
root.left.right = getNode(9);
root.right.left = getNode(4);
root.right.right = getNode(8);
System.out.println(maxproduct(root));
}
}
// This code is contributed by gfgking.
# Python Program to implement
# the above approach
# Structure of a node of binary tree
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Function to get a new node
def getNode(data):
# Allocate space
newNode = Node(data)
return newNode
# Function to get the parent
# of siblings with maximum product
def maxproduct(root):
mproduct = 10 ** -9
ans = 0;
# Checking base case
if (root == None or (root.left == None and root.right == None)):
return 0;
# Declaration of queue to run
# level order traversal
q = [];
q.append(root);
# Loop to implement level order traversal
while (len(q)):
temp = q[0];
q.pop(0);
# If both the siblings are present
# then take their product
if (temp.right and temp.left):
curr_max = temp.right.data * temp.left.data;
if (mproduct < curr_max):
mproduct = curr_max
ans = temp.data
elif (mproduct == curr_max):
# if max product is equal to
# curr_max then consider node
# which has maximum value
ans = max(ans, temp.data)
# pushing childs in the queue
if (temp.right):
q.append(temp.right)
if (temp.left):
q.append(temp.left)
return ans
# Driver Code
""" Binary tree creation
1
/ \
3 5
/ \ / \
6 9 4 8
"""
root = getNode(1);
root.left = getNode(3);
root.right = getNode(5);
root.left.left = getNode(6);
root.left.right = getNode(9);
root.right.left = getNode(4);
root.right.right = getNode(8);
print(maxproduct(root));
# This code is contributed by gfgking
// C# code to find the Parent Node
// of maximum product Siblings
// in given Binary Tree
using System;
using System.Collections.Generic;
public class GFG {
// Structure for Node
class Node {
public int data;
public Node left;
public Node right;
public Node(int data) {
this.data = data;
this.left = null;
this.right = null;
}
};
// Function to get a new node
static Node getNode(int data) {
// Allocate space
Node newNode = new Node(data);
// Put in the data
newNode.data = data;
newNode.left = newNode.right = null;
return newNode;
}
// Function to get the parent
// of siblings with maximum product
static int maxproduct(Node root) {
int mproduct = int.MinValue;
int ans = 0;
// Checking base case
if (root == null
|| (root.left == null
&& root.right == null))
return 0;
// Declaration of queue to run
// level order traversal
Queue<Node> q = new Queue<Node>();
q.Enqueue(root);
// Loop to implement level order traversal
while (q.Count!=0) {
Node temp = q.Peek();
q.Dequeue();
// If both the siblings are present
// then take their product
if (temp.right != null && temp.left != null) {
int curr_max = temp.right.data
* temp.left.data;
if (mproduct < curr_max) {
mproduct = curr_max;
ans = temp.data;
} else if (mproduct == curr_max) {
// if max product is equal to
// curr_max then consider node
// which has maximum value
ans = Math.Max(ans, temp.data);
}
}
// pushing childs in the queue
if (temp.right != null) {
q.Enqueue(temp.right);
}
if (temp.left != null) {
q.Enqueue(temp.left);
}
}
return ans;
}
// Driver Code
public static void Main(String []args) {
/*
* Binary tree creation
* 1
* / \
* 3 5
* / \ / \
* 6 9 4 8
*/
Node root = getNode(1);
root.left = getNode(3);
root.right = getNode(5);
root.left.left = getNode(6);
root.left.right = getNode(9);
root.right.left = getNode(4);
root.right.right = getNode(8);
Console.WriteLine(maxproduct(root));
}
}
// This code is contributed by shikhasingrajput
<script>
// JavaScript Program to implement
// the above approach
// Structure of a node of binary tree
class Node {
constructor(data) {
this.data = data;
this.left = null;
this.right = null;
}
};
// Function to get a new node
function getNode(data)
{
// Allocate space
let newNode
= new Node(data);
return newNode;
}
// Function to get the parent
// of siblings with maximum product
function maxproduct(root) {
let mproduct = Number.MIN_VALUE;
let ans = 0;
// Checking base case
if (root == null
|| (root.left == null
&& root.right == null))
return 0;
// Declaration of queue to run
// level order traversal
let q = [];
q.push(root);
// Loop to implement level order traversal
while (!q.length == 0) {
let temp = q[0];
q.shift();
// If both the siblings are present
// then take their product
if (temp.right && temp.left) {
let curr_max
= temp.right.data
* temp.left.data;
if (mproduct < curr_max) {
mproduct = curr_max;
ans = temp.data;
}
else if (mproduct == curr_max) {
// if max product is equal to
// curr_max then consider node
// which has maximum value
ans = Math.max(ans, temp.data);
}
}
// pushing childs in the queue
if (temp.right) {
q.push(temp.right);
}
if (temp.left) {
q.push(temp.left);
}
}
return ans;
}
// Driver Code
/* Binary tree creation
1
/ \
3 5
/ \ / \
6 9 4 8
*/
let root = getNode(1);
root.left = getNode(3);
root.right = getNode(5);
root.left.left = getNode(6);
root.left.right = getNode(9);
root.right.left = getNode(4);
root.right.right = getNode(8);
document.write(maxproduct(root) + "<br>");
// This code is contributed by Potta Lokesh
</script>
Time Complexity: O(V) where V is the number of nodes in the tree.
