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
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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 which are at level k only.
Below is the implementation of 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 levelint 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 Codeint 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 levelclass 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 leveldef 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 Codeif __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 levelusing 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 levelvar i;// Recursive Function to find product// of elements at k-th levelfunction 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 Codevar tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))";var k = 2;i = 0;document.write(productAtKthLevel(tree, k, 0));</script> |
72
Time Complexity: O(N)
