feat(work): add Kth Smallest Element in a BST solution

work/1/Medium/Tree/230.kth-smallest-element-in-a-bst.py

@@ -0,0 +1,63 @@
+"""
+230. Kth Smallest Element in a BST
+Difficulty: Medium
+https://leetcode.com/problems/kth-smallest-element-in-a-bst/
+
+──────────────────────────────────────────────────
+
+Given the root of a binary search tree, and an integer k, return the
+k^th 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 <= 10^4
+
+        • 0 <= Node.val <= 10^4
+
+
+
+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: Optional[TreeNode], k: int) -> int:
+        n = 0
+        stack = []
+        cur = root
+
+        while cur or stack:
+            while cur:
+                stack.append(cur)
+                cur = cur.left
+
+            cur = stack.pop()
+            n += 1
+            if n == k:
+                return cur.val
+            cur = cur.right