Skip to content
Related Articles

Related Articles

Product of all nodes in a Binary Tree

View Discussion
Improve Article
Save Article
  • Difficulty Level : Easy
  • Last Updated : 11 Jul, 2022
View Discussion
Improve Article
Save Article

Given a Binary Tree. The task is to write a program to find the product of all of the nodes of the given binary tree.
 

In the above binary tree, 
Product = 15*10*8*12*20*16*25 = 115200000
 

The idea is to recursively: 

  • Find the product of the left subtree.
  • Find the product of the right subtree.
  • Multiply the product of left and right subtrees with the current node’s data and return.

Below is the implementation of the above approach: 

C++




// Program to print product of all
// the nodes of a binary tree
#include <iostream>
using namespace std;
 
// Binary Tree Node
struct Node {
    int key;
    Node *left, *right;
};
 
/* utility that allocates a new Node
   with the given key */
Node* newNode(int key)
{
    Node* node = new Node;
    node->key = key;
    node->left = node->right = NULL;
    return (node);
}
 
// Function to find product of
// all the nodes
int productBT(Node* root)
{
    if (root == NULL)
        return 1;
 
    return (root->key * productBT(root->left) * productBT(root->right));
}
 
// Driver Code
int main()
{
    // Binary Tree is:
    //       1
    //      /  \
    //     2    3
    //    / \  / \
    //   4   5 6  7
    //          \
    //           8
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->right->left->right = newNode(8);
 
    int prod = productBT(root);
 
    cout << "Product of all the nodes is: "
         << prod << endl;
 
    return 0;
}


Java




   
// Java Program to print product of all
// the nodes of a binary tree
import java.util.*;
class solution
{
 
// Binary Tree Node
static class Node {
    int key;
    Node left, right;
};
 
/* utility that allocates a new Node
   with the given key */
static Node newNode(int key)
{
    Node node = new Node();
    node.key = key;
    node.left = node.right = null;
    return (node);
}
 
// Function to find product of
// all the nodes
static int productBT(Node root)
{
    if (root == null)
        return 1;
 
    return (root.key * productBT(root.left) * productBT(root.right));
}
 
// Driver Code
public static void main(String args[])
{
    // Binary Tree is:
    //       1
    //      /  \
    //     2    3
    //    / \  / \
    //   4   5 6  7
    //          \
    //           8
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.right.left.right = newNode(8);
 
    int prod = productBT(root);
 
    System.out.println( "Product of all the nodes is: "+prod);
 
}
}
//contributed by Arnab Kundu


Python3




# Python3 Program to print product of
# all the nodes of a binary tree
 
# Binary Tree Node
 
""" utility that allocates a new Node
with the given key """
class newNode:
 
    # Construct to create a new node
    def __init__(self, key):
        self.key = key
        self.left = None
        self.right = None
         
# Function to find product of
# all the nodes
def productBT( root) :
 
    if (root == None):
        return 1
 
    return (root.key * productBT(root.left) *
                       productBT(root.right))
 
# Driver Code
if __name__ == '__main__':
     
    # Binary Tree is:
    #     1
    #     / \
    #     2 3
    # / \ / \
    # 4 5 6 7
    #         \
    #         8
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.left.right = newNode(5)
    root.right.left = newNode(6)
    root.right.right = newNode(7)
    root.right.left.right = newNode(8)
 
    prod = productBT(root)
 
    print("Product of all the nodes is:", prod)
     
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)


C#




// C# Program to print product of all
// the nodes of a binary tree
using System;
 
class GFG
{
 
    // Binary Tree Node
    class Node
    {
        public int key;
        public Node left, right;
    };
 
    /* utility that allocates a new Node
    with the given key */
    static Node newNode(int key)
    {
        Node node = new Node();
        node.key = key;
        node.left = node.right = null;
        return (node);
    }
 
    // Function to find product of
    // all the nodes
    static int productBT(Node root)
    {
        if (root == null)
            return 1;
 
        return (root.key * productBT(root.left) *
                        productBT(root.right));
    }
 
    // Driver Code
    public static void Main()
    {
        // Binary Tree is:
        //     1
        //     / \
        //     2 3
        // / \ / \
        // 4 5 6 7
        //         \
        //         8
        Node root = newNode(1);
        root.left = newNode(2);
        root.right = newNode(3);
        root.left.left = newNode(4);
        root.left.right = newNode(5);
        root.right.left = newNode(6);
        root.right.right = newNode(7);
        root.right.left.right = newNode(8);
 
        int prod = productBT(root);
 
        Console.WriteLine( "Product of all " +
                        "the nodes is: " + prod);
    }
}
 
/* This code is contributed PrinciRaj1992 */


Javascript




   
<script>
// javascript Program to print product of all
// the nodes of a binary tree
// Binary Tree Node
class Node {
    constructor(val) {
        this.key = val;
        this.left = null;
        this.right = null;
    }
}
/* utility that allocates a new Node
   with the given key */
function newNode(key)
{
    var node = new Node();
    node.key = key;
    node.left = node.right = null;
    return (node);
}
 
// Function to find product of
// all the nodes
function productBT(root)
{
    if (root == null)
        return 1;
 
    return (root.key * productBT(root.left) * productBT(root.right));
}
 
// Driver Code
 
    // Binary Tree is:
    //       1
    //      /  \
    //     2    3
    //    / \  / \
    //   4   5 6  7
    //          \
    //           8
    var root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.right.left.right = newNode(8);
 
    var prod = productBT(root);
 
    document.write( "Product of all the nodes is: "+prod);
 
// This code contributed by gauravrajput1
</script>


Javascript




   
<script>
// javascript Program to print product of all
// the nodes of a binary tree
// Binary Tree Node
class Node {
    constructor(val) {
        this.key = val;
        this.left = null;
        this.right = null;
    }
}
 
/* utility that allocates a new Node
   with the given key */
function newNode(key)
{
    var node = new Node();
    node.key = key;
    node.left = node.right = null;
    return (node);
}
 
// Function to find product of
// all the nodes
function productBT(root)
{
    if (root == null)
        return 1;
 
    return (root.key * productBT(root.left) * productBT(root.right));
}
 
// Driver Code
 
    // Binary Tree is:
    //       1
    //      /  \
    //     2    3
    //    / \  / \
    //   4   5 6  7
    //          \
    //           8
    var root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.right.left.right = newNode(8);
 
    var prod = productBT(root);
 
    document.write( "Product of all the nodes is: "+prod);
 
// This code contributed by umadevi9616
</script>


Output: 
 

Product of all the nodes is: 40320

Time complexity : O(n)

As we are traversing the tree only once.

Auxiliary Space: O(h)

Here h is the height of the tree. The extra space is used in recursion call stack. In the worst case(when the tree is skewed) this can go upto O(n).
 


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!