Validate if Tree is BST in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the current node is null.

DSA Typescript

if (root [1] null) return true;

Drag options to blanks, or click blank then click option'

A!==

B===

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '===' causes incorrect null check.

Using '>' or '<' operators for null comparison.

We check if root is exactly null to handle the base case of recursion.

2fill in blank

medium

Complete the code to check if the current node's value is within the valid range.

DSA Typescript

if (root.val [1] min || root.val [2] max) return false;

Drag options to blanks, or click blank then click option'

A<

B<=

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' or '>' instead of '<=' or '>=' causes boundary errors.

Checking only one side of the range.

The node's value must be strictly greater than min and strictly less than max, so we check if it is less than or equal to min or greater than or equal to max to return false.

3fill in blank

hard

Fix the error in the recursive call to validate left subtree with updated max.

DSA Typescript

return isValidBST(root.left, min, [1]) && isValidBST(root.right, root.val, max);

Drag options to blanks, or click blank then click option'

Aroot.val

Bmin

Cmax

Droot.left.val

Attempts:

3 left

💡 Hint

Common Mistakes

Passing min or max incorrectly causes wrong validation.

Using root.left.val instead of root.val.

For the left subtree, the maximum allowed value is the current node's value.

4fill in blank

hard

Fill both blanks to correctly validate the right subtree with updated min.

DSA Typescript

return isValidBST(root.left, min, root.val) && isValidBST(root.right, [1], [2]);

Drag options to blanks, or click blank then click option'

Aroot.val

Bmin

Cmax

Droot.right.val

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping min and max values.

Using root.right.val instead of root.val.

For the right subtree, the minimum allowed value is the current node's value, and max remains unchanged.

5fill in blank

hard

Fill all three blanks to complete the function signature and initial call for BST validation.

DSA Typescript

function isValidBST(root: TreeNode | null, [1]: number = -Infinity, [2]: number = [3]): boolean {
  if (root === null) return true;
  if (root.val <= min || root.val >= max) return false;
  return isValidBST(root.left, min, root.val) && isValidBST(root.right, root.val, max);
}

Drag options to blanks, or click blank then click option'

Amin

Bmax

CInfinity

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Omitting default values causes errors.

Using wrong parameter names or default values.

The function uses min and max parameters with default values -Infinity and Infinity to check valid ranges.

Practice

1. What does it mean for a binary tree to be a Binary Search Tree (BST)?

easy

A. Every node has exactly two children

B. The tree is balanced with equal height on both sides

C. All nodes have the same value

D. Every node's left subtree has smaller values and right subtree has larger values

Solution

1. Step 1: Understand BST property

   A BST requires that for each node, all values in the left subtree are smaller and all values in the right subtree are larger.

2. Step 2: Compare options with definition

   Every node's left subtree has smaller values and right subtree has larger values matches the BST property exactly, while others describe different tree properties.

3. Final Answer:

   Every node's left subtree has smaller values and right subtree has larger values -> Option D

4. Quick Check:

   BST property = Every node's left subtree has smaller values and right subtree has larger values [OK]

Hint: Remember BST means left < node < right for all nodes [OK]

Common Mistakes:

• Confusing BST with balanced tree
• Thinking all nodes must have two children
• Assuming all values are equal

2. Which TypeScript function signature correctly represents a function to check if a binary tree is a BST?

easy

A. function isBST(root: any): void

B. function isBST(root: TreeNode | null): boolean

C. function isBST(root: TreeNode): number

D. function isBST(root: number): boolean

Solution

1. Step 1: Identify correct parameter and return types

   The function should accept a tree node or null and return a boolean indicating if it is a BST.

2. Step 2: Match options with correct signature

   function isBST(root: TreeNode | null): boolean matches this: input is TreeNode or null, output is boolean. Others have wrong types or return types.

