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Find maximum level product in Binary Tree

  • Difficulty Level : Easy
  • Last Updated : 23 Jun, 2021

Given a Binary Tree having positive and negative nodes, the task is to find maximum product level in it.
Examples: 
 

Input :               4
                    /   \
                   2    -5
                  / \    /\
                -1   3 -2  6
Output: 36
Explanation :
Product of all nodes of 0'th level is 4
Product of all nodes of 1'th level is -10
Product of all nodes of 0'th level is 36
Hence maximum product is 6

Input :          1
               /   \
              2     3
             / \     \
            4   5     8
                     /  \
                    6    7  
Output :  160
Explanation :
Product of all nodes of 0'th level is 1
Product of all nodes of 1'th level is 6
Product of all nodes of 0'th level is 160
Product of all nodes of 0'th level is 42
Hence maximum product is 160

Prerequisites: Maximum Width of a Binary Tree 
 

Approach : The idea is to do level order traversal of tree. While doing traversal, process nodes of different level separately. For every level being processed, compute product of nodes in the level and keep track of maximum product. 
 

C++




// A queue based C++ program to find maximum product
// of a level in Binary Tree
#include <bits/stdc++.h>
using namespace std;
 
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct Node {
    int data;
    struct Node *left, *right;
};
 
// Function to find the maximum product of a level in tree
// using level order traversal
int maxLevelProduct(struct Node* root)
{
    // Base case
    if (root == NULL)
        return 0;
 
    // Initialize result
    int result = root->data;
 
    // Do Level order traversal keeping track of number
    // of nodes at every level.
    queue<Node*> q;
    q.push(root);
    while (!q.empty()) {
 
        // Get the size of queue when the level order
        // traversal for one level finishes
        int count = q.size();
 
        // Iterate for all the nodes in the queue currently
        int product = 1;
        while (count--) {
 
            // Dequeue an node from queue
            Node* temp = q.front();
            q.pop();
 
            // Multiply this node's value to current product.
            product = product * temp->data;
 
            // Enqueue left and right children of
            // dequeued node
            if (temp->left != NULL)
                q.push(temp->left);
            if (temp->right != NULL)
                q.push(temp->right);
        }
 
        // Update the maximum node count value
        result = max(product, result);
    }
 
    return result;
}
 
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}
 
// Driver code
int main()
{
    struct Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->right = newNode(8);
    root->right->right->left = newNode(6);
    root->right->right->right = newNode(7);
 
    /* Constructed Binary tree is:
             1
            / \
           2   3
          / \   \
         4   5   8
                / \
               6   7 */
    cout << "Maximum level product is "
         << maxLevelProduct(root) << endl;
    return 0;
}

Java




// A queue based Java program to find
// maximum product of a level in Binary Tree
import java.util.*;
 
class GFG
{
 
/* A binary tree node has data,
pointer to left child and a
pointer to right child */
static class Node
{
    int data;
    Node left, right;
};
 
// Function to find the maximum product
// of a level in tree using
// level order traversal
static int maxLevelProduct(Node root)
{
    // Base case
    if (root == null)
        return 0;
 
    // Initialize result
    int result = root.data;
 
    // Do Level order traversal keeping track
    // of number of nodes at every level.
    Queue<Node> q = new LinkedList<>();
    q.add(root);
    while (q.size() > 0)
    {
 
        // Get the size of queue when the level order
        // traversal for one level finishes
        int count = q.size();
 
        // Iterate for all the nodes
        // in the queue currently
        int product = 1;
        while (count-->0)
        {
 
            // Dequeue an node from queue
            Node temp = q.peek();
            q.remove();
 
            // Multiply this node's value
            // to current product.
            product = product* temp.data;
 
            // Enqueue left and right children of
            // dequeued node
            if (temp.left != null)
                q.add(temp.left);
            if (temp.right != null)
                q.add(temp.right);
        }
 
        // Update the maximum node count value
        result = Math.max(product, result);
    }
    return result;
}
 
/* Helper function that allocates
a new node with the given data and
null left and right pointers. */
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = node.right = null;
    return (node);
}
 
// Driver code
public static void main(String args[])
{
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.right = newNode(8);
    root.right.right.left = newNode(6);
    root.right.right.right = newNode(7);
 
    /* Constructed Binary tree is:
            1
            / \
        2 3
        / \ \
        4 5 8
                / \
            6 7 */
    System.out.print("Maximum level product is " +
                          maxLevelProduct(root) );
}
}
 
// This code is contributed by Arnub Kundu

Python3




# Python3 program to find maximum product
# of a level in Binary Tree
 
# Helper function that allocates a new
# node with the given data and None left
# and right poers.                                    
class newNode:
 
    # Construct to create a new node
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None
 
# Function to find the maximum product
# of a level in tree using level order
# traversal
def maxLevelProduct(root):
 
    # Base case
    if (root == None):
        return 0
 
    # Initialize result
    result = root.data
 
    # Do Level order traversal keeping track
    # of number of nodes at every level.
    q = []
    q.append(root)
    while (len(q)):
 
