# LeetCode 235. Lowest Common Ancestor of a Binary Search Tree

## Description

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

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: “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, 105]`.
• `-109 <= Node.val <= 109`
• All `Node.val` are unique.
• `p != q`
• `p` and `q` will exist in the BST.

## Explanation

If root is either p or q, LCA is root.

If LCA is in either subtree, return the LCA from that subtree.

If p and q are in different subtree, return the root.

## 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':

return self.helper(root, p, q)

def helper(self, root, p, q):
if not root:
return root

if root == p or root == q:
return root

left = self.helper(root.left, p, q)
right = self.helper(root.right, p, q)

if left and right:
return root

if left:
return left

if right:
return right

``````
• Time Complexity: O(N).
• Space Complexity: O(N).