## 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`.

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
``````

Thoughts: Utilize the binary search property of BST: during the traversal: if a node is about to traverse to its left. Then record the node as successor before doing so.

Code: T: O(h) -> O(logn)

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
:type p: TreeNode
:rtype: TreeNode
"""
successor = None
while root:
if p.val < root.val:
successor = root
root = root.left
else:
root = root.right
return successor
``````

C++

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
TreeNode* suc = NULL;
while(root ){
if(root-> val > p ->val) {
suc = root;
root = root -> left;
}
else root = root -> right;
}

return suc;
}
};
``````

Java

``````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;
}
``````