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

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

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

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++

// 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; 
} 

Java

// 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));  
} 
}  

Python3

# 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) 

C#

// 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 

Output:

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)

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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

C++

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Python3

C#

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