"""

Kth Smallest Element in a BST

Given the root of a binary search tree, and an integer k, return the kth smallest value (1-indexed) of all the values of the nodes in the tree.

Example 1:

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

Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 3
Output: 3

Constraints:

The number of nodes in the tree is n.
1 <= k <= n <= 104
0 <= Node.val <= 104

Follow up: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?

"""

# 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):
        """
        :type root: TreeNode
        :type k: int
        :rtype: int
        """
        # Solution 1: Inorder Traversal
        # Time Complexity: O(n)
        # Space Complexity: O(n)
        if not root:
            return None
        stack = []
        while root or stack:
            while root:
                stack.append(root)
                root = root.left
            root = stack.pop()
            k -= 1
            if k == 0:
                return root.val
            root = root.right

        # Solution 2: Binary Search
        # Time Complexity: O(n)
        # Space Complexity: O(n)
        if not root:
            return None
        stack = []
        while root or stack:
            while root:
                stack.append(root)
                root = root.left
            root = stack.pop()
            k -= 1
            if k == 0:
                return root.val
            root = root.right

        # Solution 3: Binary Search
        # Time Complexity: O(logn)
        # Space Complexity: O(1)
        if not root:
            return None
        stack = []
        while root or stack:
            while root:
                stack.append(root)
                root = root.left
            root = stack.pop()
            k -= 1
            if k == 0:
                return root.val
            root = root.right
