Leetcode - 230. Kth Smallest Element in a BST

🧠 Approach: Inorder Traversal (Because BST is Sorted Inorder)

In a BST, an inorder traversal returns nodes in ascending order.
So we:

1. Do an inorder traversal.
2. Keep a counter as we visit nodes.
3. When the counter hits k, we’ve found the kth smallest element.

We short-circuit the recursion as soon as we find the result for efficiency.

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✅ Code (JavaScript):

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val === undefined ? 0 : val);
 *     this.left = (left === undefined ? null : left);
 *     this.right = (right === undefined ? null : right);
 * }
 */

/**
 * @param {TreeNode} root
 * @param {number} k
 * @return {number}
 */
var kthSmallest = function (root, k) {
  let count = 0;
  let result = null;

  function inorder(node) {
    if (!node || result !== null) return;

    inorder(node.left);

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

    inorder(node.right);
  }

  inorder(root);
  return result;
};

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🧪 Example:

For this BST:

     5
    / \
   3   6
  / \
 2   4
/
1

Inorder traversal: [1, 2, 3, 4, 5, 6]

• k = 3 → result = 3

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⏱ Time & Space Complexity

Type	Complexity
Time	O(h + k) in the best case (early exit), O(n) worst case
Space	O(h) for recursion stack (h = height of tree)

In a balanced BST, h = log n. In worst case (like a skewed tree), h = n.

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