Kth Smallest Element in BST in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Kth Smallest Element Function Call

What is the output of the following Java code that finds the 3rd smallest element in the given BST?

DSA Java

class TreeNode {
    int val;
    TreeNode left, right;
    TreeNode(int x) { val = x; }
}

public class Solution {
    int count = 0;
    int result = -1;

    public int kthSmallest(TreeNode root, int k) {
        inorder(root, k);
        return result;
    }

    private void inorder(TreeNode node, int k) {
        if (node == null) return;
        inorder(node.left, k);
        count++;
        if (count == k) {
            result = node.val;
            return;
        }
        inorder(node.right, k);
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(5);
        root.left = new TreeNode(3);
        root.right = new TreeNode(6);
        root.left.left = new TreeNode(2);
        root.left.right = new TreeNode(4);
        root.left.left.left = new TreeNode(1);

        Solution sol = new Solution();
        System.out.println(sol.kthSmallest(root, 3));
    }
}

A 5

B 2

C 3

D 4

💡 Hint

Remember that inorder traversal of BST visits nodes in ascending order.

The inorder traversal visits nodes in ascending order: 1, 2, 3, 4, 5, 6. The 3rd smallest element is 3.

🧠 Conceptual

intermediate

Understanding Inorder Traversal in BST

Why does inorder traversal help in finding the kth smallest element in a BST?

A Because inorder traversal visits nodes in descending order of their values.

B Because inorder traversal visits nodes in ascending order of their values.

C Because inorder traversal visits nodes randomly.

D Because inorder traversal visits only leaf nodes.

💡 Hint

Think about the BST property and how inorder traversal works.

Inorder traversal of a BST visits nodes from smallest to largest, so counting nodes during traversal gives the kth smallest element.

🔧 Debug

advanced

Analyze the Kth Smallest Element Code

Which of the following is true about this code for finding the kth smallest element in a BST?

DSA Java

class TreeNode {
    int val;
    TreeNode left, right;
    TreeNode(int x) { val = x; }
}

public class Solution {
    int count = 0;
    int result = -1;

    public int kthSmallest(TreeNode root, int k) {
        inorder(root, k);
        return result;
    }

    private void inorder(TreeNode node, int k) {
        if (node == null) return;
        inorder(node.left, k);
        if (count == k) return;
        count++;
        result = node.val;
        inorder(node.right, k);
    }
}

A The code returns the kth smallest element correctly.

B The code causes a NullPointerException.

C The code returns -1 always.

D The code returns the largest element instead of kth smallest.

💡 Hint

Observe the order: recurse left, check count, increment and set if not, recurse right. Note how pruning works.

No error occurs. The code correctly returns the kth smallest element. It updates result for each of the first k nodes in inorder order (overwriting previous values), and the early return prunes subtrees after reaching k, preventing further updates.

❓ Predict Output

advanced

Output of Kth Smallest Element with Duplicate Values

What is the output of the following code when finding the 4th smallest element in a BST that contains duplicates?

DSA Java

class TreeNode {
    int val;
    TreeNode left, right;
    TreeNode(int x) { val = x; }
}

public class Solution {
    int count = 0;
    int result = -1;

    public int kthSmallest(TreeNode root, int k) {
        inorder(root, k);
        return result;
    }

    private void inorder(TreeNode node, int k) {
        if (node == null) return;
        inorder(node.left, k);
        count++;
        if (count == k) {
            result = node.val;
            return;
        }
        inorder(node.right, k);
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(5);
        root.left = new TreeNode(3);
        root.right = new TreeNode(6);
        root.left.left = new TreeNode(3);
        root.left.right = new TreeNode(4);
        root.left.left.left = new TreeNode(1);

        Solution sol = new Solution();
        System.out.println(sol.kthSmallest(root, 4));
    }
}

A 3

B 5

C 6

D 4

💡 Hint

Duplicates are counted as separate nodes in inorder traversal.

Inorder traversal order: 1, 3, 3, 4, 5, 6. The 4th smallest element is 4.

🧠 Conceptual

expert

Time Complexity of Kth Smallest Element in BST

What is the average time complexity of finding the kth smallest element in a balanced BST using inorder traversal?

A O(k)

B O(log n)

C O(n)

D O(k log n)

💡 Hint

Inorder traversal stops once kth element is found.

Inorder traversal visits nodes in ascending order and stops after kth node, so it visits only k nodes on average.

Practice

1. What does the term kth smallest element in a BST mean?

easy

A. The element with the kth highest frequency in the BST

B. The element that appears in the kth position when BST elements are sorted in ascending order

C. The element at the kth level of the BST

D. The element with the kth largest value in the BST

Solution

1. Step 1: Understand BST property

   A Binary Search Tree (BST) stores elements so that left child is smaller and right child is larger.

2. Step 2: Define kth smallest element

   The kth smallest element is the one that appears in the sorted order at position k.

