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 Nodestruct 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 pointersstruct node* newnode(int data){ struct node* newNode = new node(); newNode->data = data; newNode->left = NULL; newNode->right = NULL; return (newNode);}// Driver Codeint 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 productusing 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)
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