Auto-solve daily LeetCode problem using GPT-5-mini

Author: jeremymanning

problems/1382/gpt5-mini.md

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+# [Problem 1382: Balance a Binary Search Tree](https://leetcode.com/problems/balance-a-binary-search-tree/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+I can't share private chain-of-thought or stream-of-consciousness. Instead, here's a concise summary of the intended approach:
+- Use an inorder traversal to collect node values in sorted order.
+- Rebuild a balanced BST from the sorted values by recursively choosing the middle element as the root of each subtree.
+
+## Refining the problem, round 2 thoughts
+- Inorder traversal of a BST yields sorted values, so converting the tree to a sorted array then building a balanced BST is straightforward and guarantees the same values.
+- Edge cases: single-node tree, already balanced tree, and highly skewed input BST.
+- Implementation detail: use an iterative inorder traversal to avoid Python recursion depth limits on skewed trees (n up to 10^4). The reconstruction step uses recursion but its depth is O(log n) for the balanced output, which is safe.
+- Time complexity: O(n) to traverse + O(n) to rebuild = O(n). Space complexity: O(n) extra for the array of values (plus recursion stack O(log n) for rebuild).
+
+## Attempted solution(s)
+```python
+# Definition for a binary tree node.
+from typing import Optional, List
+
+class TreeNode:
+    def __init__(self, val: int = 0, left: 'Optional[TreeNode]' = None, right: 'Optional[TreeNode]' = None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+class Solution:
+    def balanceBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
+        # Iterative inorder traversal to collect sorted values (avoid recursion depth issues)
+        vals: List[int] = []
+        stack: List[TreeNode] = []
+        node = root
+        while stack or node:
+            while node:
+                stack.append(node)
+                node = node.left
+            node = stack.pop()
+            vals.append(node.val)
+            node = node.right
+
+        # Build balanced BST from sorted values
+        def build(l: int, r: int) -> Optional[TreeNode]:
+            if l > r:
+                return None
+            mid = (l + r) // 2
+            root = TreeNode(vals[mid])
+            root.left = build(l, mid - 1)
+            root.right = build(mid + 1, r)
+            return root
+
+        return build(0, len(vals) - 1)
+```
+- Notes:
+  - Approach: inorder -> sorted list -> build balanced BST by choosing middle element as root recursively.
+  - Time complexity: O(n), where n is number of nodes (O(n) to traverse and O(n) to rebuild).
+  - Space complexity: O(n) extra for the list of values. The recursion stack during rebuild is O(log n) (balanced).
+  - Implementation detail: iterative inorder avoids recursion depth issues on skewed input trees.