# 235 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: β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).β

Example 1:

``````      6
/   \
2     8
/  \   /  \
0    4 7    9
/ \
3   5

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:

``````      6
/   \
2     8
/  \   /  \
0    4 7    9
/ \
3   5

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.
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 `````` ``````# 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': while (root.val - p.val) * (root.val - q.val) > 0: root = root.left if p.val < root.val else root.right return root``````