solve: kth smallest element in a bst

Author: sounmind

kth-smallest-element-in-a-bst/evan.py

@@ -0,0 +1,40 @@
+# Definition for a binary tree node.
+from typing import Optional
+
+
+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:
+        result = []
+
+        def inorderTraverse(node: Optional[TreeNode]):
+            if node is None or len(result) >= k:
+                return
+
+            inorderTraverse(node.left)
+
+            if len(result) < k:
+                result.append(node.val)
+
+            inorderTraverse(node.right)
+
+        inorderTraverse(root)
+
+        return result[k - 1]
+
+
+# Time Complexity: O(N)
+# In the worst case, we need to visit all the nodes in the tree.
+# Thus, the time complexity is O(N), where N is the number of nodes in the tree.
+
+# Space Complexity: O(N)
+# The space complexity is determined by the recursion stack and the result list.
+# 1. Recursion stack: In the worst case (unbalanced tree), the recursion stack can go up to N levels deep, so the space complexity is O(N).
+#    In the best case (balanced tree), the recursion stack depth is log(N), so the space complexity is O(log N).
+# 2. Result list: The result list stores up to k elements, so the space complexity is O(k).