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+---
+title: '235. Lowest Common Ancestor of a Binary Search Tree'
+description: Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST
+sidebar:
+  label: 'Lowest Common Ancestor of a Binary Search Tree'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Binary Search Tree</Badge>
+
+### 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.
+
+## Solution
+
+```py
+# 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
+```