        # Get the size of queue when the level
        # order traversal for one level finishes
        count = len(q)
 
        # Iterate for all the nodes in
        # the queue currently
        product = 1
        while (count):
            count -= 1
             
            # Dequeue an node from queue
            temp = q[0]
            q.pop(0)
 
            # Multiply this node's value to
            # current product.
            product = product * temp.data
 
            # Enqueue left and right children
            # of dequeued node
            if (temp.left != None):
                q.append(temp.left)
            if (temp.right != None):
                q.append(temp.right)
         
        # Update the maximum node count value
        result = max(product, result)
     
    return result
 
# Driver Code
if __name__ == '__main__':
     
    """
    Let us create Binary Tree
    shown in above example """
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.left.right = newNode(5)
    root.right.right = newNode(8)
    root.right.right.left = newNode(6)
    root.right.right.right = newNode(7)
     
    """ Constructed Binary tree is:
            1
            / \
        2 3
        / \ \
        4 5 8
                / \
            6 7 """
 
    print("Maximum level product is",
               maxLevelProduct(root))
 
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

C#




// A queue based C# program to find
// maximum product of a level in Binary Tree
using System;
using System.Collections.Generic;
 
class GFG
{
 
    /* A binary tree node has data,
    pointer to left child and a
    pointer to right child */
    class Node
    {
        public int data;
        public Node left, right;
    };
 
    // Function to find the maximum product
    // of a level in tree using
    // level order traversal
    static int maxLevelProduct(Node root)
    {
        // Base case
        if (root == null)
        {
            return 0;
        }
 
        // Initialize result
        int result = root.data;
 
        // Do Level order traversal keeping track
        // of number of nodes at every level.
        Queue<Node> q = new Queue<Node>();
        q.Enqueue(root);
        while (q.Count > 0)
        {
 
            // Get the size of queue when the level order
            // traversal for one level finishes
            int count = q.Count;
 
            // Iterate for all the nodes
            // in the queue currently
            int product = 1;
            while (count-- > 0)
            {
 
                // Dequeue an node from queue
                Node temp = q.Peek();
                q.Dequeue();
 
                // Multiply this node's value
                // to current product.
                product = product * temp.data;
 
                // Enqueue left and right children of
                // dequeued node
                if (temp.left != null)
                {
                    q.Enqueue(temp.left);
                }
                if (temp.right != null)
                {
                    q.Enqueue(temp.right);
                }
            }
 
            // Update the maximum node count value
            result = Math.Max(product, result);
        }
        return result;
    }
 
    /* Helper function that allocates
    a new node with the given data and
    null left and right pointers. */
    static Node newNode(int data)
    {
        Node node = new Node();
        node.data = data;
        node.left = node.right = null;
        return (node);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        Node root = newNode(1);
        root.left = newNode(2);
        root.right = newNode(3);
        root.left.left = newNode(4);
        root.left.right = newNode(5);
        root.right.right = newNode(8);
        root.right.right.left = newNode(6);
        root.right.right.right = newNode(7);
 
        /* Constructed Binary tree is:
            1
            / \
        2 3
        / \ \
        4 5 8
                / \
            6 7 */
        Console.Write("Maximum level product is " +
                            maxLevelProduct(root));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// A queue based Javascript program to find
// maximum product of a level in Binary Tree
/* A binary tree node has data,
pointer to left child and a
pointer to right child */
class Node
{
    constructor()
    {
        this.left = null;
        this.right = null;
        this.data = 0;
    }
};
 
// Function to find the maximum product
// of a level in tree using
// level order traversal
function maxLevelProduct(root)
{
    // Base case
    if (root == null)
    {
        return 0;
    }
    // Initialize result
    var result = root.data;
    // Do Level order traversal keeping track
    // of number of nodes at every level.
    var q = [];
    q.push(root);
    while (q.length > 0)
    {
        // Get the size of queue when the level order
        // traversal for one level finishes
        var count = q.length;
        // Iterate for all the nodes
        // in the queue currently
        var product = 1;
        while (count-- > 0)
        {
            // Dequeue an node from queue
            var temp = q[0];
            q.shift();
            // Multiply this node's value
            // to current product.
            product = product * temp.data;
            // push left and right children of
            // dequeued node
            if (temp.left != null)
            {
                q.push(temp.left);
            }
            if (temp.right != null)
            {
                q.push(temp.right);
            }
        }
        // Update the maximum node count value
        result = Math.max(product, result);
    }
    return result;
}
/* Helper function that allocates
a new node with the given data and
null left and right pointers. */
function newNode(data)
{
    var node = new Node();
    node.data = data;
    node.left = node.right = null;
    return (node);
}
 
// Driver code
var root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
root.right.right = newNode(8);
root.right.right.left = newNode(6);
root.right.right.right = newNode(7);
/* Constructed Binary tree is:
    1
    / \
2 3
/ \ \
4 5 8
        / \
    6 7 */
document.write("Maximum level product is " +
                    maxLevelProduct(root));
 
// This code is contributed by famously.
</script>

Output : 

Maximum level product is 160

Time Complexity : O(n) 
Auxiliary Space : O(n)
 

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