---
title: '230. Kth Smallest Element in a BST'
description: Given the root of a binary search tree, and an integer k, return the k^th smallest value (1-indexed) of all the values of the nodes in the tree
icon: dot
topics:
  - { name: "Tree", slug: "tree" }
  - { name: "Depth-First Search", slug: "depth-first-search" }
  - { name: "Binary Search Tree", slug: "binary-search-tree" }
  - { name: "Binary Tree", slug: "binary-tree" }
video: "5LUXSvjmGCw"
issue: "https://github.com/prdlk/leetcode/issues/112"
sidebar:
  label: 'Kth Smallest Element in a BST'
  badge: 'Medium'
---

::::warning
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?
::::

### 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 <= 10^4`
- `0 <= Node.val <= 10^4`

## Approach

```mermaid
flowchart TD
  S(["kthSmallest(root, k)"]) --> I["n = 0, stack = [], cur = root"]
  I --> W{"cur or stack?"}
  W -- no --> E(["fall off the loop — unreachable, k <= n"])
  W -- yes --> L{"cur?"}
  L -- yes --> P["stack.append(cur), cur = cur.left — push the whole left spine"]
  P --> L
  L -- no --> O["cur = stack.pop(), n += 1 — nodes pop in ascending order"]
  O --> K{"n == k?"}
  K -- yes --> R(["return cur.val"])
  K -- no --> N["cur = cur.right"]
  N --> W
```

## Solution

```py
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
        n = 0
        stack = []
        cur = root

        while cur or stack:
            while cur:
                stack.append(cur)
                cur = cur.left

            cur = stack.pop()
            n += 1
            if n == k:
                return cur.val
            cur = cur.right
```