[Swift]LeetCode285. 二叉搜索树中的中序后继节点 $ Inorder Successor in BST

Given a binary search tree and a node in it, find the in-order successor of that node in the BST.

The successor of a node p is the node with the smallest key greater than p.val.

Example 1:

Input: root = [2,1,3], p = 1
Output: 2
Explanation: 1‘s in-order successor node is 2. Note that both p and the return value is of TreeNode type.

Example 2:

Input: root = [5,3,6,2,4,null,null,1], p = 6
Output: null
Explanation: There is no in-order successor of the current node, so the answer is null.

Note:

1. If the given node has no in-order successor in the tree, return null.
2. It‘s guaranteed that the values of the tree are unique.

节点p的后续节点是键最小大于p.val的节点。

例 1:

输入: root = [2,1,3], p = 1
输出: 2
说明：1的顺序继承节点为2。请注意，p和返回值都是treenode类型。

例  2:

输入: root = [5,3,6,2,4,null,null,1], p = 6
输出: null说明：当前节点没有按顺序的后续节点，因此答案为空。

注：

1. 如果给定的节点在树中没有顺序继承者，则返回空。
2. 它保证了树的值是唯一的。

 1 public class TreeNode {
 2     public var val: Int
 3     public var left: TreeNode?
 4     public var right: TreeNode?
 5     public init(_ val: Int) {
 6         self.val = val
 7         self.left = nil
 8         self.right = nil
 9     }
10 }
11
12 class Solution {
13     func inorderSuccessor(_ root: TreeNode?,_ p: TreeNode?) -> TreeNode? {
14         var root = root
15         var res:TreeNode? = nil
16         while(root != nil)
17         {
18             if root!.val > p!.val
19             {
20                 res = root
21                 root = root?.left
22             }
23             else
24             {
25                 root = root?.right
26             }
27         }
28         return res
29     }
30 }

原文地址：https://www.cnblogs.com/strengthen/p/10686067.html