Sum of nodes at k-th level in a tree represented as string

Last Updated : 20 Dec, 2022

Given an integer ‘K’ and a binary tree in string format. Every node of a tree has a value in the range of 0 to 9. We need to find the sum of elements at the K-th level from the 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

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

Implementation:

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 
        else if (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

Javascript

<script>
 
// Javascript implementation to find sum of 
// digits of elements at k-th level 
     
// Function to find sum of digits 
// of elements at k-th level 
function sumAtKthLevel(tree, k)
{
    let level = -1; 
     
    // Initialize result 
    let sum = 0; 
    let n = tree.length; 
     
    for(let 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
let tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))"; 
let k = 2; 
 
document.write(sumAtKthLevel(tree, k)); 
 
// This code is contributed by avanitrachhadiya2155
 
</script>

Output

Time Complexity: O(n)
Auxiliary Space: O(1)

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 that are at level k only.

Following is the implementation of the same:

C++

// C++ implementation to find sum of 
// digits of elements at k-th level 
#include <bits/stdc++.h> 
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; 
}

Java

// Java implementation to find sum of 
// digits of elements at k-th level 
class GFG 
{
    static int i;
 
    // Recursive Function to find sum of digits
    // of elements at k-th level
    static int sumAtKthLevel(String tree, int k, int level)
    {
 
        if (tree.charAt(i++) == '(')
        {
 
            // if subtree is null, just like if root == null
            if (tree.charAt(i) == ')')
                return 0;
 
            int sum = 0;
 
            // Consider only level k node to be part of the sum
            if (level == k)
                sum = tree.charAt(i) - '0';
 
            // Recur for Left Subtree
            ++i;
            int leftsum = sumAtKthLevel(tree, k, level + 1);
 
            // Recur for Right Subtree
            ++i;
            int rightsum = sumAtKthLevel(tree, k, level + 1);
 
            // Taking care of ')' after left and right subtree
            ++i;
            return sum + leftsum + rightsum;
        }
        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(sumAtKthLevel(tree, k, 0));
    }
}
 
// This code is contributed by 29AjayKumar

Python

# Python implementation to find sum of 
# digits of elements at k-th level 
 
# Recursive Function to find sum of digits 
# of elements at k-th level 
def sumAtKthLevel(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 0           
         
        sum = 0
         
        # Consider only level k node to be part of the sum 
        if(level == k):
            sum = int(tree[i[0]])
             
        # Recur for Left Subtree 
        i[0] += 1
        leftsum = sumAtKthLevel(tree, k, i, level + 1) 
             
        # Recur for Right Subtree 
        i[0] += 1
        rightsum = sumAtKthLevel(tree, k, i, level + 1) 
             
        # Taking care of ')' after left and right subtree 
        i[0] += 1
        return sum + leftsum + rightsum     
     
# Driver program to test above 
tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))"
k = 2
i = [0] 
print(sumAtKthLevel(tree, k, i, 0)) 
 
# This code is contributed by SHUBHAMSINGH10

C#

// C# implementation to find sum of 
// digits of elements at k-th level 
using System;
 
class GFG 
{
    static int i;
 
    // Recursive Function to find sum of digits
    // of elements at k-th level
    static int sumAtKthLevel(String tree, int k, 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
            ++i;
            int leftsum = sumAtKthLevel(tree, k, level + 1);
 
            // Recur for Right Subtree
            ++i;
            int rightsum = sumAtKthLevel(tree, k, level + 1);
 
            // Taking care of ')' after left and right subtree
            ++i;
            return sum + leftsum + rightsum;
        }
        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(sumAtKthLevel(tree, k, 0));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
// Javascript implementation to find sum of
// digits of elements at k-th level
     
    // Recursive Function to find sum of digits
    // of elements at k-th level
    function sumAtKthLevel(tree,k,level)
    {
        if (tree[i++] == '(')
        {
  
            // if subtree is null, just like if root == null
            if (tree[i] == ')')
                return 0;
  
            let sum = 0;
  
            // Consider only level k node to be part of the sum
            if (level == k)
                sum = tree[i] - '0';
  
            // Recur for Left Subtree
            ++i;
            let leftsum = sumAtKthLevel(tree, k, level + 1);
  
            // Recur for Right Subtree
            ++i;
            let rightsum = sumAtKthLevel(tree, k, level + 1);
  
            // Taking care of ')' after left and right subtree
            ++i;
            return sum + leftsum + rightsum;
        }
        return Number.MIN_VALUE;
    }
     
     // Driver code
    let tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))";
    let k = 2;
    let i = 0;
    document.write(sumAtKthLevel(tree, k, 0));
     
    // This code is contributed by rag2127
</script>

Output

Time Complexity: O(n), the time complexity of this algorithm is O(n) as we need to traverse all the nodes of the tree in order to get the sum of digits of elements at the kth level.
Auxiliary Space: O(1), If we consider the recursive call stack then it will be O(K).

A

Ayush Jauhari

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