Root to leaf path product equal to a given number

Given a binary tree and a number, the return is true if the tree has a root-to-leaf path such that the product of all the values along that path equals the given number. The return is false if no such path can be found. 

For example, in the above tree, there exist three roots to leaf paths with the following products.

• 240 –> 10 – 8 – 3
• 400 –> 10 – 8 – 5
• 40 –> 10 – 2 – 2

Approach: The idea is to start traversing the tree recursively and divide the current node’s value from the product if it is divisible when recurring down, and check to see if the product is 1 when you reach the leaf node of the current path of the tree. 

Below is the implementation of the above approach: 

C++

// C++ program to check if there exists
// a root to leaf path product with
// given product

#include <bits/stdc++.h>
using namespace std;

// Binary Tree Node
struct node {
    int data;
    struct node* left;
    struct node* right;
};

// Function to check if there exists a path
// with given product

// Strategy: divide the node value from the product
// if it is divisible when recurring down, and check
// to see if the product is 1 when you reach leaf
// node of the current path out of tree.
bool hasPathProduct(struct node* node, int prod)
{
    // return true if we run out
    // of tree and prod==1
    if (node == NULL) {
        return (prod == 1);
    }
    else {
        bool ans = 1;

        // Check if product is divisible by
        // current node, if not we are on wrong path
        if (prod % (node->data))
            return false;

        // otherwise check both subtrees
        int subProduct = prod / node->data;

        // If we reach a leaf node and prod
        // becomes 1 then return true
        if (subProduct == 1 && node->left == NULL
            && node->right == NULL)
            return 1;

        if (node->left)
            ans = hasPathProduct(node->left, subProduct);
        if (node->right)
            ans = hasPathProduct(node->right, subProduct);

        return ans;
    }
}

/* UTILITY FUNCTIONS */
// Helper function that allocates
// a new node with the given data
// and NULL left and right pointers
struct node* newnode(int data)
{
    struct node* newNode = new node();
    newNode->data = data;
    newNode->left = NULL;
    newNode->right = NULL;

    return (newNode);
}

// Driver Code
int main()
{
    int prod = 400;

    /* Constructed binary tree is
            10
            / \
           8   2
          / \ /
         3    5 2
    */

    struct node* root = newnode(10);
    root->left = newnode(8);
    root->right = newnode(2);
    root->left->left = newnode(3);
    root->left->right = newnode(5);
    root->right->left = newnode(2);

    if (hasPathProduct(root, prod))
        cout << "YES";
    else
        cout << "NO";

    return 0;
}

Java

// Java program to check if there exists
// a root to leaf path product with
// given product
class GFG
{

    // Binary Tree Node
    static class node
    {

        int data;
        node left;
        node right;
    };

    // Function to check if there exists a path
    // with given product
    // Strategy: divide the node value from the product
    // if it is divisible when recurring down, and check
    // to see if the product is 1 when you reach leaf
    // node of the current path out of tree.
    static boolean hasPathProduct(node node, int prod)
    {
        // return true if we run out
        // of tree and prod==1
        if (node == null)
        {
            return (prod == 1);
        }
        else
        {
            boolean ans = true;

            // Check if product is divisible by
            // current node, if not we are on wrong path
            if (prod % (node.data) == 1)
            {
                return false;
            }

            // otherwise check both subtrees
            int subProduct = prod / node.data;

            // If we reach a leaf node and prod
            // becomes 1 then return true
            if (subProduct == 1 && node.left == null
                    && node.right == null)
            {
                return true;
            }

            if (node.left != null)
            {
                ans = hasPathProduct(node.left, subProduct);
            }
            if (node.right != null)
            {
                ans = hasPathProduct(node.right, subProduct);
            }

            return ans;
        }
    }

    /* UTILITY FUNCTIONS */
    // 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 = null;
        node.right = null;

        return (node);
    }

    // Driver Code
    public static void main(String[] args)
    {
        int prod = 400;

        /* Constructed binary tree is
            10
            / \
        8 2
        / \ /
        3 5 2
        */
        node root = newnode(10);
        root.left = newnode(8);
        root.right = newnode(2);
        root.left.left = newnode(3);
        root.left.right = newnode(5);
        root.right.left = newnode(2);

        if (hasPathProduct(root, prod))
        {
            System.out.println("Yes");
        }
        else
        {
            System.out.println("No");
        }
    }
}

// This code is contributed by Princi Singh

Python3

# Python3 program to check if there exists
# a root to leaf path product with
# given product

""" UTILITY FUNCTIONS """
# Helper function that allocates
# a new node with the given data
# and None left and right pointers
class newnode:
    def __init__(self, data):
        self.data = data
        self.left = self.right = None

# Function to check if there exists
# a path with given product

# Strategy: divide the node value from
# the product if it is divisible when
# recurring down, and check to see if
# the product is 1 when you reach leaf
# node of the current path out of tree.
def hasPathProduct(node, prod) :

