Day 17 of 100 days dsa coding challenge

Taking on a new challenge: solving GeeksforGeeks POTD daily and sharing my solutions! 💻🔥
The goal: sharpen problem-solving skills, level up coding, and learn something new every day. Follow my journey! 🚀

100DaysOfCode #CodingChallenge #ProblemSolving #GeeksforGeeks #DeveloperJourney

Problem:
https://www.geeksforgeeks.org/problems/bst-to-greater-sum-tree/1

Median of BST

Difficulty: Medium Accuracy: 27.43%

You are given the root of a Binary Search Tree, find the median of it.
Let the nodes of the BST, when written in ascending order (inorder traversal), be represented as V1, V2, V3, …, Vn, where n is the total number of nodes in the BST.
• If number of nodes are even: return V(n/2)
• If number of nodes are odd: return V((n+1)/2)
Examples:
Input: root = [20, 8, 22, 4, 12, N, N, N, N, 10, 14]

Output: 12
Explanation: The inorder of given BST is 4, 8, 10, 12, 14, 20, 22. Here, n = 7, so, here median will be ((7+1)/2)th value, i.e., 4th value, i.e, 12.

Input: root = [5, 4, 8, 1]

Output: 4
Explanation: The inorder of given BST is 1, 4, 5, 8. Here, n = 4(even), so, here median will be (4/2)th value, i.e., 2nd value, i.e, 4.

Constraints:
1 ≤ number of nodes ≤ 105
1 ≤ node.data ≤ 105

Solution:
class Solution:
def findMedian(self, root):
def inorder(node):
if not node:
return []
return inorder(node.left) + [node.data] + inorder(node.right)

    vals = inorder(root)
    n = len(vals)
    if n % 2 == 1:
        return vals[n // 2]
    else:
        return vals[(n // 2) - 1]