Converting Sorted Array to Balanced BST with Divide & Conquer

Day 72 of #100DaysOfCode Problem: Convert Sorted Array to Height-Balanced BST Today I learned how to efficiently convert a sorted array into a balanced Binary Search Tree using Divide & Conquer. Key Insight: Pick the middle element as the root to maintain balance. Recursively build: Left subtree from left half Right subtree from right half This ensures: Optimal height Faster search operations ⏱ Time Complexity: O(n) 📦 Space Complexity: O(log n) Consistency is the real game changer #DSA #Java #BinaryTree #LeetCode #CodingJourney

  • graphical user interface, application

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