Auxiliary Space: O(V).
- The idea is to traverse the tree in postorder fashion .
- Recursively , call the left subtree.
- Recursively , call the right subtree.
- Get , the product of left child and right child and update the max variable and max parent variable with parent root of left and right child.
- Here , max parent variable will be storing the parent node of the maximum product siblings.
Below is the Implementation of this Approach :-
C++
#include <climits>
#include <iostream>
// Structure for Node
struct Node {
int data;
Node* left;
Node* right;
Node(int data)
: data(data)
, left(nullptr)
, right(nullptr)
{
}
};
// Function to get a new node
Node* getNode(int data)
{
// Allocate space
Node* newNode = new Node(data);
// Put in the data
newNode->data = data;
newNode->left = newNode->right = nullptr;
return newNode;
}
// Initialize, parentNode with -1
int parentNode = -1;
// Initialize, maxProductVal with INT_MIN
int maxProductVal = INT_MIN;
// Forward declaration of the helper function
int helper(Node* root);
// Function to get the parent of siblings with maximum
// product
int calculateMaxProduct(Node* root)
{
// calling helper function to get ParentNode
helper(root);
return parentNode;
}
int helper(Node* root)
{
// Base Case
if (root == nullptr) {
return 0;
}
// recursively, calling left part
int left = helper(root->left);
// recursively, calling right part
int right = helper(root->right);
// multiplying left and right to get product of the root
int currProduct = left * right;
// update maxProduct
if (maxProductVal < currProduct) {
maxProductVal = currProduct;
// update the parentNode
parentNode = root->data;
}
// return root data for every node
return root->data;
}
// Driver Code
int main()
{
/*
* Binary tree creation
* 1
* / \
* 3 5
* / \ / \
* 6 9 4 8
*/
Node* root = getNode(1);
root->left = getNode(3);
root->right = getNode(5);
root->left->left = getNode(6);
root->left->right = getNode(9);
root->right->left = getNode(4);
root->right->right = getNode(8);
std::cout << calculateMaxProduct(root) << std::endl;
// Free allocated memory
delete root->left->left;
delete root->left->right;
delete root->right->left;
delete root->right->right;
delete root->left;
delete root->right;
delete root;
return 0;
}
// Java code to find the Parent Node
// of maximum product Siblings
// in given Binary Tree
import java.util.LinkedList;
import java.util.Queue;
class GFG {
// Structure for Node
static class Node {
int data;
Node left;
Node right;
public Node(int data)
{
this.data = data;
this.left = null;
this.right = null;
}
};
// Function to get a new node
public static Node getNode(int data)
{
// Allocate space
Node newNode = new Node(data);
// Put in the data
newNode.data = data;
newNode.left = newNode.right = null;
return newNode;
}
// Initialize , parentNode with -1
static int parentNode = -1;
// Initialize , maxProduct with INT_MIN
static int maxProduct = Integer.MIN_VALUE;
// Function to get the parent
// of siblings with maximum product
public static int maxproduct(Node root)
{
// calling helper function to get
// ParentNode
helper(root);
return parentNode;
}
static int helper(Node root)
{
// Base Case
if (root == null) {
return 0;
}
// recursively , calling left part
int left = helper(root.left);
// recursively , calling right part
int right = helper(root.right);
// multiplying left and right to get product
// of the root
int currProduct = left * right;
// update maxProduct
if (maxProduct < currProduct) {
maxProduct = currProduct;
// update the parentNode
parentNode = root.data;
}
// return root data for every node
return root.data;
}
// Driver Code
public static void main(String args[])
{
/*
* Binary tree creation
* 1
* / \
* 3 5
* / \ / \
* 6 9 4 8
*/
Node root = getNode(1);
root.left = getNode(3);
root.right = getNode(5);
root.left.left = getNode(6);
root.left.right = getNode(9);
root.right.left = getNode(4);
root.right.right = getNode(8);
System.out.println(maxproduct(root));
}
}
// This code is contributed by srimann7
import sys
# Class definition for Node
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Function to get a new node
def get_node(data):
# Allocate space
new_node = Node(data)
# Put in the data
new_node.data = data
new_node.left = new_node.right = None
return new_node
# Initialize parentNode with -1
parent_node = -1
# Initialize max_product_val with INT_MIN
max_product_val = -sys.maxsize - 1
