# Question

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

The in-order successor of a node is the left-most node in its right sub-tree, if exists. If not, then it is the root of the smallest left sub-tree this node is in.

```
# 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
:type p: TreeNode
:rtype: TreeNode
"""
successor = None
if p.right is not None:
node = p.right
successor = node
while node.left is not None:
node = node.left
successor = node
return successor
else:
node = root
successor = None
while True:
if node is None:
return None
if node == p:
break
elif p.val < node.val:
successor = node
node = node.left
else:
node = node.right
return successor
```