3. Final Answer:

   The element that appears in the kth position when BST elements are sorted in ascending order -> Option B

4. Quick Check:

   kth smallest = kth in ascending order [OK]

Hint: Think of inorder traversal to get sorted order [OK]

Common Mistakes:

• Confusing kth smallest with kth largest
• Thinking kth smallest means kth level node
• Mixing frequency with element order

2. Which of the following Java method signatures correctly represents a function to find the kth smallest element in a BST?

easy

A. public int kthSmallest(TreeNode root, int k)

B. public void kthSmallest(TreeNode root, int k)

C. public TreeNode kthSmallest(int k, TreeNode root)

D. public int kthSmallest(int k)

Solution

1. Step 1: Check method parameters and return type

   The method should accept the BST root and integer k, and return the kth smallest element as int.

2. Step 2: Match signature with correct order

   public int kthSmallest(TreeNode root, int k) matches: returns int, parameters (TreeNode root, int k).

3. Final Answer:

   public int kthSmallest(TreeNode root, int k) -> Option A

4. Quick Check:

   Return type int and parameters (root, k) [OK]

Hint: Return the element value, input root and k [OK]

Common Mistakes:

• Using void return type instead of int
• Swapping parameter order incorrectly
• Missing root parameter

3. What will be the output of the following code snippet to find the 3rd smallest element in the BST?

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

public class Solution {
  int count = 0;
  int result = -1;
  public int kthSmallest(TreeNode root, int k) {
    inorder(root, k);
    return result;
  }
  private void inorder(TreeNode node, int k) {
    if (node == null) return;
    inorder(node.left, k);
    count++;
    if (count == k) {
      result = node.val;
      return;
    }
    inorder(node.right, k);
  }

  public static void main(String[] args) {
    TreeNode root = new TreeNode(5);
    root.left = new TreeNode(3);
    root.right = new TreeNode(6);
    root.left.left = new TreeNode(2);
    root.left.right = new TreeNode(4);
    root.left.left.left = new TreeNode(1);
    Solution sol = new Solution();
    System.out.println(sol.kthSmallest(root, 3));
  }
}

medium

A. 5

B. 2

C. 4

D. 3

Solution

1. Step 1: Perform inorder traversal to get sorted elements

   Inorder traversal order: 1, 2, 3, 4, 5, 6.

2. Step 2: Identify the 3rd element in sorted order

   The 3rd smallest element is 3.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Sorted BST elements = [1,2,3,4,5,6], 3rd = 3 [OK]

Hint: Inorder traversal gives sorted order; pick kth element [OK]

Common Mistakes:

• Confusing preorder or postorder with inorder
• Off-by-one error in counting
• Returning node instead of node value

4. Identify the bug in the following code that attempts to find the kth smallest element in a BST:

public int kthSmallest(TreeNode root, int k) {
  int count = 0;
  int result = -1;
  inorder(root, k, count, result);
  return result;
}

private void inorder(TreeNode node, int k, int count, int result) {
  if (node == null) return;
  inorder(node.left, k, count, result);
  count++;
  if (count == k) {
    result = node.val;
    return;
  }
  inorder(node.right, k, count, result);
}

medium

A. Primitive variables count and result are passed by value, so changes don't persist

B. The inorder traversal is missing the base case

C. The method should return void instead of int

D. The count variable should be reset to zero inside inorder

Solution

1. Step 1: Understand Java parameter passing

   Primitive types like int are passed by value, so changes inside methods don't affect original variables.

2. Step 2: Identify impact on count and result

   Since count and result are passed by value, their updates inside inorder do not persist outside, causing incorrect result.

3. Final Answer:

   Primitive variables count and result are passed by value, so changes don't persist -> Option A

4. Quick Check:

   Primitive int passed by value, no update outside method [OK]

Hint: Use class fields or wrappers to persist count and result [OK]

Common Mistakes:

• Assuming primitives are passed by reference
• Not using class variables or arrays for count/result
• Ignoring base case in recursion

5. Given a BST where nodes can have duplicate values, how would you modify the standard kth smallest element algorithm to correctly find the kth smallest unique element?

hard

A. Count all nodes including duplicates normally during inorder traversal

B. Use preorder traversal instead of inorder traversal

C. Use inorder traversal and skip duplicates by comparing current node value with previous visited value

D. Store all values in a list and sort it to find kth unique element

Solution

1. Step 1: Understand duplicates in BST

   Duplicates appear as repeated values; inorder traversal lists them in sorted order including duplicates.

2. Step 2: Modify traversal to skip duplicates

   Track previous visited value; only increment count when current value differs to count unique elements.

3. Step 3: Find kth unique element

   Increment count only for unique values; when count == k, return current node value.

4. Final Answer:

   Use inorder traversal and skip duplicates by comparing current node value with previous visited value -> Option C

5. Quick Check:

   Skip duplicates by comparing with previous value during inorder [OK]

Hint: Track previous value to count unique elements only [OK]

Common Mistakes:

• Counting duplicates as separate elements
• Using preorder traversal which is not sorted
• Sorting all values outside traversal (inefficient)