# Helper function to calculate the maximum product
def helper(root):
global parent_node, max_product_val
# Base Case
if root is None:
return 0
# Recursively calling left part
left = helper(root.left)
# Recursively calling right part
right = helper(root.right)
# Multiplying left and right to get the product of the root
curr_product = left * right
# Update max_product_val
if max_product_val < curr_product:
max_product_val = curr_product
# Update the parent_node
parent_node = root.data
# Return root data for every node
return root.data
# Function to calculate the parent of siblings with the maximum product
def calculate_max_product(root):
# Calling the helper function to get parent_node
helper(root)
return parent_node
# Driver Code
if __name__ == "__main__":
"""
Binary tree creation
1
/ \
3 5
/ \ / \
6 9 4 8
"""
root = get_node(1)
root.left = get_node(3)
root.right = get_node(5)
root.left.left = get_node(6)
root.left.right = get_node(9)
root.right.left = get_node(4)
root.right.right = get_node(8)
print(calculate_max_product(root))
# Free allocated memory
del root.left.left
del root.left.right
del root.right.left
del root.right.right
del root.left
del root.right
del root
using System;
// Structure for Node
public class Node {
public int data;
public Node left;
public Node right;
public Node(int data)
{
this.data = data;
this.left = null;
this.right = null;
}
}
public class BinaryTree {
// Initialize parentNode with -1
private static int parentNode = -1;
// Initialize maxProductVal with int.MinValue
private static int maxProductVal = int.MinValue;
// Function to get the parent of siblings with maximum
// product
private static int CalculateMaxProduct(Node root)
{
// calling helper function to get parentNode
Helper(root);
return parentNode;
}
private static int Helper(Node root)
{
// Base Case
if (root == null) {
return 0;
}
// Recursively calling the left part
int left = Helper(root.left);
// Recursively calling the right part
int right = Helper(root.right);
// Multiplying left and right to get the product of
// the root
int currProduct = left * right;
// Update maxProduct
if (maxProductVal < currProduct) {
maxProductVal = currProduct;
// Update the parentNode
parentNode = root.data;
}
// Return root data for every node
return root.data;
}
// Driver Code
public static void Main()
{
/*
* Binary tree creation
* 1
* / \
* 3 5
* / \ / \
* 6 9 4 8
*/
Node root = new Node(1);
root.left = new Node(3);
root.right = new Node(5);
root.left.left = new Node(6);
root.left.right = new Node(9);
root.right.left = new Node(4);
root.right.right = new Node(8);
Console.WriteLine(CalculateMaxProduct(root));
// No need to free memory in C#, as it is managed by
// the garbage collector
}
}
// Structure for Node
class Node {
constructor(data) {
this.data = data;
this.left = null;
this.right = null;
}
}
// Function to get a new node
function getNode(data) {
const newNode = new Node(data);
newNode.data = data;
newNode.left = null;
newNode.right = null;
return newNode;
}
// Initialize parentNode with -1
let parentNode = -1;
// Initialize maxProductVal with Number.MIN_SAFE_INTEGER
let maxProductVal = Number.MIN_SAFE_INTEGER;
// Forward declaration of the helper function
function helper(root) {
// Base Case
if (root === null) {
return 0;
}
// Recursively calling left part
const left = helper(root.left);
// Recursively calling right part
const right = helper(root.right);
// Multiplying left and right to get product of the root
const currProduct = left * right;
// Update maxProduct
if (maxProductVal < currProduct) {
maxProductVal = currProduct;
// Update the parentNode
parentNode = root.data;
}
// Return root data for every node
return root.data;
}
// Function to get the parent of siblings with maximum product
function calculateMaxProduct(root) {
// Calling helper function to get parentNode
helper(root);
return parentNode;
}
// Driver Code
/*
* Binary tree creation
* 1
* / \
* 3 5
* / \ / \
* 6 9 4 8
*/
const root = getNode(1);
root.left = getNode(3);
root.right = getNode(5);
root.left.left = getNode(6);
root.left.right = getNode(9);
root.right.left = getNode(4);
root.right.right = getNode(8);
console.log(calculateMaxProduct(root));
Time Complexity: O(N) , where N is the number of nodes in the tree
Auxiliary Space : O(N) , recursion stack space
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GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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