[Swift]LeetCode235. 二叉搜索树的最近公共祖先 | Lowest Common Ancestor of a Binary Search Tree

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 = [6,2,8,4,7,9,null,3,5]

 

Example 1:

Input: root = [6,5],p = 2,q = 8
Output: 6
Explanation: The LCA of nodes  and  is .
286

Example 2:

Input: root = [6,q = 4
Output: 2
Explanation: The LCA of nodes  and  is,since a node can be a descendant of itself according to the LCA deFinition.
242

 

Note:

• All of the nodes‘ values will be unique.
• p and q are different and both values will exist in the BST.

给定一个二叉搜索树,找到该树中两个指定节点的最近公共祖先。

百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”

例如，给定如下二叉搜索树:  root = [6,5]

 

示例 1:

输入: root = [6,q = 8
输出: 6 
解释: 节点 和节点 的最近公共祖先是 
286。

示例 2:

输入: root = [6,q = 4
输出: 2
解释: 节点  和节点  的最近公共祖先是,因为根据定义最近公共祖先节点可以为节点本身。242

 

说明:

• 所有节点的值都是唯一的。
• p、q 为不同节点且均存在于给定的二叉搜索树中。

递归

 1 /**
 2  * DeFinition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func lowestCommonAncestor(_ root: TreeNode?,_ p: TreeNode?,_ q: TreeNode?) -> TreeNode? {
16         if root == nil {return nil}
17         if root!.val > max(p!.val,q!.val)
18         {
19             return lowestCommonAncestor(root!.left,p,q)
20         }
21         else if root!.val < min(p!.val,q!.val)
22         {
23             return lowestCommonAncestor(root!.right,q)
24         }
25         else 
26         {
27             return root
28         }
29     }
30 }