# 235. Lowest Common Ancestor of a Binary Search Tree

# Problem

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

# Solution

2 important things to note:

- We start at the root, because an ancestor is always higher so must always start at the root.
- it's a binary search tree.

If we're looking to find `p = 2`

, it'll be in the **left subtree** because 2 is less than 6 (binary search tree property).

```
# 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
"""
curr = root
# p and q are gurnateed to exist in the tree
# makes it run forever until we find the result
while curr:
# if p and q greater than root, it's right subtree
if p.val > curr.val and q.val > curr.val:
curr = curr.right
elif p.val < curr.val and q.val < curr.val:
# elif its left subtree
curr = curr.left
else:
# we found the split where we have found our result
return curr
# no return stateemnt outside of loop as it's guranteed to exist
# see the constraints
# > p and q will exist in the BST.
```