Leetcode 285: 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.

Note: If the given node has no in-order successor in the tree, return null.

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public int val;
 5  *     public TreeNode left;
 6  *     public TreeNode right;
 7  *     public TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public TreeNode InorderSuccessor(TreeNode root, TreeNode p) {
12         // clarify questions: 1. what if P is null? 2. what if P doesn‘t exist in the tree? 3. what if p is the last node in inorder                 // traverse?
13         // idea: since it‘s a BST, we need to use the feature of BST
14
15         if (root == null || p == null) return null;
16
17         if (root.val <= p.val)
18         {
19             return InorderSuccessor(root.right, p);
20         }
21         else
22         {
23             var left = InorderSuccessor(root.left, p);
24
25             return left == null ? root : left;
26         }
27     }
28 }

原文地址：https://www.cnblogs.com/liangmou/p/8370477.html