Leetcode #450: Delete Node in a BST
In this guide, we solve Leetcode #450 Delete Node in a 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 a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Binary Search Tree, Binary Tree
Intuition
The input is a tree, so recursive decomposition is a natural fit.
We can compute the answer by combining results from left and right subtrees.
Approach
Use DFS and pass the required state through recursive calls.
Combine child results to compute the answer for each node.
Steps:
- Pick traversal order.
- Recurse with state.
- Combine results from children.
Example
Input: root = [5,3,6,2,4,null,7], key = 3
Output: [5,4,6,2,null,null,7]
Explanation: Given key to delete is 3. So we find the node with value 3 and delete it.
One valid answer is [5,4,6,2,null,null,7], shown in the above BST.
Please notice that another valid answer is [5,2,6,null,4,null,7] and it's also accepted.
Python Solution
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
if root is None:
return None
if root.val > key:
root.left = self.deleteNode(root.left, key)
return root
if root.val < key:
root.right = self.deleteNode(root.right, key)
return root
if root.left is None:
return root.right
if root.right is None:
return root.left
node = root.right
while node.left:
node = node.left
node.left = root.left
root = root.right
return root
Complexity
The time complexity is O(n). The space complexity is O(h).
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.