# Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
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# 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).”
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# Example 1:
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# 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.
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# Example 2:
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# 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.
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# Example 3:
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# Input: root = [2,1], p = 2, q = 1
# Output: 2
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# Constraints:
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# 	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.
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#


# 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':
        # 二叉搜索树 
        if p.val < root.val and q.val < root.val:
            return self.lowestCommonAncestor(root.left, p, q)
        if p.val > root.val and q.val > root.val:
            return self.lowestCommonAncestor(root.right, p, q)
        return root
