---
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
icon: dot
topics:
  - { name: "Tree", slug: "tree" }
  - { name: "Depth-First Search", slug: "depth-first-search" }
  - { name: "Binary Search Tree", slug: "binary-search-tree" }
  - { name: "Binary Tree", slug: "binary-tree" }
  - { name: "Binary Lifting", slug: "binary-lifting" }
  - { name: "Lowest Common Ancestor", slug: "lowest-common-ancestor" }
video: "gs2LMfuOR9k"
issue: "https://github.com/prdlk/leetcode/issues/113"
sidebar:
  label: 'Lowest Common Ancestor of a Binary Search Tree'
  badge: 'Medium'
---

### 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
```