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++ 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 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 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# 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 |
<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> |
14
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++ 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 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 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# 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 |
<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> |
14
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).