---
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'
---
Binary Search Tree
### 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.
## Approach
```mermaid
flowchart TD
S(["lowestCommonAncestor(root, p, q)"]) --> I["cur = root"]
I --> W{"cur?"}
W -- no --> E(["fall off the loop — unreachable, p and q exist in the BST"])
W -- yes --> A{"p.val < cur.val and q.val < cur.val?"}
A -- yes --> L["cur = cur.left — both sit in the left subtree"]
L --> W
A -- no --> B{"p.val > cur.val and q.val > cur.val?"}
B -- yes --> R["cur = cur.right — both sit in the right subtree"]
R --> W
B -- no --> C(["return cur — the paths to p and q split here"])
```
## 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
```
## Explanation