Solution Explanation

The problem asks to find the k-th smallest element in a Binary Search Tree (BST). We can solve this using two main approaches: iterative inorder traversal and a more optimized approach using the properties of a BST.

Approach 1: Iterative Inorder Traversal

This approach leverages the property of inorder traversal in a BST: Inorder traversal visits nodes in ascending order. We use an iterative approach (using a stack) to perform inorder traversal efficiently. We stop the traversal when we've encountered the k-th smallest element.

Algorithm:

1. Initialize an empty stack stk.
2. While the current node root is not NULL or the stack is not empty:
   • If root is not NULL: push root onto the stack and move to its left child (root = root.left).
   • Otherwise (if root is NULL): pop the top node from the stack. Decrement k. If k becomes 0, the popped node's value is the k-th smallest element, so return its value. Then, move to the right child of the popped node (root = root.right).

Code (Python):

class Solution:
    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
        stk = []
        while root or stk:
            if root:
                stk.append(root)
                root = root.left
            else:
                root = stk.pop()
                k -= 1
                if k == 0:
                    return root.val
                root = root.right

Time Complexity Analysis: O(H + k), where H is the height of the BST. In the worst case (a skewed tree), H can be equal to N (number of nodes), resulting in O(N) time complexity. However, for balanced BSTs, H is log(N), and the complexity becomes closer to O(log N + k). In the best case, the k-th smallest element might be near the root, leading to O(k) time. The k factor comes from the potential need to traverse up to k nodes during the inorder traversal.

Space Complexity Analysis: O(H) in the worst case (skewed tree), which becomes O(log N) for a balanced BST, due to the stack's size being proportional to the height of the tree.

Approach 2: Optimized BST Approach (Using Node Counts)

This approach is more sophisticated and leverages the inherent structure of a BST. We pre-compute the number of nodes in the subtree rooted at each node. Then, we can efficiently navigate the tree to find the k-th smallest element.

Algorithm:

1. Create a BST class that precomputes the node counts for each node in the tree.
2. Implement a kthSmallest method within the BST class to efficiently find the k-th smallest element using the precomputed node counts.

Code (Python):

from collections import Counter
 
class BST:
    def __init__(self, root):
        self.cnt = Counter()
        self.root = root
        self.count(root)
 
    def kthSmallest(self, k):
        node = self.root
        while node:
            if self.cnt[node.left] == k - 1:
                return node.val
            if self.cnt[node.left] < k - 1:
                k -= self.cnt[node.left] + 1
                node = node.right
            else:
                node = node.left
        return 0
 
    def count(self, root):
        if root is None:
            return 0
        n = 1 + self.count(root.left) + self.count(root.right)
        self.cnt[root] = n
        return n
 
 
class Solution:
    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
        bst = BST(root)
        return bst.kthSmallest(k)

Time Complexity Analysis: O(N) for pre-computation of node counts and O(log N) for finding the k-th smallest element (in a balanced tree). For a skewed tree, the search can still take O(N) time.

Space Complexity Analysis: O(N) to store the node counts in the cnt dictionary (hash map).

Conclusion:

Approach 1 (iterative inorder traversal) is generally simpler to implement. Approach 2 offers better performance if you need to repeatedly query for k-th smallest elements on the same BST. The choice depends on the specific application and whether repeated queries are expected. Both approaches are provided with examples in several programming languages.