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feat: refactored docs structure
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-77
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---
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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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issue: "https://github.com/prdlk/leetcode/issues/112"
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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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---
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title: '235. Lowest Common Ancestor of a Binary Search Tree'
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description: Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST
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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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- { name: "Binary Lifting", slug: "binary-lifting" }
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- { name: "Lowest Common Ancestor", slug: "lowest-common-ancestor" }
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video: "gs2LMfuOR9k"
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issue: "https://github.com/prdlk/leetcode/issues/113"
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sidebar:
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label: 'Lowest Common Ancestor of a Binary Search Tree'
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badge: 'Medium'
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---
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### Example 1:
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- Input: `root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8`
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- Output: `6`
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- Explanation: The LCA of nodes `2` and `8` is `6`.
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### Example 2:
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- Input: `root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4`
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- Output: `2`
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- Explanation: The LCA of nodes `2` and `4` is `2`, since a node can be a descendant of itself according to the LCA definition.
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### Example 3:
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- Input: `root = [2,1], p = 2, q = 1`
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- Output: `2`
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### Constraints:
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- The number of nodes in the tree is in the range [2, 10^5].
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- `-10^9 <= Node.val <= 10^9`
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- All Node.val are unique.
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- `p != q`
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- p and q will exist in the BST.
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## Approach
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```mermaid
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flowchart TD
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S(["lowestCommonAncestor(root, p, q)"]) --> I["cur = root"]
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I --> W{"cur?"}
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W -- no --> E(["fall off the loop — unreachable, p and q exist in the BST"])
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W -- yes --> A{"p.val < cur.val and q.val < cur.val?"}
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A -- yes --> L["cur = cur.left — both sit in the left subtree"]
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L --> W
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A -- no --> B{"p.val > cur.val and q.val > cur.val?"}
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B -- yes --> R["cur = cur.right — both sit in the right subtree"]
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R --> W
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B -- no --> C(["return cur — the paths to p and q split here"])
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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, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def lowestCommonAncestor(
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self, root: "TreeNode", p: "TreeNode", q: "TreeNode"
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) -> "TreeNode":
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cur = root
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while cur:
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if p.val < cur.val and q.val < cur.val:
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cur = cur.left
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elif p.val > cur.val and q.val > cur.val:
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cur = cur.right
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else:
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return cur
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```
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-69
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---
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title: '98. Validate Binary Search Tree'
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description: Given the root of a binary tree, determine if it is a valid binary search tree (BST)
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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: "s6ATEkipzow"
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issue: "https://github.com/prdlk/leetcode/issues/111"
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sidebar:
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label: 'Validate Binary Search Tree'
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badge: 'Medium'
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---
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### Example 1:
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- Input: `root = [2,1,3]`
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- Output: `true`
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### Example 2:
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- Input: `root = [5,1,4,null,null,3,6]`
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- Output: `false`
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- Explanation: The `root` node's value is `5` but its right child's value is `4`.
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### Constraints:
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- The number of nodes in the tree is in the range [1, 10^4].
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- `-2^31 <= Node.val <= 2^31 - 1`
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## Approach
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```mermaid
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flowchart TD
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S(["isValidBST(root)"]) --> C["valid(root, float('-inf'), float('inf')) — the root may hold any value"]
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C --> E(["return that result"])
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subgraph valid["valid(node, low, high)"]
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V1{"not node?"}
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V1 -- yes --> V2(["return True"])
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V1 -- no --> V3{"not (low < node.val < high)?"}
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V3 -- yes --> V4(["return False — node breaks a bound set by an ancestor"])
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V3 -- no --> V5["valid(node.left, low, node.val) and valid(node.right, node.val, high) — each child tightens one bound"]
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V5 --> V6(["return that result"])
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end
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C --> V1
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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 isValidBST(self, root: Optional[TreeNode]) -> bool:
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def valid(node, low, high):
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if not node:
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return True
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if not (low < node.val < high):
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return False
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return valid(node.left, low, node.val) and valid(node.right, node.val, high)
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return valid(root, float("-inf"), float("inf"))
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```
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@@ -1,5 +0,0 @@
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---
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title: Binary Search Tree
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description: 'Each node holds a value that is larger than its left subtree and smaller than its right subtree.'
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hidden: true
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---
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import { defineMeta } from "blume";
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export default defineMeta({ title: "Binary Search Tree", order: 13, icon: "git-fork", display: "page" });
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