docs: rename category folders to remove parentheses and update heap meta title

apps/docs/content/binary-search-tree/230-kth-smallest-element-in-a-bst.mdx

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+---
+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
+sidebar:
+  label: 'Kth Smallest Element in a BST'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Tree</Badge>
+
+::::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
+```
+
+## Explanation
+
+<YouTube id="5LUXSvjmGCw" title="Kth Smallest Element in a BST - Leetcode 230 - Python" />

apps/docs/content/binary-search-tree/235-lowest-common-ancestor-of-a-binary-search-tree.mdx

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+---
+title: '235. Lowest Common Ancestor of a Binary Search Tree'
+description: Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST
+sidebar:
+  label: 'Lowest Common Ancestor of a Binary Search Tree'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Tree</Badge>
+
+### Example 1:
+- Input: `root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8`
+- Output: `6`
+- Explanation: The LCA of nodes `2` and `8` is `6`.
+
+### Example 2:
+- Input: `root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4`
+- Output: `2`
+- Explanation: The LCA of nodes `2` and `4` is `2`, since a node can be a descendant of itself according to the LCA definition.
+
+### Example 3:
+- Input: `root = [2,1], p = 2, q = 1`
+- Output: `2`
+
+### Constraints:
+
+- The number of nodes in the tree is in the range [2, 10^5].
+- `-10^9 <= Node.val <= 10^9`
+- All Node.val are unique.
+- `p != q`
+- p and q will exist in the BST.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["lowestCommonAncestor(root, p, q)"]) --> I["cur = root"]
+  I --> W{"cur?"}
+  W -- no --> E(["fall off the loop — unreachable, p and q exist in the BST"])
+  W -- yes --> A{"p.val < cur.val and q.val < cur.val?"}
+  A -- yes --> L["cur = cur.left — both sit in the left subtree"]
+  L --> W
+  A -- no --> B{"p.val > cur.val and q.val > cur.val?"}
+  B -- yes --> R["cur = cur.right — both sit in the right subtree"]
+  R --> W
+  B -- no --> C(["return cur — the paths to p and q split here"])
+```
+
+## Solution
+
+```py
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+
+class Solution:
+    def lowestCommonAncestor(
+        self, root: "TreeNode", p: "TreeNode", q: "TreeNode"
+    ) -> "TreeNode":
+        cur = root
+        while cur:
+            if p.val < cur.val and q.val < cur.val:
+                cur = cur.left
+            elif p.val > cur.val and q.val > cur.val:
+                cur = cur.right
+            else:
+                return cur
+```
+
+## Explanation
+
+<YouTube id="gs2LMfuOR9k" title="Lowest Common Ancestor of a Binary Search Tree - Leetcode 235 - Python" />

apps/docs/content/binary-search-tree/98-validate-binary-search-tree.mdx

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+---
+title: '98. Validate Binary Search Tree'
+description: Given the root of a binary tree, determine if it is a valid binary search tree (BST)
+sidebar:
+  label: 'Validate Binary Search Tree'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Tree</Badge>
+
+### Example 1:
+- Input: `root = [2,1,3]`
+- Output: `true`
+
+### Example 2:
+- Input: `root = [5,1,4,null,null,3,6]`
+- Output: `false`
+- Explanation: The `root` node's value is `5` but its right child's value is `4`.
+
+### Constraints:
+
+- The number of nodes in the tree is in the range [1, 10^4].
+- `-2^31 <= Node.val <= 2^31 - 1`
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["isValidBST(root)"]) --> C["valid(root, float('-inf'), float('inf')) — the root may hold any value"]
+  C --> E(["return that result"])
+  subgraph valid["valid(node, low, high)"]
+    V1{"not node?"}
+    V1 -- yes --> V2(["return True"])
+    V1 -- no --> V3{"not (low < node.val < high)?"}
+    V3 -- yes --> V4(["return False — node breaks a bound set by an ancestor"])
+    V3 -- no --> V5["valid(node.left, low, node.val) and valid(node.right, node.val, high) — each child tightens one bound"]
+    V5 --> V6(["return that result"])
+  end
+  C --> V1
+```
+
+## 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 isValidBST(self, root: Optional[TreeNode]) -> bool:
+        def valid(node, low, high):
+            if not node:
+                return True
+
+            if not (low < node.val < high):
+                return False
+
+            return valid(node.left, low, node.val) and valid(node.right, node.val, high)
+
+        return valid(root, float("-inf"), float("inf"))
+```
+
+## Explanation
+
+<YouTube id="s6ATEkipzow" title="Validate Binary Search Tree - Depth First Search - Leetcode 98" />

apps/docs/content/binary-search-tree/meta.ts

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+import { defineMeta } from "blume";
+
+export default defineMeta({ title: "Binary Search Tree", order: 13 });