From 6c18301aa66f176a32b31c9a42f61f089db523ff Mon Sep 17 00:00:00 2001
From: Prad Nukala <prad@sonr.io>
Date: Sat, 5 Sep 2026 16:33:03 -0400
Subject: [PATCH] feat(work): add Lowest Common Ancestor of BST solution

---
 ...common-ancestor-of-a-binary-search-tree.py | 70 +++++++++++++++++++
 1 file changed, 70 insertions(+)
 create mode 100644 work/1/Medium/Tree/235.lowest-common-ancestor-of-a-binary-search-tree.py

diff --git a/work/1/Medium/Tree/235.lowest-common-ancestor-of-a-binary-search-tree.py b/work/1/Medium/Tree/235.lowest-common-ancestor-of-a-binary-search-tree.py
new file mode 100644
index 0000000..591bb05
--- /dev/null
+++ b/work/1/Medium/Tree/235.lowest-common-ancestor-of-a-binary-search-tree.py
@@ -0,0 +1,70 @@
+"""
+235. Lowest Common Ancestor of a Binary Search Tree
+Difficulty: Medium
+https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
+
+──────────────────────────────────────────────────
+
+Given a binary search tree (BST), find the lowest common ancestor
+(LCA) node of two given nodes in the BST.
+
+According to the definition of LCA on Wikipedia: &ldquo;The lowest
+common ancestor is defined between two nodes p and q as the lowest
+node in T that has both p and q as descendants (where we allow a node
+to be a descendant of itself).&rdquo;
+
+
+
+Example 1:
+
+Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
+Output: 6
+Explanation: The LCA of nodes 2 and 8 is 6.
+
+Example 2:
+
+Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
+Output: 2
+Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a
+descendant of itself according to the LCA definition.
+
+Example 3:
+
+Input: root = [2,1], p = 2, q = 1
+Output: 2
+
+
+
+Constraints:
+
+        • The number of nodes in the tree is in the range [2, 10^5].
+
+        • -10^9 <= Node.val <= 10^9
+
+        • All Node.val are unique.
+
+        • p != q
+
+        • p and q will exist in the BST.
+"""
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+
+class Solution:
+    def lowestCommonAncestor(
+        self, root: "TreeNode", p: "TreeNode", q: "TreeNode"
+    ) -> "TreeNode":
+        cur = root
+        while cur:
+            if p.val < cur.val and q.val < cur.val:
+                cur = cur.left
+            elif p.val > cur.val and q.val > cur.val:
+                cur = cur.right
+            else:
+                return cur