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Medium

Validate Binary Search Tree

A medium BST problem included in Apna College, Love Babbar 450, Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.

Topic
BST
Sheets
3
Core for
2 roles
Platform
LeetCode
Practice in the compilerWatch this solved step by step

The problem

Given the root of a binary tree, determine if it is a valid binary search tree. A valid BST is defined as: the left subtree of a node contains only nodes with values strictly less than the node's value, and the right subtree contains only nodes with values strictly greater than the node's value. Both subtrees must also be BSTs.

Example 1

Input
[2,1,3]
Output
true
Why
Root 2 has left child 1 (less than 2) and right child 3 (greater than 2). Both subtrees are valid.

Example 2

Input
[5,1,4,null,null,3,6]
Output
false
Why
Root 5 has right child 4, but 4 is less than 5, violating the BST property.

Example 3

Input
[1]
Output
true
Why
A single node is always a valid BST.

Constraints

  • The number of nodes is in the range [1, 104].
  • -231 <= Node.val <= 231 - 1

How to think about it

Updated 2026-09-09

Checking only that each node is bigger than its left child and smaller than its right child is the classic trap: it lets invalid values slip deep into subtrees. The true BST condition is that every node must fit inside an inherited open interval (low, high) tightened by its ancestors, or equivalently, an inorder traversal must be strictly increasing.

Approaches, worst first

  1. Collect and check sorted

    time O(n) · space O(n)

    Run a full inorder traversal into an array, then scan to confirm strict ascending order. Clean and quick to verify, but it allocates heap memory for all nodes and cannot stop early on the first violation it encounters.

  2. Inorder walk with previous pointer

    time O(n) · space O(h)

    Traverse inorder iteratively or recursively while holding only the previous node's value. Return false the instant a value does not strictly exceed the predecessor. Prunes early and needs space only for the recursion stack.

  3. Recursive range narrowingWrite this one

    time O(n) · space O(h)

    Pass an allowed range (minVal, maxVal) down the call tree, starting with infinite bounds. Check the root against the current bounds, then branch left narrowing the upper limit and right narrowing the lower limit. It stops as soon as an anomaly is seen.

Where people lose marks · 3
  • Comparing only immediate children. A tree with root 10, right child 15, and 15 having left child 6 satisfies local child checks but violates the whole-tree invariant.
  • Allowing duplicate values. A strict BST requires left < node < right; checking `<=` mistakenly accepts duplicates.
  • Using 32-bit signed integer limits for initial bounds. Because Node.val can reach -2^31 or 2^31 - 1, a root equal to a limit causes false rejections; use 64-bit integers or null sentinel bounds.

The theory behind it

BST — the ground this problem stands on. All BST problems

What BST is

A binary search tree is a binary tree that enforces a strict ordering rule at every node. Every value stored in a node's left branch must be smaller than the node itself, and every value in its right branch must be larger. Because this rule holds true everywhere down the tree, searching for a value does not require checking every branch. At each step, a single comparison lets you discard an entire half of the remaining nodes.

When to reach for it

Reach for a binary search tree when problems mention sorted tree structures, finding the kth smallest element, checking tree validity, or searching within a range of values. Signals include queries for the next greater element in a dynamic set, finding the lowest common ancestor in a sorted tree, or converting sorted arrays into height-balanced trees. When data must remain sorted while supporting continuous insertions and lookups, BST properties are the target.

How the pattern works

Use the ordering property to prune entire subtrees. When looking for a key, move left if the target is smaller than the current node, or move right if it is larger, stopping when values match or when reaching a null link. For validating a tree, do not merely compare a node to its immediate children; pass valid lower and upper value bounds down through recursive calls. Walking a valid binary search tree in-order visits all values in strictly increasing numerical order.

What each operation costs

OperationTime
search, insert, or delete on balanced treeO(log n)
search, insert, or delete on degenerate line treeO(n)
in-order traversal visiting all nodes in sorted orderO(n)
What usually goes wrong with BST
  • Validating a tree by checking only immediate children against the parent, missing violations where a node deep in the left subtree is larger than an ancestor higher up.
  • Allowing duplicate values on the wrong side when the problem specification requires strictly smaller values on the left and strictly greater on the right.
  • Deleting a node with two children by removing it directly instead of replacing its value with its in-order predecessor or successor and deleting that leaf instead.

Which roles need this problem

BST is a core topic for these 2 roles — if you're targeting one of them, this problem is early in your path, not optional.

SDE / Backend EngineerDatabase Engineer

Secondary for 6 more roles, including Full-Stack Developer, Android Developer, iOS Developer.

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