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.

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.

1、题目描述¶

给你一个二叉搜索树和其中的某一个结点，请你找出该结点在树中顺序后继的节点。

结点 p 的后继是值比 p.val 大的结点中键值最小的结点。

示例 1:

输入: root = [2,1,3], p = 1
输出: 2
解析: 这里 1 的顺序后继是 2。请注意 p 和返回值都应是 TreeNode 类型。

示例 2:

输入: root = [5,3,6,2,4,null,null,1], p = 6
输出: null
解析: 因为给出的结点没有顺序后继，所以答案就返回 null 了。

注意:

• 假如给出的结点在该树中没有顺序后继的话，请返回 null
• 我们保证树中每个结点的值是唯一的