Lowest Common Ancestor of a BST

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).”

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

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         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.

# 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
        """
        if p.val > q.val:
            # return self.lowestCommonAncestor(root, q, p)
            p, q = q, p
        if root in [None, p, q]:
            return root
        if q.val < root.val:
            return self.lowestCommonAncestor(root.left, p, q)
        elif p.val > root.val:
            return self.lowestCommonAncestor(root.right, p, q)
        else:
            return root
            # left = self.lowestCommonAncestor(root.left, p, q)
            # right = self.lowestCommonAncestor(root.right, p, q)
            # if left and right:
            #     return root
            # elif left:
            #     return left
            # else:
            #     return right