Kth Smallest Element in BST in DSA Java

Mental Model

A binary search tree keeps smaller values on the left and bigger on the right, so the smallest values appear first if we look in order. We find the kth smallest by counting nodes in order.

Analogy: Imagine a library where books are arranged by size from smallest on the left shelf to largest on the right. To find the kth smallest book, you start from the smallest shelf and count books one by one until you reach k.

      5
     / \
    3   7
   / \   \
  2   4   8
 ↑ (root)

Dry Run Walkthrough

Input: BST: 5 -> 3 -> 7 -> 2 -> 4 -> 8, find 3rd smallest element

Goal: Find the 3rd smallest value in the BST by visiting nodes in ascending order

Step 1: Traverse left from 5 to 3

5 -> [3 ↑] -> 7 -> 2 -> 4 -> 8

Why: Left subtree has smaller values, so start there to find smallest nodes

Step 2: Traverse left from 3 to 2

5 -> 3 -> 7 -> [2 ↑] -> 4 -> 8

Why: Keep going left to reach the smallest node

Step 3: Visit node 2 (count=1)

Visited 2 (1st smallest)

Why: First smallest node found

Step 4: Back to 3 and visit it (count=2)

Visited 3 (2nd smallest)

Why: Next smallest after 2 is 3

Step 5: Traverse right from 3 to 4

5 -> 3 -> 7 -> 2 -> [4 ↑] -> 8

Why: Check right child for next smallest

Step 6: Visit node 4 (count=3)

Visited 4 (3rd smallest)

Why: Found the kth smallest node

Result:

2 -> 3 -> [4] -> 5 -> 7 -> 8
Answer: 4

Annotated Code

DSA Java

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

public class KthSmallestInBST {
    private int count = 0;
    private 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); // traverse left subtree

        count++;
        if (count == k) {
            result = node.val; // found kth smallest
            return;
        }

        inorder(node.right, k); // traverse right subtree
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(5);
        root.left = new TreeNode(3);
        root.right = new TreeNode(7);
        root.left.left = new TreeNode(2);
        root.left.right = new TreeNode(4);
        root.right.right = new TreeNode(8);

        KthSmallestInBST solver = new KthSmallestInBST();
        int k = 3;
        int kthSmallest = solver.kthSmallest(root, k);
        System.out.println(kthSmallest);
    }
}

inorder(node.left, k); // traverse left subtree

advance to smaller nodes first to maintain ascending order

count++;

increment count when visiting a node to track order

if (count == k) { result = node.val; return; }

stop traversal and record value when kth smallest is found

inorder(node.right, k); // traverse right subtree

visit larger nodes after left and current node

Output

4

Complexity Analysis

Time: O(h + k) because we traverse down the tree height h and visit k nodes in order

Space: O(h) due to recursion stack for tree height h

vs Alternative: Naive approach sorts all nodes O(n log n), this uses BST property to stop early at kth node

Edge Cases

k is 1 (smallest element)

Returns the smallest node value by visiting leftmost node

DSA Java

if (count == k) { result = node.val; return; }

k equals number of nodes (largest element)

Visits all nodes in order and returns largest value

DSA Java

if (count == k) { result = node.val; return; }

Empty tree (root is null)

Returns -1 as no nodes exist

DSA Java

if (node == null) return;

When to Use This Pattern

When asked for the kth smallest or kth largest in a BST, use inorder traversal because it visits nodes in sorted order.

Common Mistakes

Mistake: Not stopping traversal after finding kth smallest, causing unnecessary work
Fix: Add a return after setting result when count == k to stop recursion

Mistake: Using preorder or postorder traversal which does not guarantee sorted order
Fix: Use inorder traversal to visit nodes in ascending order

Summary

Finds the kth smallest value in a binary search tree by counting nodes in inorder traversal.

Use when you need the kth smallest or kth largest element efficiently in a BST.

The key insight is that inorder traversal of BST visits nodes in ascending order.

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)