BST Search Operation in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to compare the target value with the current node's value.

DSA Java

if (target [1] node.value) {
    return search(node.left, target);
}

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 traversal.

Using '>' leads to searching the wrong subtree.

In a BST, if the target is less than the current node's value, we search the left subtree.

2fill in blank

medium

Complete the code to return true if the target is found at the current node.

DSA Java

if (node.value [1] target) {
    return true;
}

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 results.

Using '!=' returns true for non-matching values.

We return true when the current node's value equals the target.

3fill in blank

hard

Fix the error in the recursive call to search the right subtree.

DSA Java

if (target [1] node.value) {
    return search(node.right, target);
}

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

A>

B<

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' causes searching the wrong subtree.

Using '==' or '!=' here is incorrect.

If the target is greater than the current node's value, we search the right subtree.

4fill in blank

hard

Fill both blanks to complete the base case and recursive step for BST search.

DSA Java

if (node [1] null) {
    return false;
} else if (target [2] node.value) {
    return search(node.left, target);
}

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

A==

B!=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' for null check causes errors.

Using '>' instead of '<' for left subtree search is wrong.

The base case checks if the node is null (not found). If target is less than node value, search left.

5fill in blank

hard

Fill all three blanks to complete the BST search method with all cases.

DSA Java

if (node == null) {
    return false;
} else if (target [1] node.value) {
    return search(node.left, target);
} else if (target [2] node.value) {
    return search(node.right, target);
} else if (node.value [3] target) {
    return true;
}

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

A<

B>

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up comparison operators leads to wrong traversal.

Using '!=' instead of '==' for equality check causes errors.

If target is less, go left; if greater, go right; if equal, return true.

Practice

1. What is the main purpose of the search operation in a Binary Search Tree (BST)?

easy

A. To insert a new value into the BST

B. To find whether a given value exists in the BST

C. To delete a value from the BST

D. To traverse all nodes in the BST

Solution

1. Step 1: Understand the search operation

   The search operation checks if a specific value is present in the BST by comparing it with nodes.

2. Step 2: Differentiate from other operations

   Insertion adds new nodes, deletion removes nodes, and traversal visits all nodes. Search only checks presence.

3. Final Answer:

   To find whether a given value exists in the BST -> Option B

4. Quick Check:

   Search = Find value presence [OK]

Hint: Search checks presence, not insertion or deletion [OK]

Common Mistakes:

• Confusing search with insert or delete
• Thinking search returns all nodes
• Assuming search modifies the tree

2. Which of the following is the correct Java method signature for a BST search function that returns a boolean indicating if a value exists?

easy

A. public boolean search(int value)

B. public void search(int value)

C. public int search(int value)

D. public String search(int value)

Solution

1. Step 1: Identify return type for search

   Search should return true or false, so return type must be boolean.

2. Step 2: Check method signature correctness

   Method takes an int value to search and returns boolean. public boolean search(int value) matches this.

3. Final Answer:

   public boolean search(int value) -> Option A

4. Quick Check:

   Return type boolean for search [OK]

Hint: Search returns true/false, so use boolean return type [OK]

Common Mistakes:

• Using void return type for search
• Returning int or String instead of boolean
• Incorrect method parameter types

3. Consider the following BST and the search code snippet:

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

public boolean search(Node root, int key) {
  if (root == null) return false;
  if (root.data == key) return true;
  if (key < root.data) return search(root.left, key);
  else return search(root.right, key);
}

// BST structure:
//       10
//      /  \
//     5    15
//    / \     \
//   3   7     20

What is the output of search(root, 7)?

medium

A. null

B. false

C. true

D. Runtime error

Solution

1. Step 1: Trace search for key 7 starting at root 10

   7 < 10, so search left subtree (node 5).

2. Step 2: Trace search at node 5

   7 > 5, so search right subtree (node 7).

3. Step 3: Trace search at node 7

   Node data equals key (7), so return true.

4. Final Answer:

   true -> Option C

5. Quick Check:

   7 found in BST = true [OK]

Hint: Follow BST rules left/right to find key [OK]

Common Mistakes:

• Stopping search too early
• Confusing left and right subtree directions
• Assuming search returns node instead of boolean

4. The following Java method is intended to search a BST for a key. What is the bug?

public boolean search(Node root, int key) {
  if (root == null) return false;
  if (root.data == key) return true;
  if (key > root.data) return search(root.left, key);
  else return search(root.right, key);
}

medium

A. The recursive calls for left and right subtrees are swapped

B. The base case is missing

C. The method should return int instead of boolean

D. The method does not handle null keys

Solution

1. Step 1: Analyze recursive calls for BST search

   For BST, if key < root.data, search left; if key > root.data, search right.

2. Step 2: Identify swapped calls

   Code calls search(root.left, key) when key > root.data, which is incorrect; left and right calls are swapped.

3. Final Answer:

   The recursive calls for left and right subtrees are swapped -> Option A

4. Quick Check:

   Left/right calls must match key comparison [OK]

Hint: Left subtree for smaller keys, right for larger keys [OK]

Common Mistakes:

• Swapping left and right recursive calls
• Missing base case for null root
• Returning wrong data type

5. Given a BST with nodes inserted in this order: 15, 10, 20, 8, 12, 17, 25. Which sequence of nodes will the search operation visit when searching for 12?

hard

A. 10 -> 15 -> 12

B. 15 -> 20 -> 17 -> 12

C. 15 -> 10 -> 8 -> 12

D. 15 -> 10 -> 12

Solution

1. Step 1: Build BST structure from insertion order

   Insert 15 as root; 10 left of 15; 20 right of 15; 8 left of 10; 12 right of 10; 17 left of 20; 25 right of 20.

2. Step 2: Trace search path for key 12

   Start at 15: 12 < 15, go left to 10; 12 > 10, go right to 12; found key.

3. Final Answer:

   15 -> 10 -> 12 -> Option D

4. Quick Check:

   Search path follows BST rules [OK]

Hint: Follow left if smaller, right if larger until found [OK]

Common Mistakes:

• Visiting nodes not on search path
• Confusing insertion order with search path
• Assuming search visits all nodes