Why BST Over Plain Binary Tree in DSA Java - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a value into a Binary Search Tree (BST).

DSA Java

if (root == null) {
    root = new Node([1]);
}

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

Aroot

Bvalue

Cnull

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'root' instead of 'value' inside the new Node constructor.

Assigning null instead of a new node.

The new node is created with the given value when the root is null.

2fill in blank

medium

Complete the code to decide whether to insert the value in the left subtree of a BST node.

DSA Java

if (value [1] root.data) {
    root.left = insert(root.left, value);
}

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 subtree insertion.

Using '==' does not handle insertion properly.

In a BST, values less than the current node go to the left subtree.

3fill in blank

hard

Fix the error in the code to search a value in a BST.

DSA Java

if (root == null || root.data [1] value) {
    return root;
}

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

A!=

B<

C>

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' or '>' instead of '==' causes wrong search results.

Using '!=' returns wrong nodes.

The search returns the node if the current node's data equals the value.

4fill in blank

hard

Fill both blanks to complete the BST search logic for left and right subtree traversal.

DSA Java

if (value [1] root.data) {
    return search(root.left, value);
} else if (value [2] root.data) {
    return search(root.right, value);
}

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

A<

B>

C==

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping '<' and '>' operators.

Using '==' in these conditions instead of comparison operators.

If the value is less than the node's data, search left subtree; if greater, search right subtree.

5fill in blank

hard

Fill all three blanks to complete the code that finds the minimum value node in a BST.

DSA Java

Node current = root;
while (current.[1] != null) {
    current = current.[2];
}
return current.[3];

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

Aleft

Bright

Cdata

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' to find minimum.

Returning the node instead of its data.

The minimum value in a BST is found by going left until no more left child exists, then returning the data.

Practice

1. Why is a Binary Search Tree (BST) preferred over a plain Binary Tree for searching elements?

easy

A. Because BST does not require pointers to child nodes

B. Because BST uses less memory than a plain Binary Tree

C. Because BST nodes can have more than two children

D. Because BST keeps elements in sorted order allowing faster search

Solution

1. Step 1: Understand the structure difference

   A plain Binary Tree has no order, so searching requires checking many nodes. BST keeps nodes sorted so left child is smaller and right child is larger.

2. Step 2: Analyze search efficiency

   Because of sorting, BST allows searching by comparing and moving left or right, reducing search time to O(log n) on average, unlike plain Binary Tree which can be O(n).

3. Final Answer:

   Because BST keeps elements in sorted order allowing faster search -> Option D

4. Quick Check:

   Sorted structure = faster search [OK]

Hint: BST is sorted, so search is faster than plain tree [OK]

Common Mistakes:

• Thinking BST uses less memory
• Confusing number of children per node
• Assuming BST has no pointers

2. Which of the following is the correct way to declare a node class for a BST in Java?

easy

A. class Node { int data; Node left; Node right; }

B. class Node { int data; Node left, right; void insert(int val); }

C. class Node { int data; Node left; Node right; Node parent; }

D. class Node { int data; Node left; Node right; int height; }

Solution

1. Step 1: Identify minimal BST node structure

   A BST node must store data and references to left and right child nodes. class Node { int data; Node left; Node right; } shows this minimal structure.

2. Step 2: Check other options for extra or missing parts

   class Node { int data; Node left, right; void insert(int val); } adds a method which is not part of declaration only. class Node { int data; Node left; Node right; Node parent; } adds parent pointer which is optional. class Node { int data; Node left; Node right; int height; } adds height used in AVL trees, not basic BST.

