[Leetcode] 0235. Lowest Common Ancestor of a Binary Search Tree

/ Leetcode, Technical Article / By

Explore diverse LeetCode solutions in Python, C++, JavaScript, SQL, and TypeScript. Ideal for interview prep, learning, and code practice in multiple programming languages. Github Repo Link

Table of contents

Medium

Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

 

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.

Python

				
					# 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

related articles