Kth Smallest Element in BST in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a Binary Search Tree (BST)?

A Binary Search Tree is a tree where each node has at most two children. For every node, all nodes in its left subtree have smaller values, and all nodes in its right subtree have larger values.

beginner

What does 'Kth smallest element in BST' mean?

It means finding the element that would appear in position K if all elements of the BST were sorted in ascending order.

beginner

Which traversal method helps to find the Kth smallest element in a BST?

Inorder traversal visits nodes in ascending order for a BST, so it helps to find the Kth smallest element by counting nodes during traversal.

intermediate

What is the time complexity of finding the Kth smallest element using inorder traversal?

The time complexity is O(H + K), where H is the height of the tree, because we traverse down to the leftmost node and then visit K nodes in order.

beginner

Why is inorder traversal suitable for finding the Kth smallest element in a BST?

Because inorder traversal visits nodes in sorted order (smallest to largest), so counting nodes during this traversal directly gives the Kth smallest element.

Which traversal order visits BST nodes in ascending order?

A Preorder

B Postorder

C Inorder

D Level order

Inorder traversal visits left subtree, then node, then right subtree, which results in ascending order for BST.

What does the 'K' represent in 'Kth smallest element'?

A The value of the element

B The height of the BST

C The number of nodes in the BST

D The position in sorted order

'K' is the position of the element when all BST elements are sorted from smallest to largest.

If a BST has 10 nodes, what is the smallest possible K to find the smallest element?

A 0

B 1

C 10

D 11

K starts from 1 for the smallest element, so K=1 gives the smallest element.

What is the main idea to find the Kth smallest element in a BST?

A Use inorder traversal and count nodes

B Use preorder traversal and count nodes

C Use postorder traversal and count nodes

D Use level order traversal and count nodes

Inorder traversal visits nodes in ascending order, so counting nodes during inorder traversal finds the Kth smallest.

What is the space complexity of recursive inorder traversal for a BST of height H?

A O(H)

B O(1)

C O(N)

D O(log N)

Recursive calls stack up to the height of the tree, so space complexity is O(H).

Explain how inorder traversal helps find the Kth smallest element in a BST.

Think about the order nodes are visited and how counting helps.

Describe the steps to find the Kth smallest element in a BST using recursion.

Focus on the recursive calls and counting logic.

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)