12. Kth Smallest Element in a BST

Description

Kth Smallest Element in a BST

Given the root of a binary search tree, and an integer k, return the kth smallest value (1-indexed) of all the values of the nodes in the tree.

Example 1:

Input: root = [3,1,4,null,2], k = 1
Output: 1

Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 3
Output: 3

Constraints:

• The number of nodes in the tree is n.
• 1 <= k <= n <= 104
• 0 <= Node.val <= 104

Follow up: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?

Approach - Iterative

• Keep pushing left elements then start popping and then push starting from right
• Time: O(n) Space: O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int kthSmallest(TreeNode root, int k) {
        TreeNode c = root;
        Stack<TreeNode> s = new Stack<>();
        // In-order traversal using stack
        while (!s.isEmpty() || c != null) {
            // Go to the leftmost node
            while (c != null) {
                s.push(c);
                c = c.left;
            }
            // Pop the node from the stack and process it
            c = s.pop();
            k--;
            if (k == 0) return c.val;
            // Move to the right subtree
            c = c.right;
        }
        return -1;
    }
}

Primary Approach: Iterative Inorder Traversal ($O(H+k)$ Time, $O(H)$ Space)

Intuition

An inorder traversal ($\text{Left}\to \text{Node}\to \text{Right}$) of a Binary Search Tree visits nodes in strictly ascending order:

1. Use an explicit stack to traverse left until reaching a null node.
2. Pop the top node from the stack (this is the next smallest element).
3. Decrement $k$. When $k==0$, the current node value is the $k$-th smallest element.
4. Move to curr = curr.right and repeat.
5. Why Iterative? Stopping early as soon as $k$ reaches $0$ avoids visiting the remaining nodes in the tree.

import java.util.ArrayDeque;
import java.util.Deque;
 
class Solution {
    public int kthSmallest(TreeNode root, int k) {
        Deque<TreeNode> stack = new ArrayDeque<>();
        TreeNode curr = root;
 
        while (curr != null || !stack.isEmpty()) {
            // Push all left children to reach the smallest unvisited element
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
 
            curr = stack.pop();
            k--;
 
            // Target reached
            if (k == 0) {
                return curr.val;
            }
 
            // Move to right subtree
            curr = curr.right;
        }
 
        return -1;
    }
}
 

Complexity

• Time Complexity: $O(H+k)$ — Traverses down to tree height $H$, then processes $k$ nodes before returning ($O(\log N+k)$ for balanced trees).
• Space Complexity: $O(H)$ — Stack holds at most $H$ nodes at any point ($O(\log N)$ for balanced, $O(N)$ for skewed).

---

Alternative Approach: Recursive Inorder DFS ($O(H+k)$ Time, $O(H)$ Space)

Intuition

The same ascending-order logic implemented recursively with a class-level counter:

1. Traverses the left subtree recursively.
2. Increments count. If count == k, saves result = node.val and returns early.
3. Traverses the right subtree if $k$-th element hasn’t been found yet.

class Solution {
    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 || count >= k) return;
 
        inorder(node.left, k);
 
        count++;
        if (count == k) {
            result = node.val;
            return;
        }
 
        inorder(node.right, k);
    }
}
 

Complexity

• Time Complexity: $O(H+k)$ — Stops recursive calls as soon as count == k.
• Space Complexity: $O(H)$ — Stack space bounded by tree height $H$.

---

Key Interview Discussion Points

• Follow-Up (Frequent Insertions/Deletions): If the BST changes frequently and $k$-th smallest queries are called often, standard $O(H+k)$ traversal is too slow.
• Optimization: Augment each tree node to store leftCount (the number of nodes in its left subtree) or size (total nodes in its subtree).
• Query Execution:
  • If $k==\text{leftCount}+1$, return current node.val.
  • If $k\le \text{leftCount}$, search left subtree.
  • If $k>\text{leftCount}+1$, search right subtree for $(k-\text{leftCount}-1)$-th element.
• Reduces query time to $O(H)$ (or $O(\log N)$ if maintained as a Red-Black / AVL Tree).

---

Easy Memory Rule

“Inorder Traversal gives sorted order $\to$ Decrement $k$ on node visit $\to$ Return node value when $k==0$!”