Kth Smallest Element in BST in DSA Go - 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 the Kth position if all the elements of the BST were sorted in ascending order.

beginner

Why does an inorder traversal of a BST give elements in sorted order?

Because inorder traversal visits nodes in the order: left child, current node, right child. In a BST, left child nodes are smaller and right child nodes are larger, so this order produces sorted elements.

intermediate

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

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

advanced

How can you optimize finding the Kth smallest element if the BST is frequently queried?

By augmenting each node with the count of nodes in its left subtree, you can decide whether to go left, right, or return the current node without full traversal, reducing time complexity to O(H).

intermediate

What is the role of a stack in iterative inorder traversal to find the Kth smallest element?

The stack keeps track of nodes to visit next, simulating the recursive call stack. It helps traverse the tree without recursion, visiting nodes in sorted order until the Kth element is found.

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

A Level order traversal

B Preorder traversal

C Postorder traversal

D Inorder traversal

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

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

A The largest element

B The root element

C The smallest element

D The median element

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

What data structure is commonly used to implement iterative inorder traversal?

A Queue

B Stack

C Linked list

D Hash map

A stack is used to keep track of nodes during iterative inorder traversal.

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

A O(N)

B O(1)

C O(log N)

D O(N log N)

In a skewed BST, height H can be N, so time complexity is O(N).

How can augmenting BST nodes with subtree counts help in finding the Kth smallest element?

A It helps skip unnecessary nodes

B It sorts the tree

C It balances the tree

D It stores the Kth element directly

Subtree counts allow deciding which subtree to explore, skipping nodes that are not needed.

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

Think about the order nodes are visited in a BST.

Describe one optimization to improve the efficiency of finding the Kth smallest element in a BST.

Consider adding extra information to nodes.

Practice

1. What does the term Kth Smallest Element in BST mean?

easy

A. The element with the highest frequency in the BST

B. The element at the root of the BST

C. The largest element in the BST

D. The element that is the Kth smallest when BST elements are sorted in ascending order

Solution

1. Step 1: Understand BST property

   A Binary Search Tree (BST) stores elements so that left child nodes are smaller and right child nodes are larger than the parent node.

2. Step 2: Define Kth smallest element

   The Kth smallest element is the element that appears in the Kth position when all elements are sorted in ascending order.

3. Final Answer:

   The element that is the Kth smallest when BST elements are sorted in ascending order -> Option D

4. Quick Check:

   Kth smallest = Kth element in sorted order [OK]

Hint: Think in sorted order of BST elements [OK]

Common Mistakes:

• Confusing Kth smallest with root element
• Thinking it means largest element
• Mixing frequency with order

2. Which of the following Go function signatures correctly represents a function to find the Kth smallest element in a BST node structure named TreeNode?

easy

A. func kthSmallest(root *TreeNode) int

B. func kthSmallest(root *TreeNode, k int) int

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

D. func kthSmallest(k int, root *TreeNode) int

Solution

1. Step 1: Check parameter order and types

   The function should take the root pointer of type *TreeNode and an integer k to specify the Kth element.

2. Step 2: Check return type

   The function returns an int representing the value of the Kth smallest element.

3. Final Answer:

   func kthSmallest(root *TreeNode, k int) int -> Option B

4. Quick Check:

   Correct parameters and return type [OK]

Hint: Root pointer first, then k, return int [OK]

Common Mistakes:

• Swapping parameter order
• Passing root by value instead of pointer
• Missing k parameter
• Wrong return type

3. Given the BST below and k = 3, what will the output of kthSmallest(root, 3) be?

      5
     / \
    3   7
   / \   \
  2   4   8

medium

A. 4

B. 5

C. 3

D. 7

Solution

1. Step 1: Perform in-order traversal

   In-order traversal visits nodes in ascending order: 2, 3, 4, 5, 7, 8.

2. Step 2: Identify the 3rd element

   The 3rd element in this order is 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   In-order 3rd element = 4 [OK]

Hint: In-order traversal gives sorted order [OK]

Common Mistakes:

• Counting root as first without traversal
• Confusing pre-order or post-order with in-order
• Off-by-one counting errors

4. Consider this Go function snippet to find the Kth smallest element in a BST:

func kthSmallest(root *TreeNode, k int) int {
    count := 0
    var result int
    var inorder func(node *TreeNode)
    inorder = func(node *TreeNode) {
        if node == nil {
            return
        }
        inorder(node.Left)
        count++
        if count == k {
            result = node.Val
            return
        }
        inorder(node.Right)
    }
    inorder(root)
    return result
}

What is the main bug in this code?

medium

A. The function does not stop traversal after finding the Kth element

B. The count variable is not incremented correctly

C. The result variable is not returned

D. The inorder function is not called

Solution

1. Step 1: Analyze traversal stopping condition

   The code sets result when count == k but does not stop further recursion, so it continues traversing unnecessarily.

2. Step 2: Identify missing early return

   After finding the Kth element, the function should stop traversing to avoid overwriting result or extra work.

3. Final Answer:

   The function does not stop traversal after finding the Kth element -> Option A

4. Quick Check:

   Traversal continues after Kth found [OK]

Hint: Stop traversal immediately after finding Kth element [OK]

Common Mistakes:

• Not stopping recursion after finding result
• Incorrect count increment placement
• Returning wrong variable

5. You want to find the Kth smallest element in a BST but the tree is very large and you want to optimize the search. Which approach below best improves efficiency?

hard

A. Convert the BST to a sorted array and then pick the Kth element

B. Perform a full in-order traversal every time to find the Kth element

C. Store the count of nodes in each subtree and use it to skip branches

D. Use breadth-first search to find the Kth smallest element

Solution

1. Step 1: Understand subtree counts

   By storing the number of nodes in each subtree, we can decide if the Kth smallest is in left subtree, current node, or right subtree without full traversal.

2. Step 2: Use counts to skip branches

   If left subtree has more than or equal to k nodes, recurse left; else adjust k and recurse right, reducing unnecessary visits.

3. Final Answer:

   Store the count of nodes in each subtree and use it to skip branches -> Option C

4. Quick Check:

   Subtree counts optimize search [OK]

Hint: Use subtree node counts to jump branches [OK]

Common Mistakes:

• Doing full traversal every time
• Converting to array wastes space/time
• Using BFS which is not sorted order