Root to leaf path product equal to a given number

Given a binary tree and a number, return true if the tree has a root-to-leaf path such that product of all the values along that path equals the given number. Return 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 leaf node of the current path of the tree.

Below is the implementation of the above approach:

C++

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// 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* node = (struct node*)
        malloc(sizeof(struct node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;
  
    return (node);
}
  
// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

YES

Time Complexity : O(n)



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