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# 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`.

**Example 1:**

```
Input: root = [2,1,3], p = 1

  2
 / \
1   3

Output: 2
```

**Example 2:**

```
Input: root = [5,3,6,2,4,null,null,1], p = 6

      5
     / \
    3   6
   / \
  2   4
 /   
1

Output: null
```

## Solution

```
/**
 * 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) {
        TreeNode succ = null;
        while (root != null) {
            if (p.val < root.val) {
                succ = root;
                root = root.left;
            } else {
                root = root.right;
            }
        }
        return succ;
    }
}
```