Validate if Tree is BST in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the current node is null in the BST validation.

DSA C#

if (node [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 base case.

Using comparison operators like '<' or '>' with null causes errors.

We check if the node is equal to null to stop recursion and return true for empty subtree.

2fill in blank

medium

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

DSA C#

if (node.Value [1] min || node.Value [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 incorrect validation.

Checking only one side of the range misses invalid cases.

We check if node's value is less than or equal to min or greater than or equal to max to invalidate BST.

3fill in blank

hard

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

DSA C#

return IsValidBST(node.Left, min, [1]) && IsValidBST(node.Right, node.Value, max);

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

Anode.Value

Bmin

Cmax

Dnode.Left.Value

Attempts:

3 left

💡 Hint

Common Mistakes

Passing min or max instead of node.Value causes incorrect range checks.

Using node.Left.Value may cause null reference errors.

For the left subtree, max should be updated to current node's value to maintain BST property.

4fill in blank

hard

Fill both blanks to update the recursive call for the right subtree with correct min value.

DSA C#

return IsValidBST(node.Left, min, node.Value) && IsValidBST(node.Right, [1], [2]);

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

Anode.Value

Bmax

Cnode.Right.Value

Dmin

Attempts:

3 left

💡 Hint

Common Mistakes

Passing min as min for right subtree causes incorrect validation.

Using node.Right.Value may cause null reference errors.

For the right subtree, min should be updated to current node's value and max remains the same.

5fill in blank

hard

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

DSA C#

public bool IsValidBST(TreeNode [1], int [2], int [3]) {
    if ([1] == null) return true;
    if ([1].Value <= [2] || [1].Value >= [3]) return false;
    return IsValidBST([1].Left, [2], [1].Value) && IsValidBST([1].Right, [1].Value, [3]);
}

// Initial call example:
// bool result = IsValidBST(root, int.MinValue, int.MaxValue);

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

Anode

Bmin

Cmax

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using inconsistent parameter names causes confusion and errors.

Using 'root' inside method signature instead of 'node' is misleading.

The method uses 'node' as parameter name and 'min' and 'max' as range boundaries.

Practice

1. What is the main property that defines a Binary Search Tree (BST)?

easy

A. For each node, all nodes in the left subtree are smaller and all nodes in the right subtree are larger

B. Each node has exactly two children

C. All nodes have unique values but no order is required

D. The tree is always balanced

Solution

1. Step 1: Understand BST property

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

2. Step 2: Compare options with BST definition

   For each node, all nodes in the left subtree are smaller and all nodes in the right subtree are larger matches the BST property exactly, while others describe different tree properties.

3. Final Answer:

   For each node, all nodes in the left subtree are smaller and all nodes in the right subtree are larger -> Option A

4. Quick Check:

   BST property = For each node, all nodes in the left subtree are smaller and all nodes in the right subtree are larger [OK]

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

Common Mistakes:

• Confusing BST with balanced tree
• Thinking all nodes must have two children
• Ignoring the ordering property

2. Which of the following C# method signatures is correct to check if a binary tree is a BST?

easy

A. bool IsBST(Node root, int min, int max)

B. void IsBST(Node root)

C. int IsBST(Node root)

D. bool IsBST(Node root, Node parent)

Solution

1. Step 1: Identify parameters needed for BST validation

   To validate BST, we need to check node values within a range (min, max) as we traverse.

2. Step 2: Match method signature with required parameters

   bool IsBST(Node root, int min, int max) includes root node and min/max range, suitable for recursive BST check.

3. Final Answer:

   bool IsBST(Node root, int min, int max) -> Option A

4. Quick Check:

   Method with min/max range = bool IsBST(Node root, int min, int max) [OK]

Hint: Use min and max range parameters for BST validation [OK]

Common Mistakes:

• Using void return type instead of bool
• Missing min and max parameters for range checks
• Passing parent node instead of value ranges

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

class Node {
  public int val;
  public Node left, right;
  public Node(int v) { val = v; left = right = null; }
}

bool IsBST(Node node, int min, int max) {
  if (node == null) return true;
  if (node.val <= min || node.val >= max) return false;
  return IsBST(node.left, min, node.val) && IsBST(node.right, node.val, max);
}

var root = new Node(10);
root.left = new Node(5);
root.right = new Node(15);
root.left.left = new Node(1);
root.left.right = new Node(12);

Console.WriteLine(IsBST(root, int.MinValue, int.MaxValue));

medium

A. True

B. False

C. Compilation error

D. Runtime exception

Solution

1. Step 1: Trace the tree structure and BST property

   The tree has root 10, left child 5, right child 15. Left child's right child is 12, which is greater than 10 but in left subtree.

2. Step 2: Check BST validation logic

   IsBST will fail because 12 is in left subtree of 10 but greater than 10, violating BST rules.

3. Final Answer:

   False -> Option B

4. Quick Check:

   Node 12 in left subtree breaks BST = False [OK]

Hint: Check if left subtree nodes are less than root [OK]

Common Mistakes:

• Assuming all nodes are valid without checking subtree ranges
• Ignoring node 12's position violating BST
• Confusing left and right subtree values

4. Identify the bug in the following C# code that checks if a binary tree is a BST:

bool IsBST(Node node) {
  if (node == null) return true;
  if (node.left != null && node.left.val > node.val) return false;
  if (node.right != null && node.right.val < node.val) return false;
  return IsBST(node.left) && IsBST(node.right);
}

medium

A. It should return int instead of bool

B. It uses incorrect comparison operators

C. It does not handle null nodes correctly

D. It does not check the entire subtree values, only immediate children

Solution

1. Step 1: Analyze the checks performed

   The code only compares node values with immediate left and right children, not entire subtrees.

2. Step 2: Understand BST validation requirement

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

3. Final Answer:

   It does not check the entire subtree values, only immediate children -> Option D

4. Quick Check:

   Check full subtree values, not just children [OK]

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

Common Mistakes:

• Only comparing immediate children values
• Ignoring subtree min/max constraints
• Assuming null handling is incorrect

5. Given a binary tree, how would you modify the standard BST validation to allow duplicate values only on the right subtree?

hard

A. Change the comparison to allow node.val <= max for right subtree and node.val > min for left subtree

B. Allow duplicates only if node.val >= min for left subtree and node.val < max for right subtree

C. Allow duplicates only if node.val > min for left subtree and node.val <= max for right subtree

D. Disallow duplicates completely by strict less than and greater than comparisons

Solution

1. Step 1: Understand duplicate placement rule

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

2. Step 2: Adjust comparison operators accordingly

   Left subtree: node.val > min (strictly less), right subtree: node.val <= max (allow equal or greater).

3. Final Answer:

   Allow duplicates only if node.val > min for left subtree and node.val <= max for right subtree -> Option C

4. Quick Check:

   Duplicates on right subtree use <= max [OK]

Hint: Use <= for right subtree, > for left subtree to allow duplicates right only [OK]

Common Mistakes:

• Allowing duplicates on left subtree
• Using <= on left subtree incorrectly
• Disallowing duplicates completely when allowed