Validate if Tree is BST in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

function isBST(node) {
  if (node [1] null) {
    return true;
  }
  // further checks
}

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 base case handling.

Using logical operators instead of comparison.

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

2fill in blank

medium

Complete the code to check if the left child's value is less than the current node's value.

DSA Javascript

if (node.left !== null && node.left.val [1] node.val) {
  return false;
}

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

A===

B<

C>=

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of >= causes wrong validation.

Using equality check instead of comparison.

For a BST, left child's value must be less than the current node's value. If it is greater or equal, return false.

3fill in blank

hard

Fix the error in the recursive call to validate the right subtree with updated min value.

DSA Javascript

return isBST(node.left, min, node.val) && isBST(node.right, [1], max);

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

Anode.val

Bmin

Cmax

Dnode.left.val

Attempts:

3 left

💡 Hint

Common Mistakes

Passing min instead of node.val causes incorrect range checks.

Passing max instead of min for right subtree.

For the right subtree, the minimum allowed value is the current node's value.

4fill in blank

hard

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

DSA Javascript

function isBST(node, [1] = -Infinity, [2] = Infinity) {
  if (node === null) return true;
  // validation logic
}

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

Amin

Bmax

Cvalue

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect parameter names causes confusion.

Swapping min and max values.

The function uses min and max to keep track of valid value ranges for nodes.

5fill in blank

hard

Fill all four blanks to complete the BST validation condition inside the function.

DSA Javascript

if (node.val [1] [2] || node.val [3] [4]) {
  return false;
}

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

A<=

B>=

Cmin

Dmax

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators.

Swapping min and max in conditions.

The node's value must be greater than min and less than max to be valid in BST.

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. Every node's left child is smaller and right child is larger than the node

C. All nodes have the same value

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

Solution

1. Step 1: Understand BST property

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

2. Step 2: Compare options with definition

   Every node's left child is smaller and right child is larger than the node correctly states this property. Other options describe different tree properties.

3. Final Answer:

   Every node's left child is smaller and right child is larger than the node -> Option B

4. Quick Check:

   BST property = Every node's left child is smaller and right child is larger than the node [OK]

Hint: BST means left < node < right for every node [OK]

Common Mistakes:

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

2. Which of the following JavaScript function signatures is correct to check if a binary tree is a BST?

easy

A. function isBST(node, min, max) {}

B. function isBST(node, left, right, parent) {}

C. function checkBST(root) {}

D. function validateBST(tree, minVal) {}

Solution

1. Step 1: Identify common BST validation parameters

   To validate BST, we usually pass the current node and allowed min and max values to check constraints.

2. Step 2: Match function signatures

   function isBST(node, min, max) {} matches this pattern with node, min, and max parameters. Others have extra or missing parameters.

3. Final Answer:

   function isBST(node, min, max) {} -> Option A

4. Quick Check:

   Correct BST function signature = function isBST(node, min, max) {} [OK]

Hint: Use min and max limits to check BST recursively [OK]

Common Mistakes:

• Adding unnecessary parameters
• Missing min or max limits
• Using wrong parameter names

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

class Node {
  constructor(val, left = null, right = null) {
    this.val = val;
    this.left = left;
    this.right = right;
  }
}

function isBST(node, min = -Infinity, max = Infinity) {
  if (!node) return true;
  if (node.val <= min || node.val >= max) return false;
  return isBST(node.left, min, node.val) && isBST(node.right, node.val, max);
}

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

medium

A. undefined

B. true

C. false

D. TypeError

Solution

1. Step 1: Analyze tree structure

   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 left child of 15 but 6 < 10, which violates BST property because all right subtree nodes must be > 10.

3. Final Answer:

   false -> Option C

4. Quick Check:

   6 in right subtree < 10 breaks BST = false [OK]

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

Common Mistakes:

• Ignoring subtree value constraints
• Assuming all children just compare with parent
• Missing base case for null nodes

4. Find the bug in this BST validation function and select the fix:

function isBST(node) {
  if (!node) return true;
  if (node.left && node.left.val >= node.val) return false;
  if (node.right && node.right.val <= node.val) return false;
  return isBST(node.left) && isBST(node.right);
}

medium

A. Add min and max parameters to track value ranges

B. Change >= to > and <= to < in comparisons

C. Return false if node is null

D. Remove recursive calls to isBST

Solution

1. Step 1: Understand current checks

   Function only compares immediate children with current node, missing subtree constraints.

2. Step 2: Identify fix

   To correctly validate BST, function must track allowed min and max values for each subtree, not just immediate children.

3. Final Answer:

   Add min and max parameters to track value ranges -> Option A

4. Quick Check:

   Use min/max limits to validate BST correctly [OK]

Hint: Use min/max limits, not just immediate child values [OK]

Common Mistakes:

• Only checking immediate children
• Ignoring subtree value constraints
• Not using recursion properly

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 node.val <= min and node.val >= max checks in the recursive function.

hard

A. Allow duplicates by changing left subtree check to node.val < min and right subtree check to node.val > max

B. Allow duplicates on right by changing left subtree check to node.val <= min and right subtree check to node.val > max

C. Allow duplicates on right by changing left subtree check to node.val <= min and right subtree check to node.val >= max

D. Allow duplicates on right by changing left subtree check to node.val < min and right subtree check to node.val >= max

Solution

1. Step 1: Understand duplicates in BST

   Allowing duplicates only on right means left subtree nodes must be strictly less than current node, right subtree nodes can be greater or equal.

2. Step 2: Adjust min/max checks accordingly

   Left subtree check should be node.val < min (strictly less), right subtree check should be node.val >= max (allow equal).

3. Final Answer:

   Allow duplicates on right by changing left subtree check to node.val < min and right subtree check to node.val >= max -> Option D

4. Quick Check:

   Duplicates only right: left < node, right >= node [OK]

Hint: Strict less on left, allow equal on right for duplicates [OK]

Common Mistakes:

• Allowing duplicates on both sides
• Using non-strict inequalities on left subtree
• Confusing min and max checks