docs(docs): add documentation for 235 Lowest Common Ancestor of a Binary Search Tree

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Prad Nukala
2026-09-05 16:32:59 -04:00
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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
```