LeetCode 110 | Balanced Binary Tree 🔥 | One-Pass DFS Explained | O(n) Solution скачать в хорошем качестве

LeetCode 110 | Balanced Binary Tree 🔥 | One-Pass DFS Explained | O(n) Solution

In this video, we solve LeetCode 110 – Balanced Binary Tree using an efficient one-pass DFS approach 🚀 A binary tree is considered height-balanced if, for every node, the height difference between its left and right subtree is at most 1. 💡 Key Highlights: • Bottom-up DFS strategy ✅ • Using -1 to detect imbalance early • Clean and optimized Java solution ⏱ Time Complexity: O(n) 📦 Space Complexity: O(h) (recursion stack) Solution: https://leetcode.com/problems/balance... This problem is perfect for understanding tree recursion, height calculation, and DFS optimization — a must-know concept for coding interviews! 👉 Like 👍 | Share 🔁 | Subscribe 🔔 for daily LeetCode POTD solutions. #codewithkhan #codinginterview #dsa #leetcode #problemsolving #competitiveprogramming #leetcodepotd #algorithms #datastructures #dailycoding #LeetCode110 #BalancedBinaryTree #TreeDFS #BinaryTree #DataStructures #CodingInterview #Java #LeetCodePOTD #DSA #CodeWithKhan #Algorithm #OofN