---
link: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
tags: tree, bst, easy
---

# 235. Lowest Common Ancestor of a Binary Search Tree

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

According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): “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**).”

Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]


**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.

**Note:**

- All of the nodes' values will be unique.
- p and q are different and both values will exist in the BST.

## Solution: Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        self.lca = None
        self.helper(root, p, q)
        return self.lca
    
    def helper(self, node, p, q):
        if node is None:
            return
        
        if p.val < node.val and q.val < node.val:
            self.helper(node.left, p, q)
        elif p.val > node.val and q.val > node.val:
            self.helper(node.right, p, q)
        else:
            self.lca = node
        
    def helper_p_q_may_not_exist(self, node, p, q):
        if node is None:
            return False, False
        
        l_found_p, l_found_q = (
            self.helper(node.left, p, q)
            if (p.val < node.val or q.val < node.val) else
            (False, False)
        )
        r_found_p, r_found_q = (
            self.helper(node.right, p, q)
            if (p.val > node.val or q.val > node.val) else
            (False, False)
        )
        n_found_p, n_found_q = node == p, node == q
        
        found_p = (l_found_p or r_found_p or n_found_p)
        found_q = (l_found_q or r_found_q or n_found_q)
                
        if found_p and found_q:
            self.lca = node
            return False, False
        
        return found_p, found_q

## Solution: Java