Validate if Tree is BST in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does BST stand for and what is its main property?

BST stands for Binary Search Tree. Its main property is that for every node, all values in the left subtree are smaller, and all values in the right subtree are larger.

beginner

What is the simplest way to check if a binary tree is a BST?

Perform an in-order traversal and check if the values are in strictly increasing order.

intermediate

Why can't we just check the left and right child values to validate BST property?

Because the BST property applies to the entire subtree, not just immediate children. A node in the left subtree must be less than the current node, even if it's deeper down.

intermediate

What role do minimum and maximum value limits play in BST validation?

They help ensure that all nodes in a subtree fall within valid value ranges, maintaining the BST property throughout the tree.

beginner

Show the base case for a recursive BST validation function.

If the current node is null (empty), return true because an empty tree is a valid BST.

Which traversal method helps check if a tree is a BST by producing sorted values?

A In-order traversal

B Pre-order traversal

C Post-order traversal

D Level-order traversal

In-order traversal visits nodes in ascending order for BSTs, so checking if the output is sorted confirms the BST property.

What should be returned when the node is null during BST validation?

A null

B false

C true

D undefined

An empty tree or subtree is considered a valid BST, so the function should return true.

If a node's value is not within the allowed min and max range during validation, what does it mean?

A The tree is not a BST

B The tree is a BST

C The node is a leaf

D The node is the root

Violating the min-max range means the BST property is broken, so the tree is not a BST.

Which of these is NOT a valid step in validating a BST?

A Check if right subtree nodes are greater than current node

B Check if left child is greater than right child

C Check if left subtree nodes are less than current node

D Use min and max limits to validate subtrees

Checking if left child is greater than right child is not relevant; BST property depends on subtree values, not direct child comparison.

What is the time complexity of validating a BST using min-max recursion?

A O(1)

B O(log n)

C O(n^2)

D O(n)

Each node is visited once, so the time complexity is linear, O(n).

Explain how to validate if a binary tree is a BST using recursion and min-max limits.

Think about how each subtree must fit within a range of allowed values.

Describe why an in-order traversal output can be used to check if a tree is a BST.

Consider the sorted order property of BST in in-order traversal.

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