> For the complete documentation index, see [llms.txt](https://maksimdan.gitbook.io/interview-practice-problems/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://maksimdan.gitbook.io/interview-practice-problems/leetcode_sessions/285-inorder-successor-in-bst.md).

# 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`.

**The Idea:** The problem can be summed up into two cases. In the first case when `p` has a right child, then the next in order successor is the smallest value of that right subtree. Otherwise, we have to ascend down from the root of the tree and find `p` using plain old binary search. In this case, maintain a previous node that follows `root.left`, since this will be the the first largest node that is larger than root once `p` is found.

**Complexity:** O(logn) time and O(1) space

```python
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def inorderSuccessor(self, root, p):
        """
        :type root: TreeNode - the actual root of the tree
        :type p: TreeNode - find the next in order successor of this node
        :rtype: TreeNode
        """
        if not root or not p:
            return None

        # case 1
        if p.right:
            p = p.right
            while p.left:
                p = p.left
            return p

        # case 2
        prev = None
        while root != p:
            if p.val > root.val:
                root = root.right
            else:
                prev = root
                root = root.left
        return prev
```


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