What is the brute force approach and complexity for Validate Binary Search Tree?

For every node, scan its left subtree for the max and its right subtree for the min with time complexity O(N²) and space complexity O(H).

What is the optimal approach and complexity for Validate Binary Search Tree?

Each node must sit strictly between a low and a high bound set by its ancestors with optimal time complexity O(N) and space complexity O(H).

How do you identify the pattern for Validate Binary Search Tree?

Look for key signals in the prompt: "left < node < right" indicates A rule for each node that covers its whole subtree, not only the children.; "strictly bigger / strictly smaller" indicates Equal values fail, so both bounds are strict.; "is this tree valid" indicates One bad node anywhere makes the answer false..

Validate Binary Search Tree

Medium·LeetCode #98 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

You are given the root of a binary tree. Decide whether it is a valid binary search tree.

Every node must be strictly bigger than every value in its left subtree, and strictly smaller than every value in its right subtree. The rule covers the whole subtree, not only the direct children. Equal values are not allowed.

Input

The root of a binary tree (between 1 and 10^4 nodes). Values can be any 32-bit int, including the smallest and largest int.

Output

true if the tree is a valid binary search tree, otherwise false.

Example 1

Input: root = [2, 1, 3]

Output: true

1 is smaller than 2, and 3 is bigger than 2.

Example 2

Input: root = [5, 1, 4, null, null, 3, 6]

Output: false

The root is 5, but its right child is 4, which is not bigger than 5.

Constraints

• The number of nodes is between 1 and 10^4
• −2^31 ≤ Node.val ≤ 2^31 − 1

Pattern Recognition Clues

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

“left < node < right”

→ A rule for each node that covers its whole subtree, not only the children.

“strictly bigger / strictly smaller”

→ Equal values fail, so both bounds are strict.

“is this tree valid”

→ One bad node anywhere makes the answer false.

Approach Complexity Ladder

Brute Force — scan each node's whole subtreesO(N²)O(H)

For every node, scan its left subtree for the max and its right subtree for the min

Better — the inorder values must strictly increaseO(N)O(N)

Write the inorder walk into a list, then check that the list strictly increases

Optimal — pass a (low, high) range down the treeO(N)O(H)

Each node must sit strictly between a low and a high bound set by its ancestors

Brute Force Intuition

For each node, find the biggest value in its left subtree and the smallest value in its right subtree, and check both against the node. Do this for every node. Each check scans a whole subtree, so the same values are scanned again and again.

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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