7-235. Lowest Common Ancestor of a Binary Search Tree

题目描述：

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.

代码实现：

# 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':
        if p.val < root.val and q.val < root.val:
            return self.lowestCommonAncestor(root.left, p, q) # 在一个函数中调用自身时，必须要加self
        elif p.val > root.val and q.val > root.val:
            return self.lowestCommonAncestor(root.right, p, q)
        else:
            return root

分析：

二叉树和二叉搜索树的区别见博客：https://blog.csdn.net/u012292754/article/details/87474802

在二叉搜索树中，左子树上所有节点的值均小于它的根节点的值，右子树中所有节点的值均大于它的根节点的值。

这是一条非常有用的性质，相当于事先对二叉树做了一定的排序。

因此在二叉搜索树中，任意两个节点与根节点之间的位置关系只能有以下三种情况：

第一种情况：这两个节点都在根节点的左子树中（即p.val < root.val and q.val < root.val），此时只要继续深入根节点的左子树中查找即可；

第二种情况：这两个节点都在根节点的右子树中（即p.val > root.val and q.val > root.val），此时只要继续深入根节点的右子树中查找即可；

第三种情况：这两个节点有一个是根节点或这两个节点横跨在根节点的两侧（一个在左子树中，一个在右子树中），此时最近公共祖先必为根节点本身。

以上三种情况即为上述代码的实现思路。