Validate if Tree is BST in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Java

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 wrong base case.

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

We check if the current node is null using '==' to return true for an empty subtree.

2fill in blank

medium

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

DSA Java

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 incorrect validation.

Mixing up min and max comparisons.

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

DSA Java

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 causes errors.

For the left subtree, the maximum allowed value is the current node's value, so we pass root.val as max.

4fill in blank

hard

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

DSA Java

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

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

Aroot.val

Bmax

Cmin

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 (root.val), and the maximum remains max.

5fill in blank

hard

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

DSA Java

public boolean isValidBST([1] root) {
    return isValidBST(root, [2], [3]);
}

private boolean isValidBST(TreeNode root, Integer min, Integer max) {
    if (root == null) return true;
    if ((min != null && root.val <= min) || (max != null && 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'

ATreeNode

Bnull

Dint

Attempts:

3 left

💡 Hint

Common Mistakes

Using primitive int instead of Integer causes null issues.

Passing 0 instead of null for min or max.

The public method takes a TreeNode root and calls the private helper with null min and max to start without 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 correctly states this property, while others describe different tree characteristics.

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 Java method signatures is correct to check if a binary tree is a BST?

easy

A. public int isBST(Node root, int min, int max)

B. public boolean isBST(Node root, int min, int max)

C. public void isBST(Node root)

D. public boolean isBST(Node root, Node min, Node max)

Solution

1. Step 1: Identify correct return type and parameters

   Checking BST returns true/false, so return type should be boolean. Also, min and max values are needed as integers to validate node values.

2. Step 2: Match method signature

   public boolean isBST(Node root, int min, int max) matches this: boolean return type and int min, max parameters. Others have wrong return types or parameter types.

3. Final Answer:

   public boolean isBST(Node root, int min, int max) -> Option B

4. Quick Check:

   Return boolean with int min/max params [OK]

Hint: Use boolean return and int min/max for range checks [OK]

Common Mistakes:

• Using void return type for a check method
• Using Node type for min/max instead of int
• Returning int instead of boolean

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

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

public class Test {
  static boolean isBST(Node root, int min, int max) {
    if (root == null) return true;
    if (root.val <= min || root.val >= max) return false;
    return isBST(root.left, min, root.val) && isBST(root.right, root.val, max);
  }

  public static void main(String[] args) {
    Node 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);
    System.out.println(isBST(root, Integer.MIN_VALUE, Integer.MAX_VALUE));
  }
}

medium

A. Runtime exception

B. true

C. Compilation error

D. false

Solution

1. Step 1: Trace the tree nodes and BST conditions

   The node with value 12 is in the left subtree of 10 but greater than 10, violating BST property.

2. Step 2: Check isBST calls

   When checking root.left.right (12), it fails the condition root.val <= min or root.val >= max because 12 >= 10 is true, so returns false.

3. Final Answer:

   false -> Option D

4. Quick Check:

   Node 12 violates BST rule, so false [OK]

Hint: Check if any left child > root or right child < root [OK]

Common Mistakes:

• Ignoring subtree value ranges
• Assuming all nodes in left subtree are less than root only
• Not updating min/max during recursion

4. Identify the bug in this BST validation code snippet:

boolean isBST(Node root) {
  if (root == null) return true;
  if (root.left != null && root.left.val > root.val) return false;
  if (root.right != null && root.right.val < root.val) return false;
  return isBST(root.left) && isBST(root.right);
}

medium

A. It does not handle null root properly

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

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

D. It should return int instead of boolean

Solution

1. Step 1: Analyze the checks performed

   The code only compares root with its immediate left and right children, not the entire subtree.

2. Step 2: Understand BST validation requirement

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

3. Final Answer:

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

4. Quick Check:

   Check full subtree ranges, not just children [OK]

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

Common Mistakes:

• Only comparing immediate children values
• Ignoring recursive min/max range checks
• Assuming local checks are enough

5. Given a binary tree, which approach correctly validates if it is a BST considering edge cases like duplicate values and integer limits?

hard

A. Use min and max integer range parameters updated during recursion

B. Use in-order traversal to check if values are strictly increasing

C. Check only immediate children values recursively

D. Compare each node with root only, ignoring subtrees

Solution

1. Step 1: Understand edge cases in BST validation

   Duplicates and integer limits require careful range checks, not just value comparisons.

2. Step 2: Evaluate approaches

   Using min and max parameters updated during recursion correctly enforces BST property for all nodes and handles edge cases.

3. Final Answer:

   Use min and max integer range parameters updated during recursion -> Option A

4. Quick Check:

   Range checks with min/max handle edge cases [OK]

Hint: Use min/max range checks recursively for full validation [OK]

Common Mistakes:

• Ignoring duplicates or equal values
• Not updating min/max during recursion
• Only checking immediate children