Leetcode #285: Inorder Successor in BST
In this guide, we solve Leetcode #285 Inorder Successor in BST in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given the root of a binary search tree and a node p in it, return the in-order successor of that node in the BST. If the given node has no in-order successor in the tree, return null.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Tree, Depth-First Search, Binary Search Tree, Binary Tree
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: root = [2,1,3], p = 1
Output: 2
Explanation: 1's in-order successor node is 2. Note that both p and the return value is of TreeNode type.
Python Solution
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def inorderSuccessor(self, root: TreeNode, p: TreeNode) -> Optional[TreeNode]:
ans = None
while root:
if root.val > p.val:
ans = root
root = root.left
else:
root = root.right
return ans
Complexity
The time complexity is , where is the height of the binary search tree. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.