Sum 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 sum of elements at K-th level from root. The root is at level 0.
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 : 14
Its tree representation is shown below
TREE
Elements at level k = 2 are 6, 4, 1, 3
sum of the digits of these elements = 6+4+1+3 = 14 


Input : tree = "(8(3(2()())(6(5()())()))(5(10()())(7(13()())())))" 
        k = 3
Output : 9
Elements at level k = 3 are 5, 1 and 3
sum of digits of these elements = 5+1+3 = 9

1. Input 'tree' in string format and level k
2. Initialize level = -1 and sum = 0
3. for each character 'ch' in 'tree'
   3.1  if ch == '(' then
        --> level++
   3.2  else if ch == ')' then
        --> level--
   3.3  else
        if level == k then
           sum = sum + (ch-'0')
4. Print sum

C++

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// C++ implementation to find sum of
// digits of elements at k-th level
#include <bits/stdc++.h>
using namespace std;
  
// Function to find sum of digits
// of elements at k-th level
int sumAtKthLevel(string tree, int k)
{
    int level = -1;
    int sum = 0;  // Initialize result
    int n = tree.length();
  
    for (int i=0; i<n; i++)
    {
        // increasing level number
        if (tree[i] == '(')
            level++;
  
        // decreasing level number
        else if (tree[i] == ')')
            level--;
  
        else
        {
            // check if current level is
            // the desired level or not
            if (level == k)
                sum += (tree[i]-'0');
        }
    }
  
    // required sum
    return sum;
}
  
// Driver program to test above
int main()
{
    string tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))";
    int k = 2;
    cout << sumAtKthLevel(tree, k);
    return 0;
}

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Java

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// Java implementation to find sum of 
// digits of elements at k-th level 
class GfG { 
  
// Function to find sum of digits 
// of elements at k-th level 
static int sumAtKthLevel(String tree, int k) 
    int level = -1
    int sum = 0; // Initialize result 
    int n = tree.length(); 
  
    for (int i=0; i<n; i++) 
    
        // increasing level number 
        if (tree.charAt(i) == '('
            level++; 
  
        // decreasing level number 
        else if (tree.charAt(i) == ')'
            level--; 
  
        else
        
            // check if current level is 
            // the desired level or not 
            if (level == k) 
                sum += (tree.charAt(i)-'0'); 
        
    
  
    // required sum 
    return sum; 
  
// Driver program to test above 
public static void main(String[] args) 
    String tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))"
    int k = 2
    System.out.println(sumAtKthLevel(tree, k)); 
}

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Python3

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# Python3 implementation to find sum of 
# digits of elements at k-th level 
  
# Function to find sum of digits 
# of elements at k-th level 
def sumAtKthLevel(tree, k) :
  
    level = -1
    sum = 0 # Initialize result 
    n = len(tree) 
  
    for i in range(n):
          
        # increasing level number 
        if (tree[i] == '(') :
            level += 1
  
        # decreasing level number 
        elif (tree[i] == ')'): 
            level -= 1
  
        else:
          
            # check if current level is 
            # the desired level or not 
            if (level == k) :
                sum += (ord(tree[i]) - ord('0')) 
          
    # required sum 
    return sum
  
# Driver Code 
if __name__ == '__main__':
    tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))"
    k = 2
    print(sumAtKthLevel(tree, k))
  
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

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C#

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// C# implementation to find sum of 
// digits of elements at k-th level 
  
using System;
class GfG { 
  
// Function to find sum of digits 
// of elements at k-th level 
static int sumAtKthLevel(string tree, int k) 
    int level = -1; 
    int sum = 0; // Initialize result 
    int n = tree.Length; 
  
    for (int i = 0; i < n; i++) 
    
        // increasing level number 
        if (tree[i] == '('
            level++; 
  
        // decreasing level number 
        else if (tree[i] == ')'
            level--; 
  
        else
        
            // check if current level is 
            // the desired level or not 
            if (level == k) 
                sum += (tree[i]-'0'); 
        
    
  
    // required sum 
    return sum; 
  
// Driver code
public static void Main() 
    string tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))"
    int k = 2; 
    Console.Write(sumAtKthLevel(tree, k)); 
}
  
// This code is contributed by Ita_c

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Output:

14

Time Complexity: O(n)

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

Following is the C++ implementation of the same:

// C++ implementation to find sum of
// digits of elements at k-th level
#include
using namespace std;

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

if(tree[i++]=='(‘)
{

// if subtree is null, just like if root == NULL
if(tree[i] == ‘)’)
return 0;

int sum=0;

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

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

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

// Taking care of ‘)’ after left and right subtree
++i;
return sum+leftsum+rightsum;
}
}

// Driver program to test above
int main()
{
string tree = “(0(5(6()())(4()(9()())))(7(1()())(3()())))”;
int k = 2;
int i=0;
cout << sumAtKthLevel(tree, k,i,0); return 0; } Time Complexity: O(n)

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