"""
235. Lowest Common Ancestor of a Binary Search Tree
Difficulty: Medium
https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

──────────────────────────────────────────────────

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



Constraints:

        • The number of nodes in the tree is in the range [2, 10^5].

        • -10^9 <= Node.val <= 10^9

        • All Node.val are unique.

        • p != q

        • p and q will exist in the BST.
"""

# 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":
        cur = root
        while cur:
            if p.val < cur.val and q.val < cur.val:
                cur = cur.left
            elif p.val > cur.val and q.val > cur.val:
                cur = cur.right
            else:
                return cur