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

Recall & Review

beginner

What is the Kth smallest element in a Binary Search Tree (BST)?

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

beginner

Which traversal method is best suited to find the Kth smallest element in a BST?

Inorder traversal, because it visits nodes in ascending order for BSTs.

beginner

In the context of finding the Kth smallest element, what does the inorder traversal do?

It visits nodes from smallest to largest, so counting nodes during inorder traversal helps identify the Kth smallest element.

intermediate

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

O(H + K), where H is the height of the tree, because traversal stops once the Kth element is found.

advanced

How can you optimize repeated queries for the Kth smallest element in a BST?

By augmenting each node with the count of nodes in its left subtree, allowing O(H) time queries without full traversal.

Which traversal order visits BST nodes in ascending order?

A Postorder

B Preorder

C Level order

D Inorder

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

What is the first step to find the Kth smallest element in a BST?

A Perform inorder traversal

B Perform preorder traversal

C Perform postorder traversal

D Perform level order traversal

Inorder traversal visits nodes in ascending order, making it suitable to find the Kth smallest element.

If K is 1, what element does the Kth smallest element represent in a BST?

A The smallest element

B The largest element

C The root element

D The median element

The 1st smallest element is the smallest element in the BST.

What data structure can help optimize multiple Kth smallest queries in a BST?

A Hash map

B Stack

C Augmented BST with subtree counts

D Queue

Augmenting BST nodes with counts of left subtree nodes allows quick navigation to the Kth smallest element.

What is the worst-case time complexity to find the Kth smallest element in an unbalanced BST using inorder traversal?

A O(log N)

B O(N)

C O(1)

D O(K log N)

In the worst case (unbalanced tree), inorder traversal may visit all nodes, so O(N).

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 one way to optimize repeated Kth smallest element queries in a BST.

Consider adding extra information to nodes.

Practice

1. What does the Kth Smallest Element in BST problem ask you to find?

easy

A. The height of the Binary Search Tree

B. The element that is the Kth smallest value in the Binary Search Tree

C. The largest element in the Binary Search Tree

D. The sum of all elements in the Binary Search Tree

Solution

1. Step 1: Understand the problem statement

   The problem asks to find the element that is the Kth smallest in the BST, meaning the element that would appear in position K if the BST elements were sorted.

2. Step 2: Clarify what Kth smallest means in BST context

   Since BST stores elements in sorted order when traversed in-order, the Kth smallest element is the Kth element visited in an in-order traversal.

3. Final Answer:

   The element that is the Kth smallest value in the Binary Search Tree -> Option B

4. Quick Check:

   Kth smallest element = Kth element in in-order traversal [OK]

Hint: Think in-order traversal gives sorted order [OK]

Common Mistakes:

• Confusing Kth smallest with Kth largest
• Thinking it's about tree height
• Summing elements instead of finding one

2. Which of the following C++ function signatures is correct for a function that finds the Kth smallest element in a BST node structure defined as struct TreeNode { int val; TreeNode* left; TreeNode* right; };?

easy

A. int kthSmallest(TreeNode* root, int k);

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

C. int kthSmallest(TreeNode root, int* k);

D. int kthSmallest(TreeNode* root);

Solution

1. Step 1: Check parameter types for BST traversal

   The function should receive a pointer to the root node to traverse the BST properly, so TreeNode* root is correct.

2. Step 2: Check parameters for K value

   K is an integer input, so passing it as int k is correct. The function returns the Kth smallest value as an integer.

3. Final Answer:

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

4. Quick Check:

   Pointer to root + int k parameter = correct signature [OK]

Hint: Use pointer for tree root, int for K [OK]

Common Mistakes:

• Passing root by value instead of pointer
• Missing K parameter
• Wrong return type or missing return

3. Consider the following BST and code snippet. What will be the output?

struct TreeNode {
  int val;
  TreeNode* left;
  TreeNode* right;
  TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

int count = 0;
int result = -1;

void inorder(TreeNode* root, int k) {
  if (!root) return;
  inorder(root->left, k);
  count++;
  if (count == k) {
    result = root->val;
    return;
  }
  inorder(root->right, k);
}

int main() {
  TreeNode* root = new TreeNode(3);
  root->left = new TreeNode(1);
  root->right = new TreeNode(4);
  root->left->right = new TreeNode(2);
  inorder(root, 2);
  std::cout << result << std::endl;
  return 0;
}

medium

A. 2

B. 1

C. 3

D. 4

Solution

1. Step 1: Perform in-order traversal on the BST

   The BST elements in in-order are visited as: 1, 2, 3, 4.

2. Step 2: Identify the 2nd element visited

   The 1st element visited is 1, the 2nd is 2, so result will be set to 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   In-order 2nd element = 2 [OK]

Hint: In-order traversal visits nodes in sorted order [OK]

Common Mistakes:

• Confusing pre-order or post-order traversal
• Off-by-one error in counting
• Not stopping traversal after finding Kth element

4. The following code attempts to find the Kth smallest element in a BST but does not work correctly. What is the main issue?

int count = 0;
int result = -1;

void inorder(TreeNode* root, int k) {
  if (!root) return;
  inorder(root->left, k);
  if (count == k) {
    result = root->val;
    return;
  }
  count++;
  inorder(root->right, k);
}

medium

A. The function uses global variables which is not allowed

B. The function does not handle null root correctly

C. The result variable is not updated inside the function

D. The check for count == k is before incrementing count, so it misses the Kth element

Solution

1. Step 1: Analyze the order of count checks and increments

   The code checks if count == k before incrementing count, so it skips setting result when count reaches k.

2. Step 2: Understand impact on finding Kth element

   Because count is checked before increment, the Kth element is never assigned to result, causing incorrect behavior.

3. Final Answer:

   The check for count == k is before incrementing count, so it misses the Kth element -> Option D

4. Quick Check:

   Check count after increment, not before [OK]

Hint: Increment count before checking if equals K [OK]

Common Mistakes:

• Checking count before incrementing
• Ignoring early return after finding Kth element
• Assuming global variables cause error

5. You want to find the Kth smallest element in a BST efficiently without using extra space for storing all elements. Which approach is best and why? A) Perform an in-order traversal and store all elements in a vector, then return the Kth element. B) Use a recursive in-order traversal with a counter to stop once the Kth element is found. C) Convert the BST to a sorted array using any traversal and then binary search for the Kth element. D) Use a breadth-first traversal and count nodes until K is reached.

hard

A. Store all elements then pick Kth; uses extra space

B. Convert to array then binary search; unnecessary overhead

C. Recursive in-order traversal with counter; stops early

D. Breadth-first traversal; does not guarantee sorted order

Solution

1. Step 1: Evaluate space and time efficiency of each approach

   Store all elements then pick Kth; uses extra space to store all elements, which is not efficient. Convert to array then binary search; unnecessary overhead also uses extra space and unnecessary binary search. Breadth-first traversal; does not guarantee sorted order, so it won't find the Kth smallest correctly.

2. Step 2: Identify the best approach

   Recursive in-order traversal with counter; stops early uses recursive in-order traversal which visits nodes in sorted order and stops once the Kth element is found, saving time and space.

3. Final Answer:

   Recursive in-order traversal with counter; stops early -> Option C

4. Quick Check:

   In-order traversal + early stop = efficient Kth smallest [OK]

Hint: Use in-order traversal with counter and stop early [OK]

Common Mistakes:

• Using extra space unnecessarily
• Using traversal that doesn't preserve order
• Not stopping traversal after finding Kth element