Product of nodes at k-th level in a tree represented as string using Recursion

Prerequisite: Product of nodes at k-th level in a tree represented as string
Given an integer ‘K’ and a binary tree in string format. Every node of a tree has value in range from 0 to 9. We need to find product of elements at K-th level from the root. The root is at level 0.
Note: Tree is given in the form: (node value(left subtree)(right subtree))
Examples:

Input: Tree = “(0(5(6()())(4()(9()())))(7(1()())(3()())))” 
k = 2 
Output: 72 
Explanation:
Its tree representation is shown below 

Elements at level k = 2 are 6, 4, 1, 3 
product of these elements = 6 * 4 * 1 * 3 = 72 

Input: Tree = “(8(3(2()())(6(5()())()))(5(10()())(7(13()())())))” 
k = 3 
Output: 15 
Elements at level k = 3 are 5, 1 and 3 
product of these elements = 5 * 1 * 3 = 15 

Approach: The idea is to treat the string as a tree without actually creating one, and simply traverse the string recursively in Postorder Fashion and consider nodes that are at level k only.

Below is the implementation of the above approach: 

C++

// C++ implementation to find product
// of elements at k-th level

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

// Recursive Function to find product
// of elements at k-th level
int productAtKthLevel(string tree,
            int k, int& i, int level){

    if (tree[i++] == '(') {

        // if subtree is null,
        // just like if root == NULL
        if (tree[i] == ')')
            return 1;

        int product = 1;

        // Consider only level k node
        // to be part of the product
        if (level == k)
            product = tree[i] - '0';

        // Recur for Left Subtree
        int leftproduct = productAtKthLevel(
                    tree, k, ++i, level + 1);

        // Recur for Right Subtree
        int rightproduct = productAtKthLevel(
                tree, k, ++i, level + 1);

        // Taking care of ')' after
        // left and right subtree
        ++i;
        return product * leftproduct *
                       rightproduct;
    }
}

// Driver Code
int main()
{
    string tree = "(0(5(6()())(4()"
    "(9()())))(7(1()())(3()())))";
    int k = 2;
    int i = 0;

    cout << productAtKthLevel(tree, k, i, 0);

    return 0;
}

Java

// Java implementation to find
// product of elements at k-th level

class GFG {
    static int i;

    // Recursive Function to find product
    // of elements at k-th level
    static int productAtKthLevel(
        String tree, int k, int level){

        if (tree.charAt(i++) == '(') {

            // if subtree is null,
            // just like if root == null
            if (tree.charAt(i) == ')')
                return 1;

            int product = 1;

            // Consider only level k node
            // to be part of the product
            if (level == k)
                product = tree.charAt(i) - '0';

            // Recur for Left Subtree
            ++i;
            int leftproduct = productAtKthLevel(
                            tree, k, level + 1);

            // Recur for Right Subtree
            ++i;
            int rightproduct = productAtKthLevel(
                            tree, k, level + 1);

            // Taking care of ')' after
            // left and right subtree
            ++i;
            return product * leftproduct
              * rightproduct;
        }
        return Integer.MIN_VALUE;
    }

    // Driver Code
    public static void main(String[] args)
    {
        String tree = "(0(5(6()())(4()"
        + "(9()())))(7(1()())(3()())))";
        int k = 2;
        i = 0;
        System.out.print(
            productAtKthLevel(tree, k, 0)
        );
    }
}

Python

# Python implementation to find product of
# digits of elements at k-th level

# Recursive Function to find product
# of elements at k-th level
def productAtKthLevel(tree, k, i, level):

    if(tree[i[0]]=='('):
        i[0]+= 1
        # if subtree is null,
        # just like if root == NULL
        if(tree[i[0]] == ')'):
            return 1

        product = 1
        # Consider only level k node
        # to be part of the product
        if(level == k):
            product = int(tree[i[0]])

        # Recur for Left Subtree
        i[0]+= 1
        leftproduct = productAtKthLevel(tree,
                            k, i, level + 1)

        # Recur for Right Subtree
        i[0]+= 1
        rightproduct = productAtKthLevel(tree,
                            k, i, level + 1)

        # Taking care of ')' after left and right subtree
        i[0]+= 1
        return product * leftproduct * rightproduct

# Driver Code
if __name__ == "__main__":
    tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))"
    k = 2
    i =[0]
    print(productAtKthLevel(tree, k, i, 0))

C#

// C# implementation to find product
// of elements at k-th level

using System;

class GFG {
    static int i;

    // Recursive Function to find product
    // of elements at k-th level
    static int productAtKthLevel(
        String tree, int k, int level){

        if (tree[i++] == '(') {

            // if subtree is null,
            // just like if root == null
            if (tree[i] == ')')
                return 1;

            int product = 1;

            // Consider only level k node
            // to be part of the product
            if (level == k)
                product = tree[i] - '0';

            // Recur for Left Subtree
            ++i;
            int leftproduct = productAtKthLevel(
                            tree, k, level + 1);

            // Recur for Right Subtree
            ++i;
            int rightproduct =
            productAtKthLevel(tree, k, level + 1);

            // Taking care of ')' after
            // left and right subtree
            ++i;
            return product *
              leftproduct * rightproduct;
        }
        return int.MinValue;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        String tree = "(0(5(6()())(4()"
        +"(9()())))(7(1()())(3()())))";
        int k = 2;
        i = 0;
        Console.Write(productAtKthLevel(tree, k, 0));
    }
}

Javascript

<script>

// JavaScript implementation to find product
// of elements at k-th level
var i;
// Recursive Function to find product
// of elements at k-th level
function productAtKthLevel( tree, k, level){
    if (tree[i++] == '(') {
        // if subtree is null,
        // just like if root == null
        if (tree[i] == ')')
            return 1;
        var product = 1;
        // Consider only level k node
        // to be part of the product
        if (level == k)
            product = tree[i] - '0';
        // Recur for Left Subtree
        ++i;
        var leftproduct = productAtKthLevel(
                        tree, k, level + 1);
        // Recur for Right Subtree
        ++i;
        var rightproduct =
        productAtKthLevel(tree, k, level + 1);
        // Taking care of ')' after
        // left and right subtree
        ++i;
        return product *
          leftproduct * rightproduct;
    }
    return int.MinValue;
}
// Driver Code
var tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))";
var k = 2;
i = 0;
document.write(productAtKthLevel(tree, k, 0));


</script>

Output:

72

Time Complexity: O(N)
Auxiliary space: O(N), for recursive stack space.