3. Final Answer:

   function isBST(root: TreeNode | null): boolean -> Option B

4. Quick Check:

   Correct signature = function isBST(root: TreeNode | null): boolean [OK]

Hint: Input is root node or null, output is true/false [OK]

Common Mistakes:

• Using wrong parameter type like number
• Returning number instead of boolean
• Using void return type

3. What will be the output of the following TypeScript code snippet?

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val: number, left: TreeNode | null = null, right: TreeNode | null = null) {
    this.val = val;
    this.left = left;
    this.right = right;
  }
}

function isBST(root: TreeNode | null, min: number = -Infinity, max: number = Infinity): boolean {
  if (!root) return true;
  if (root.val <= min || root.val >= max) return false;
  return isBST(root.left, min, root.val) && isBST(root.right, root.val, max);
}

const tree = new TreeNode(10, new TreeNode(5), new TreeNode(15, new TreeNode(6), new TreeNode(20)));
console.log(isBST(tree));

medium

A. SyntaxError

B. true

C. false

D. RuntimeError

Solution

1. Step 1: Analyze the tree structure

   The root is 10, left child 5 (valid), right child 15 with left child 6 and right child 20.

2. Step 2: Check BST property on right subtree

   Node 6 is in right subtree of 10 but less than 10, violating BST property. So function returns false.

3. Final Answer:

   false -> Option C

4. Quick Check:

   6 in right subtree < 10 = false [OK]

Hint: Check if left < node < right recursively [OK]

Common Mistakes:

• Ignoring min/max constraints in recursion
• Assuming all subtrees are BST without checks
• Confusing node values in subtrees

4. Identify the bug in this TypeScript function to check if a tree is BST:

function isBST(root: TreeNode | null): boolean {
  if (!root) return true;
  if (root.left && root.left.val > root.val) return false;
  if (root.right && root.right.val < root.val) return false;
  return isBST(root.left) && isBST(root.right);
}

medium

A. It does not check the entire subtree values against min/max limits

B. It uses wrong comparison operators for left and right children

C. It returns true for null nodes incorrectly

D. It does not handle empty tree input

Solution

1. Step 1: Understand the function checks

   The function only compares immediate children values with current node, not entire subtree ranges.

2. Step 2: Explain why this is a bug

   BST requires all nodes in left subtree < current node and all in right subtree > current node, not just direct children.

3. Final Answer:

   It does not check the entire subtree values against min/max limits -> Option A

4. Quick Check:

   Missing min/max range check = It does not check the entire subtree values against min/max limits [OK]

Hint: Check full subtree ranges, not just immediate children [OK]

Common Mistakes:

• Only comparing direct children values
• Ignoring recursive min/max constraints
• Assuming null nodes cause errors

5. Given a binary tree where nodes can have duplicate values, which modification to the BST validation logic allows duplicates only on the right subtree? Choose the correct condition to replace root.val <= min in the validation function:

function isBST(root: TreeNode | null, min: number = -Infinity, max: number = Infinity): boolean {
  if (!root) return true;
  if (root.val <= min || root.val >= max) return false;
  return isBST(root.left, min, root.val) && isBST(root.right, root.val, max);
}

hard

A. Change to root.val < min to allow duplicates on right

B. Change to root.val <= min and root.val < max

C. Change to root.val < min and root.val <= max

D. Change to root.val <= min and root.val <= max

Solution

1. Step 1: Understand duplicates on right subtree rule

   Duplicates allowed only on right means left subtree nodes must be strictly less than current node, right subtree nodes greater than or equal.

2. Step 2: Adjust min/max comparisons accordingly

   Replace root.val <= min with root.val < min. This fails only if strictly less than min (allows equal to min on right recursion), keeping root.val >= max for strict upper bound.

3. Final Answer:

   Change to root.val < min to allow duplicates on right -> Option A

4. Quick Check:

   Duplicates right subtree = Change to root.val < min [OK]

Hint: Use strict less on left, less or equal on right for duplicates [OK]

Common Mistakes:

• Allowing duplicates on left subtree
• Using <= on both sides
• Not changing min/max checks