LeetCode 1382 | Balance a Binary Search Tree | Java | Medium | In-Order Traversal | CodeByTushu
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In today’s video, we solve LeetCode 1382 — Balance a Binary Search Tree using In-Order Traversal and Divide and Conquer in Java.
We are given the root of an arbitrary binary search tree that may be completely skewed or unbalanced, and we must construct and return a height-balanced BST containing the exact same node values. The foundational property of any binary search tree is that an in-order traversal (left, root, right) visits all nodes in strictly non-decreasing sorted order.
By performing an in-order depth-first traversal, we first flatten the tree values into a sorted list in linear time. Next, using a recursive divide-and-conquer strategy similar to binary search, we select the middle element of the sorted list as the subtree root and recursively build the left and right subtrees from the left and right halves. This guarantees that subproblem sizes differ by at most one at every step, creating a perfectly balanced BST in optimal O(n) time and O(n) auxiliary space.
🔥 Problem Highlights
✅ Leveraged in-order traversal to extract BST values in sorted order in O(n) time
✅ Applied divide and conquer by picking the median element as root at each level
✅ Guaranteed minimal tree height of ceil(log2(n + 1)) strictly satisfying height balance
✅ Avoided complex AVL or Red-Black tree pointer tree rotations
✅ Clean, modular Java implementation with clear helper methods
✅ Achieved optimal O(n) time and O(n) auxiliary space complexity
⏱️ TIMESTAMPS
00:00 Intro
00:58 Problem Understanding & Constraints
01:55 BST Inorder Traversal & Divide & Conquer
03:10 Two-Phase Rebalancing Strategy
04:25 Java Code Walkthrough
05:58 Execution & Test Cases
07:31 Time Complexity Analysis (O(n))
08:17 Space Complexity Analysis (O(n))
09:01 Outro & Wrap-up
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/February/9-February/1382-Balance-a-Binary-Search-Tree
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/9-February/1382-Balance-a-Binary-Search-Tree/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/9-February/1382-Balance-a-Binary-Search-Tree/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/9-February/1382-Balance-a-Binary-Search-Tree/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/9-February/1382-Balance-a-Binary-Search-Tree/solution.js
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🔥 Perfect For
LeetCode Medium Problems
Java DSA Preparation
Binary Search Tree Invariants
In-Order DFS Traversal
Divide and Conquer Tree Reconstruction
Time Complexity Optimization
Coding Interview Preparation
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💬 Question of the Day
Why does an in-order traversal of a binary search tree always yield elements in strictly sorted order?
1️⃣ Because the left subtree contains strictly smaller elements and the right subtree contains strictly greater elements than the root
2️⃣ Because binary search trees require an internal sort operation on every traversal
3️⃣ Because nodes are indexed by their insertion timestamp
4️⃣ Because in-order traversal sorts elements during the backtrack step
#LeetCode1382 #LeetCode #BalanceABinarySearchTree #Java #JavaDSA #Tree #BinaryTree #DFS #Recursion #Array #BinarySearchTree #DSA #CodingInterview #ProblemSolving #LeetCodeMedium #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
In today’s video, we solve LeetCode 1382 — Balance a Binary Search Tree using In-Order Traversal and Divide and Conquer in Java.
We are given the root of an arbitrary binary search tree that may be completely skewed or unbalanced, and we must construct and return a height-balanced BST containing the exact same node values. The foundational property of any binary search tree is that an in-order traversal (left, root, right) visits all nodes in strictly non-decreasing sorted order.
By performing an in-order depth-first traversal, we first flatten the tree values into a sorted list in linear time. Next, using a recursive divide-and-conquer strategy similar to binary search, we select the middle element of the sorted list as the subtree root and recursively build the left and right subtrees from the left and right halves. This guarantees that subproblem sizes differ by at most one at every step, creating a perfectly balanced BST in optimal O(n) time and O(n) auxiliary space.
🔥 Problem Highlights
✅ Leveraged in-order traversal to extract BST values in sorted order in O(n) time
✅ Applied divide and conquer by picking the median element as root at each level
✅ Guaranteed minimal tree height of ceil(log2(n + 1)) strictly satisfying height balance
✅ Avoided complex AVL or Red-Black tree pointer tree rotations
✅ Clean, modular Java implementation with clear helper methods
✅ Achieved optimal O(n) time and O(n) auxiliary space complexity
⏱️ TIMESTAMPS
00:00 Intro
00:58 Problem Understanding & Constraints
01:55 BST Inorder Traversal & Divide & Conquer
03:10 Two-Phase Rebalancing Strategy
04:25 Java Code Walkthrough
05:58 Execution & Test Cases
07:31 Time Complexity Analysis (O(n))
08:17 Space Complexity Analysis (O(n))
09:01 Outro & Wrap-up
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/February/9-February/1382-Balance-a-Binary-Search-Tree
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/9-February/1382-Balance-a-Binary-Search-Tree/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/9-February/1382-Balance-a-Binary-Search-Tree/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/9-February/1382-Balance-a-Binary-Search-Tree/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/9-February/1382-Balance-a-Binary-Search-Tree/solution.js
🔗 LeetCode Source Code
📌 Admin type the link
🌐 All LeetCode Solutions
📌 https://codebytushu.com/Leetcode/
📱 Join CodeByTushu on WhatsApp
Get coding updates, LeetCode solutions, daily questions and more.
https://whatsapp.com/channel/0029Vb8k5mCDeON2Xwlkxe40
❤️ Donate
📌 https://codebytushu.com/Leetcode/donate.php
🔥 Perfect For
LeetCode Medium Problems
Java DSA Preparation
Binary Search Tree Invariants
In-Order DFS Traversal
Divide and Conquer Tree Reconstruction
Time Complexity Optimization
Coding Interview Preparation
Problem Solving
🌐 Follow CodeByTushu
▶️ YouTube: https://www.youtube.com/@codebytushu
📸 Instagram: https://instagram.com/codebytushu/
💼 LinkedIn: https://www.linkedin.com/company/codebytushu/
🌍 Website: https://codebytushu.com
🐙 GitHub: https://github.com/CodeByTushu
❤️ Don’t Forget To
👍 Like | 💬 Comment | 🔔 Subscribe
💬 Question of the Day
Why does an in-order traversal of a binary search tree always yield elements in strictly sorted order?
1️⃣ Because the left subtree contains strictly smaller elements and the right subtree contains strictly greater elements than the root
2️⃣ Because binary search trees require an internal sort operation on every traversal
3️⃣ Because nodes are indexed by their insertion timestamp
4️⃣ Because in-order traversal sorts elements during the backtrack step
#LeetCode1382 #LeetCode #BalanceABinarySearchTree #Java #JavaDSA #Tree #BinaryTree #DFS #Recursion #Array #BinarySearchTree #DSA #CodingInterview #ProblemSolving #LeetCodeMedium #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