    # return true if we run out
    # of tree and prod==1
    if (node == None) :
        return (prod == 1)

    else :

        ans = 1

        # Check if product is divisible by
        # current node, if not we are on wrong path
        if (prod % (node.data)) :
            return False

        # otherwise check both subtrees
        subProduct = prod // node.data

        # If we reach a leaf node and prod
        # becomes 1 then return true
        if (subProduct == 1 and
            node.left == None and
            node.right == None) :
            return 1

        if (node.left) :
            ans = hasPathProduct(node.left,
                                 subProduct)
        if (node.right) :
            ans = hasPathProduct(node.right,
                                 subProduct)

        return ans

# Driver Code
if __name__ == '__main__':
    prod = 400

    """ Constructed binary tree is
            10
            / \
        8 2
        / \ /
        3 5 2
    """
    root = newnode(10)
    root.left = newnode(8)
    root.right = newnode(2)
    root.left.left = newnode(3)
    root.left.right = newnode(5)
    root.right.left = newnode(2)

    if (hasPathProduct(root, prod)) :
        print("YES" )
    else:
        print("NO")

# This code is contributed
# by SHUBHAMSINGH10

C#

// C# program to check if there exists
// a root to leaf path product with
// given product
using System;

class GFG
{

    // Binary Tree Node
    public class node
    {

        public int data;
        public node left;
        public node right;
    };

    // Function to check if there exists a path
    // with given product
    // Strategy: divide the node value from the product
    // if it is divisible when recurring down, and check
    // to see if the product is 1 when you reach leaf
    // node of the current path out of tree.
    static bool hasPathProduct(node node, int prod)
    {
        // return true if we run out
        // of tree and prod==1
        if (node == null)
        {
            return (prod == 1);
        }
        else
        {
            bool ans = true;

            // Check if product is divisible by
            // current node, if not we are on wrong path
            if (prod % (node.data) == 1)
            {
                return false;
            }

            // otherwise check both subtrees
            int subProduct = prod / node.data;

            // If we reach a leaf node and prod
            // becomes 1 then return true
            if (subProduct == 1 && node.left == null
                    && node.right == null)
            {
                return true;
            }

            if (node.left != null)
            {
                ans = hasPathProduct(node.left, subProduct);
            }
            if (node.right != null)
            {
                ans = hasPathProduct(node.right, subProduct);
            }

            return ans;
        }
    }

    /* UTILITY FUNCTIONS */
    // 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 = null;
        node.right = null;

        return (node);
    }

    // Driver Code
    public static void Main(String[] args)
    {
        int prod = 400;

        /* Constructed binary tree is
            10
            / \
        8 2
        / \ /
        3 5 2
        */
        node root = newnode(10);
        root.left = newnode(8);
        root.right = newnode(2);
        root.left.left = newnode(3);
        root.left.right = newnode(5);
        root.right.left = newnode(2);

        if (hasPathProduct(root, prod))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
    }
}

// This code is contributed by Princi Singh

Javascript

<script>
// javascript program to check if there exists
// a root to leaf path product with
// given product    // Binary Tree Node
    class Node {
        constructor(val) {
            this.data = val;
            this.left = null;
            this.right = null;
        }
    }

    // Function to check if there exists a path
    // with given product
    // Strategy: divide the node value from the product
    // if it is divisible when recurring down, and check
    // to see if the product is 1 when you reach leaf
    // node of the current path out of tree.
    function hasPathProduct( node , prod)
    {
        // return true if we run out
        // of tree and prod==1
        if (node == null)
        {
            return (prod == 1);
        }
        else
        {
            var ans = true;

            // Check if product is divisible by
            // current node, if not we are on wrong path
            if (prod % (node.data) == 1)
            {
                return false;
            }

            // otherwise check both subtrees
            var subProduct = prod / node.data;

            // If we reach a leaf node and prod
            // becomes 1 then return true
            if (subProduct == 1 && node.left == null
                    && node.right == null)
            {
                return true;
            }

            if (node.left != null)
            {
                ans = hasPathProduct(node.left, subProduct);
            }
            if (node.right != null)
            {
                ans = hasPathProduct(node.right, subProduct);
            }

            return ans;
        }
    }

    /* UTILITY FUNCTIONS */
    // 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 = null;
        node.right = null;

        return (node);
    }

    // Driver Code

        var prod = 400;

        /* Constructed binary tree is
            10
            / \
        8 2
        / \ /
        3 5 2
        */
        var root = newnode(10);
        root.left = newnode(8);
        root.right = newnode(2);
        root.left.left = newnode(3);
        root.left.right = newnode(5);
        root.right.left = newnode(2);

        if (hasPathProduct(root, prod))
        {
            document.write("Yes");
        }
        else
        {
            document.write("No");
        }

// This code contributed by Rajput-Ji
</script>

Output

YES

Complexity Analysis:

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