3. Final Answer:

   class Node { int data; Node left; Node right; } -> Option A

4. Quick Check:

   Minimal node = data + left + right [OK]

Hint: Basic BST node has data and two child pointers only [OK]

Common Mistakes:

• Adding unnecessary fields like parent or height
• Confusing method declarations with node fields
• Using incorrect syntax for multiple variables

3. What will be the output of the following Java code snippet that inserts values into a BST and then performs an inorder traversal?

class Node {
  int data;
  Node left, right;
  Node(int d) { data = d; left = right = null; }
}

class BST {
  Node root;
  void insert(int val) { root = insertRec(root, val); }
  Node insertRec(Node root, int val) {
    if (root == null) return new Node(val);
    if (val < root.data) root.left = insertRec(root.left, val);
    else root.right = insertRec(root.right, val);
    return root;
  }
  void inorder(Node root) {
    if (root != null) {
      inorder(root.left);
      System.out.print(root.data + " ");
      inorder(root.right);
    }
  }
}

public class Main {
  public static void main(String[] args) {
    BST tree = new BST();
    tree.insert(10);
    tree.insert(5);
    tree.insert(15);
    tree.insert(7);
    tree.inorder(tree.root);
  }
}

medium

A. 10 5 7 15

B. 5 7 10 15

C. 7 5 10 15

D. 15 10 7 5

Solution

1. Step 1: Trace insertion order and BST structure

   Insert 10 as root. Insert 5 to left of 10. Insert 15 to right of 10. Insert 7 to right of 5.

2. Step 2: Perform inorder traversal

   Inorder visits left subtree, root, then right subtree. So nodes visited in order: 5, 7, 10, 15.

3. Final Answer:

   5 7 10 15 -> Option B

4. Quick Check:

   Inorder traversal = sorted values [OK]

Hint: Inorder traversal of BST prints sorted values [OK]

Common Mistakes:

• Confusing preorder or postorder with inorder
• Misplacing 7 in the tree structure
• Ignoring recursion order in traversal

4. The following code attempts to insert a value into a BST but causes a runtime error. What is the error?

class Node {
  int data;
  Node left, right;
  Node(int d) { data = d; }
}

class BST {
  Node root;
  void insert(int val) { root = insertRec(root, val); }
  Node insertRec(Node root, int val) {
    if (root == null) return new Node(val);
    if (val < root.data) root.left = insertRec(root.left, val);
    else root.right = insertRec(root.right, val);
    return root;
  }
}

medium

A. Compilation error due to missing return statement

B. StackOverflowError due to infinite recursion

C. No error, code runs fine

D. NullPointerException because left and right are not initialized in Node constructor

Solution

1. Step 1: Check Node constructor initialization

   In Java, Node left and right are automatically initialized to null by default. The constructor only sets data, which is sufficient.

2. Step 2: Analyze insertRec logic

   The base case handles null roots. Recursive calls like root.left = insertRec(root.left, val) are safe since insertRec(null) returns new Node without dereferencing null. All paths return a value. No runtime errors occur.

3. Final Answer:

   No error, code runs fine -> Option C

4. Quick Check:

   Default null + null base case = safe execution [OK]

Hint: Java defaults references to null; recursion base case prevents issues [OK]

Common Mistakes:

• Thinking uninitialized fields cause immediate NPE (defaults handle it)
• Suspecting infinite recursion (base case exists)
• Overlooking that all code paths return

5. You have a plain Binary Tree and want to convert it into a BST without changing its structure (only node values). Which approach is correct?

hard

A. Perform inorder traversal to collect values, sort them, then assign back during inorder traversal

B. Insert all nodes into a new BST and discard the old tree

C. Swap left and right children of all nodes to sort the tree

D. Replace each node value with the sum of its children

Solution

1. Step 1: Understand the problem constraints

   The tree structure must remain the same, so only node values can change. This means rearranging values to satisfy BST property without changing links.

2. Step 2: Use inorder traversal to collect and assign values

   Inorder traversal of BST gives sorted values. Collect all node values from the plain tree, sort them, then assign them back to nodes during inorder traversal to maintain structure but satisfy BST order.

3. Final Answer:

   Perform inorder traversal to collect values, sort them, then assign back during inorder traversal -> Option A

4. Quick Check:

   Sort values and reassign inorder = convert to BST [OK]

Hint: Sort values and reassign inorder to fix tree values [OK]

Common Mistakes:

• Changing tree structure instead of values
• Swapping children does not sort values
• Replacing values with sums breaks BST property