235. Lowest Common Ancestor of a Binary Search Tree

Edited by Charles

Medium · Python · Binary Tree

Problem

Read the original problem on LeetCode.

The official page contains the problem statement, examples, and constraints.

Explanation, examples & other approaches

Core idea

Recursion:

Use the property of BST.

Original study notes follow, including intermediate snippets and earlier attempts.

Recursion:

Use the property of BST.

1. if p.val and q.val are less than the root.val, the nodes of p and q are on the left subtree.
2. if p.val and q.val are greater than the root.val, the nodes of p and q are on the right subtree.
3. all left cases are p.val ≤ root.val ≤ q.val or p.val ≥ root.val ≥ q.val.
4. the constraints show that the root, p and q are always valid.

# 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':
        if p.val < root.val and q.val < root.val:
            return self.lowestCommonAncestor(root.left, p, q)

        if p.val > root.val and q.val > root.val:
            return self.lowestCommonAncestor(root.right, p, q)

        return root

Iteration:

Same idea as the recursion solution.

Study record

Field	Value
Attempts	1
Mastery	❚
The Latest Attempt	January 18, 2026

Python solution

# 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

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