BST Search Operation in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of BST Search for Existing Value

What is the output of the following Java code that searches for a value in a Binary Search Tree (BST)?

DSA Java

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

class BST {
    Node root;
    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);
    }
}

public class Main {
    public static void main(String[] args) {
        BST tree = new BST();
        tree.root = new Node(10);
        tree.root.left = new Node(5);
        tree.root.right = new Node(15);
        tree.root.left.left = new Node(3);
        tree.root.left.right = new Node(7);
        boolean found = tree.search(tree.root, 7);
        System.out.println(found);
    }
}

A Compilation error

B false

C true

D Runtime exception

💡 Hint

Think about whether the value 7 exists in the BST.

The search method recursively checks nodes. Since 7 is present in the BST, the method returns true and prints it.

❓ Predict Output

intermediate

Output of BST Search for Non-Existing Value

What is the output of the following Java code that searches for a value not present in the BST?

DSA Java

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

class BST {
    Node root;
    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);
    }
}

public class Main {
    public static void main(String[] args) {
        BST tree = new BST();
        tree.root = new Node(10);
        tree.root.left = new Node(5);
        tree.root.right = new Node(15);
        tree.root.left.left = new Node(3);
        tree.root.left.right = new Node(7);
        boolean found = tree.search(tree.root, 8);
        System.out.println(found);
    }
}

A true

B false

C Compilation error

D Runtime exception

💡 Hint

Check if 8 is present in the BST nodes.

The search method returns false because 8 is not found in the BST.

🧠 Conceptual

advanced

Understanding BST Search Path

In a BST, when searching for the value 12 starting from the root node 10, which path will the search follow?

A 10 -> 15

B 10 -> 5 -> 7

C 10 -> 15 -> 12

D 10 -> 5 -> 3

💡 Hint

Remember BST property: left child < parent < right child.

Since 12 > 10, search moves right to 15. Then 12 < 15, so it would go left if 12 existed. But since 12 is not in the tree, search ends at 15.

🔧 Debug

advanced

Identify the Bug in BST Search Method

What is the bug in the following BST search method that causes incorrect search results?

DSA Java

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);
}

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

B The base case is missing to stop recursion.

C The method does not return any value.

D The comparison uses '==' instead of '.equals()'.

💡 Hint

Check which subtree to search when key is greater or smaller than root data.

In BST, if key > root.data, search should go right subtree, but code incorrectly goes left, causing wrong results.

🚀 Application

expert

Number of Recursive Calls in BST Search

Given a BST with nodes inserted in this order: 20, 10, 30, 5, 15, 25, 35. How many recursive calls will the search method make when searching for the value 15?

A 5

B 4

C 2

D 3

💡 Hint

Count each node visited including the root until the value is found.

Search path: 20 (call 1) -> 10 (call 2) -> 15 (call 3). Found at third call.

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