Leetcode 235. Lowest Common Ancestor of a Binary Search Tree

文章作者：Tyan
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1. Description

2. Solution

思路：由于树是有序的，因此先对p, q排序，p小于q，因此在树中p一定在于最低祖先结点的左子树（包含当前结点），q位于右子树（包含当前结点），比较p, q结点与当前结点的值，找到这样的结点即为最低最先结点。

• Version 1

# 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, p, q):
        if p.val > q.val:
            p, q = q, p

        if root.val >= p.val and root.val <= q.val:
            return root

        if root.val > q.val:
            return self.lowestCommonAncestor(root.left, p, q)

        if root.val < p.val:
            return self.lowestCommonAncestor(root.right, p, q)

Reference

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