What is the brute force approach and complexity for Delete Node in a BST?

Inorder list of the same node objects, drop the key's node, then relink them into a balanced BST with time complexity O(N) and space complexity O(N).

How do you identify the pattern for Delete Node in a BST?

Look for key signals in the prompt: "delete a node from a BST" indicates Find it with the BST rule, then fix the links so the order still holds.; "two children" indicates One value must replace the node: the inorder successor (or the predecessor).; "return the (possibly updated) root" indicates The parent must store whatever the recursive call returns..

Delete Node in a BST

Medium·LeetCode #450 Pro OnlyMedium

LeetCode

Problem Statement

You are given the root of a binary search tree and a key.

Delete the node that holds key, then return the root of the tree. The tree must still be a valid BST. If key is not in the tree, the values stay the same. When the deleted node has two children, more than one shape can be correct, so any valid answer is accepted.

Input

The root of a binary search tree (between 0 and 10^4 nodes, all values unique), and an integer key.

Output

The root of a valid BST that holds the same values as before, minus key.

Example 1

Input: root = [5, 3, 6, 2, 4, null, 7], key = 3

Output: [5, 4, 6, 2, null, null, 7]

The node 3 has two children. Its inorder successor is 4, the smallest value bigger than 3. Copy 4 into that node, then remove the old 4. [5, 2, 6, null, 4, null, 7] is also accepted.

Example 2

Input: root = [5, 3, 6, 2, 4, null, 7], key = 0

Output: [5, 3, 6, 2, 4, null, 7]

No node holds 0, so nothing changes.

Example 3

Input: root = [], key = 0

Output: []

An empty tree stays empty.

Constraints

• The number of nodes is between 0 and 10^4
• −10^5 ≤ Node.val ≤ 10^5
• Each node value is unique
• root is a valid binary search tree
• −10^5 ≤ key ≤ 10^5
• Follow-up: can you do it in O(height of the tree) time?

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“delete a node from a BST”

→ Find it with the BST rule, then fix the links so the order still holds.

“two children”

→ One value must replace the node: the inorder successor (or the predecessor).

“return the (possibly updated) root”

→ The parent must store whatever the recursive call returns.

Approach Complexity Ladder

Brute Force — collect the nodes, drop one, rebuild a balanced treeO(N)O(N)

Inorder list of the same node objects, drop the key's node, then relink them into a balanced BST

Optimal — walk to the key, then fix one pathO(H)O(H)

Find the key by comparing, then handle 0, 1 or 2 children; with 2 children, copy the successor up and delete it below

Brute Force Intuition

Walk the tree in order to collect the nodes. Remove the node whose value is key. Then rebuild a balanced BST from the sorted nodes that are left, changing only their child pointers. No new nodes are created, but every node is relinked, even the ones far from key.

Recognition rule for this sub-pattern:

Left < node < right. Inorder is sorted; a valid range (low, high) is passed down.

Remember this · Tree DFS

Trigger

Paths, heights, diameters, validating structure, ancestors, anything "for every subtree".

Think

If my left and right children already gave me their answers, how do I build mine?

Key question

What does my recursive function return, and what do I pass down?

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