心路历程：

这道题可以完全按照二叉树的公共祖先来做，但是由于题目中给了二分搜索树的条件，因此可以通过值的大小简化左右子树的递归搜索。

解法一：按照二分搜索树的性质

# 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':
        def dfs(node):
            if not node or node == p or node == q:  return node
            if node.val > p.val and node.val > q.val:  return dfs(node.left)
            elif node.val < p.val and node.val < q.val: return dfs(node.right)
            else: return node
        return dfs(root)

解法二：依然用后序遍历搜索

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        def dfs(node):
            if not node or node == p or node == q: return node
            lv = dfs(node.left)
            rv = dfs(node.right)
            if lv and rv: return node
            elif lv and not rv: return lv
            elif not lv and rv: return rv
            else: return None
        return dfs(root)