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# 285.Inorder-Successor-in-BST

## 285. Inorder Successor in BST

## 题目地址

<https://leetcode.com/problems/inorder-successor-in-bst/>

## 题目描述

```
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:
If the given node has no in-order successor in the tree, return null.
It's guaranteed that the values of the tree are unique.
```

## 代码

### Approach 1: Iterative Inorder Traversal

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        if (p.right != null) {
        p = p.right;
        while (p.left != null) {
          p = p.left;
        }
        return p;
      }

      // the successor is somewhere upper in the tree
      ArrayDeque<TreeNode> stack = new ArrayDeque<>();
      int inorder = Integer.MIN_VALUE;

      // inorder traversal : left -> node -> right
      while (!stack.isEmpty() || root != null) {
        // 1. go left till you can
        while (root != null) {
          stack.push(root);
          root = root.left;
        }
        // 2. all logic around the node
        root = stack.pop();
        // if the previous node was equal to p
          // then the current node is its successor
        if (inorder == p.val) return root;
        inorder = root.val;

        // 3. go one step right
        root = root.right;
      }

      return null;
    }
}
```


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