Problem description:

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).”

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.
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.
Example 3:

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Input: root = [2,1], p = 2, q = 1
Output: 2

Solution:

Since BST is sorted, we can check if both p, q value is greater/smaller than root.val. If one of p, q is greater but the other one is smaller, then current root must be the LCA. Because p, q would be in its left and right subtree.

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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':
        while root:
            if root.val > p.val and root.val > q.val:
                root = root.left
            elif root.val < p.val and root.val < q.val:
                root = root.right
            else:
                break
        return root

time complexity: $O(logn)$
space complexity: $O(1)$
reference:
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