# Leetcode: Inorder Successor in BST

Inorder Successor in BST

Similar Problems:

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
```

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

• Solution:

Intuitive Idea would have too many details.

```## If p has right child, we can easily find the target in its sub-tree
## Otherwise: Use DFS to find all parents of p, from root down to p
##    If p is the left-subtree of q, q is the target
##    Otherwise keep looking up. If we get the root, return null
##
```
```## Blog link: https://code.dennyzhang.com/inorder-successor-in-bst
## Basic Ideas:
##
## Complexity:
# 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
"""
if p is None or root is None: return None
res = None
while root:
if root.val > p.val:
# go to left
res = root
root = root.left
else:
# go to right
root = root.right
return res
```

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