Day-20_Kth_smallest_element_in_a_BST.py

'''
	Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.

Note:
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

Example 1:

Input: root = [3,1,4,null,2], k = 1
   3
  / \
 1   4
  \
   2
Output: 1
Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 3
       5
      / \
     3   6
    / \
   2   4
  /
 1
Output: 3
Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?

   Hide Hint #1  
Try to utilize the property of a BST.
   Hide Hint #2  
Try in-order traversal. (Credits to @chan13)
   Hide Hint #3  
What if you could modify the BST node's structure?
   Hide Hint #4  
The optimal runtime complexity is O(height of BST).

'''

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def kthSmallest(self, root, k):
        def inorder(root):
            return inorder(root.left) + [root.val] + inorder(root.right) if root else []
        return inorder(root)[k - 1]