How do you solve Validate Binary Search Tree on LeetCode?

Validate Binary Search Tree (LeetCode #98) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(h) space complexity. Key insight: A valid BST must have all left descendants less than the node and all right descendants greater.

What is the brute force solution for Validate Binary Search Tree?

The brute force approach for Validate Binary Search Tree is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach checks every node in the tree to see if it adheres to the properties of a binary search tree (BST). This involves traversing the entire tree and comparing each node's value with all other nodes, which can be inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Validate Binary Search Tree?

Always clarify the definition of a BST before starting to code. Think about edge cases and how they affect your logic. Practice writing both recursive and iterative solutions to improve your flexibility.

What are common mistakes when solving Validate Binary Search Tree?

Failing to consider edge cases like duplicates or single-node trees. Not properly managing the min and max bounds during recursion.

#98

Validate Binary Search Tree

Medium

TreeDepth-First SearchBinary Search TreeBinary TreeTree TraversalRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(h)

💡

Intuition

Time O(n)Space O(h)

The optimal approach uses a recursive function that checks each node's value against a range defined by its ancestors. This avoids the need to store all values and checks the BST properties in a single pass.

⚙️

Algorithm

3 steps

1Step 1: Define a helper function that takes a node and a range (min and max values).
2Step 2: For each node, check if its value is within the range.
3Step 3: Recursively check the left subtree with updated max and the right subtree with updated min.

solution.py15 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8def isValidBST(root):
9    def validate(node, low=float('-inf'), high=float('inf')):
10        if not node:
11            return True
12        if not (low < node.val < high):
13            return False
14        return validate(node.left, low, node.val) and validate(node.right, node.val, high)
15    return validate(root)

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(h) due to the recursion stack, where h is the height of the tree.

1A valid BST must have all left descendants less than the node and all right descendants greater.
2Recursive validation can be more efficient than storing values for comparison.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.