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 belowElements 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++
// 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:
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 <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; } |
Time Complexity: O(n)
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