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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program to check if there exists ` `// a root to leaf path product with ` `// given product ` ` `  `#include ` `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; ` `} `

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 `

Output:

```YES
```

Time Complexity : O(n)

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