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# Find the parent node of maximum product Siblings in given Binary Tree

• Difficulty Level : Medium
• Last Updated : 10 Dec, 2021

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 5.

Approach: 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 ` `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 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

 `// 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 q = ``new` `LinkedList();`   `        ``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.`

## Python3

 `# 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#

 `// 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 q = ``new` `Queue();`   `        ``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`

## Javascript

 ``

Output

`3`

Time Complexity: O(V) where V is the number of nodes in the tree.
Auxiliary Space: O(V).

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