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77 lines
2.0 KiB
Plaintext
77 lines
2.0 KiB
Plaintext
---
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title: '230. Kth Smallest Element in a BST'
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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
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icon: dot
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topics:
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- { name: "Tree", slug: "tree" }
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- { name: "Depth-First Search", slug: "depth-first-search" }
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- { name: "Binary Search Tree", slug: "binary-search-tree" }
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- { name: "Binary Tree", slug: "binary-tree" }
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video: "5LUXSvjmGCw"
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sidebar:
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label: 'Kth Smallest Element in a BST'
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badge: 'Medium'
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---
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::::warning
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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?
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::::
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### Example 1:
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- Input: `root = [3,1,4,null,2], k = 1`
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- Output: `1`
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### Example 2:
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- Input: `root = [5,3,6,2,4,null,null,1], k = 3`
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- Output: `3`
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### Constraints:
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- The number of nodes in the tree is n.
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- `1 <= k <= n <= 10^4`
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- `0 <= Node.val <= 10^4`
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## Approach
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```mermaid
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flowchart TD
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S(["kthSmallest(root, k)"]) --> I["n = 0, stack = [], cur = root"]
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I --> W{"cur or stack?"}
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W -- no --> E(["fall off the loop — unreachable, k <= n"])
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W -- yes --> L{"cur?"}
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L -- yes --> P["stack.append(cur), cur = cur.left — push the whole left spine"]
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P --> L
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L -- no --> O["cur = stack.pop(), n += 1 — nodes pop in ascending order"]
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O --> K{"n == k?"}
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K -- yes --> R(["return cur.val"])
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K -- no --> N["cur = cur.right"]
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N --> W
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```
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## Solution
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```py
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
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n = 0
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stack = []
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cur = root
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while cur or stack:
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while cur:
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stack.append(cur)
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cur = cur.left
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cur = stack.pop()
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n += 1
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if n == k:
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return cur.val
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cur = cur.right
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```
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