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[235] Lowest Common Ancestor of a Binary Search Tree

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/

algorithms
Easy (42.17%)
Source Code: 235.lowest-common-ancestor-of-a-binary-search-tree.py
Total Accepted: 271.1K
Total Submissions: 614.2K
Testcase Example: '[6,2,8,0,4,7,9,null,null,3,5]\n2\n8'

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]

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.

python

# 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 not root: return None

        if root.val == p.val or root.val == q.val: return root
        elif p.val > root.val > q.val or q.val > root.val > p.val: return root
        elif p.val > root.val and q.val > root.val:
            return self.lowestCommonAncestor(root.right, p, q)
        else:
            return self.lowestCommonAncestor(root.left, p, q)