Java LeetCode 145 Binary Tree Postorder Traversal Solution

Day 34 of #100DaysOfLeetCode 💻✅ Solved #110. Balanced Binary Tree on LeetCode using Java. Approach: • Used a bottom-up recursive approach to calculate height • Returned -1 immediately if any subtree is unbalanced • Compared left and right subtree heights at each node • Checked if the height difference is greater than 1 • Stopped early to optimize unnecessary computations Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) ✓ Memory: 45.33 MB (Beats 95.41% submissions) Key Learning: ✓ Understood how to combine height calculation with balance checking ✓ Learned early termination technique in recursion ✓ Improved problem-solving for tree-based recursive problems Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinaryTree #Recursion #TreeProblems #ProblemSolving #CodingJourney #100DaysOfCode

Day 36 of #100DaysOfLeetCode 💻✅ Solved #111. Minimum Depth of Binary Tree on LeetCode using Java. Approach: • Used recursive approach to calculate minimum depth • Returned 0 when root is null (base case) • If one subtree is null, avoided taking minimum of 0 (handled edge case carefully) • Returned 1 + minimum depth of valid subtree • Ensured correct handling of skewed trees Performance: ✓ Runtime: 6 ms ✓ Memory: 82.20 MB Key Learning: ✓ Understood difference between minimum depth and maximum depth logic ✓ Learned importance of handling null subtree cases correctly ✓ Improved confidence in solving binary tree recursion problems Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinaryTree #Recursion #TreeProblems #ProblemSolving #CodingJourney #100DaysOfCode

Day 56 of #100DaysOfLeetCode 💻✅ Solved #205. Isomorphic Strings problem in Java. Approach: • Used two arrays to track character mappings for both strings • Traversed both strings simultaneously • Checked if the mapping values at current characters are equal • If not equal, returned false (mapping mismatch) • Updated both arrays with the current index + 1 • This ensures consistent one-to-one mapping between characters Performance: ✓ Runtime: 8 ms (Beats 72.19% submissions) ✓ Memory: 44.16 MB (Beats 22.15% submissions) Key Learning: ✓ Learned how to maintain mapping consistency between two strings ✓ Practiced using arrays for character indexing ✓ Strengthened understanding of string pattern matching problems Learning one problem every single day 🚀 #Java #LeetCode #DSA #Strings #ProblemSolving #CodingJourney #100DaysOfCode

🚀 Day 98 - #100DaysOfCode Today’s problem was all about string comparison and operations simulation in Java. 💡 Problem Insight: Given a list of operations like "++X", "X++", "--X", "X--", we need to compute the final value of X after performing all operations. ⚠️ One key learning today: In Java, always use .equals() for string comparison instead of ==. Using == compares references, not actual content — a very common mistake! 🧠 Approach: Initialize x = 0 Traverse through each operation Increment or decrement based on the operation string 📌 What I Improved Today: Better understanding of string handling in Java Avoiding common pitfalls in comparisons Writing cleaner conditional logic #Java #CodingJourney #LeetCode #100DaysOfCode #ProblemSolving #Consistency

🚀 Day 58 / 100 – LeetCode Challenge ✅ Problem Solved: Binary Tree Postorder Traversal 📊 Difficulty: Easy 💻 Language: Java Today’s problem focused on understanding tree traversal, specifically Postorder Traversal (Left → Right → Root). 🔍 Key Learnings: Mastered recursive approach for tree traversal Understood how DFS (Depth First Search) works in binary trees Strengthened problem-solving with clean and efficient logic ⚡ Performance: Runtime: 0 ms (Beats 100%) 🔥 Memory: 42.96 MB 💡 Approach: Used a simple recursive helper function: Traverse left subtree Traverse right subtree Add root node value This problem may be easy, but it builds a strong foundation for advanced tree problems 💪 📈 Consistency is key — 58 days down, 42 more to go! #LeetCode #100DaysOfCode #Java #DataStructures #CodingJourney #BinaryTree #DSA #Learning #Consistency

Day 47 of #100DaysOfLeetCode 💻✅ Solved #101. Symmetric Tree problem in Java. Approach: • Used recursion to compare the left and right subtrees of the root • If both nodes are null, the tree is symmetric at that level • If one node is null and the other is not, the tree is not symmetric • Compared the values of both nodes • Recursively checked left.left with right.right and left.right with right.left to verify mirror structure Performance: ✓ Runtime: 0 ms (Beats 100% submissions) ✓ Memory: 43.62 MB (Beats 48.60% submissions) Key Learning: ✓ Practiced mirror comparison in binary trees ✓ Learned how recursion can verify symmetry between two subtrees ✓ Strengthened understanding of tree traversal and structural comparison Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinaryTrees #ProblemSolving #CodingJourney #100DaysOfCode

Refactoring for Clarity: Array Manipulation in Java 👨💻 I’ve just finished a practical exercise on array manipulation. While the goal was simple (summing two arrays), I used it as an opportunity to apply improvements. - Method decomposition: Separated concerns into specialized methods (Read, Calculate, Display). - Input validation: Built a robust loop to handle invalid inputs and prevent crashes. - Data Formatting: Used printf to create a clear, readable results table. Small improvements in logic and organization make a huge difference in software quality. #Java #Algorithms #CleanCode #Backend #LearningToCode

Day 44 of #100DaysOfLeetCode 💻✅ Solved Very Simple #169. Majority Element problem in Java. Approach: • Sorted the given array using Arrays.sort() • Observed that the majority element appears more than n/2 times • After sorting, the majority element will always be present at index n/2 • Returned the element at nums[n/2] Performance: ✓ Runtime: 7 ms (Beats 42.86% submissions) ✓ Memory: 55.96 MB (Beats 16.21% submissions) Key Learning: ✓ Understood how sorting can help identify the majority element ✓ Learned the property that the majority element always occupies the middle position after sorting ✓ Practiced array manipulation and problem solving in Java Learning one problem every single day 🚀 #Java #LeetCode #DSA #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

Day 38 of #100DaysOfLeetCode 💻✅ Solved #222. Count Complete Tree Nodes problem in Java. Approach: • Used recursion to traverse the binary tree • Counted the current node and recursively counted nodes in left and right subtrees • Added the counts to get the total number of nodes • Returned 0 when the node is null to stop recursion Performance: ✓ Runtime: 0 ms (Beats 100% submissions) ✓ Memory: 44 MB Key Learning: ✓ Practiced recursion with binary trees ✓ Understood how tree traversal helps count nodes in a tree ✓ Improved recursive thinking while solving tree problems Learning one problem every single day 🚀 #Java #LeetCode #DSA #ProblemSolving #CodingJourney #100DaysOfCode

Day 42 of #100DaysOfLeetCode 💻✅ Solved #704. Binary Search problem in Java. Approach: • Since the array is already sorted, applied the Binary Search algorithm • Initialized two pointers: left at the start and right at the end of the array • Calculated the middle index using (left + right) / 2 • If the middle element equals the target, returned the index • If the middle element is smaller than the target, moved the left pointer to mid + 1 • If the middle element is greater than the target, moved the right pointer to mid - 1 • Continued until the target was found or the search space became empty Performance: ✓ Runtime: 0 ms (Beats 100% submissions) 🚀 ✓ Memory: 48.32 MB (Beats 68.26% submissions) Key Learning: ✓ Practiced Binary Search with O(log n) time complexity ✓ Learned how pointer adjustments reduce the search space efficiently ✓ Strengthened understanding of searching algorithms in sorted arrays Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinarySearch #ProblemSolving #CodingJourney #100DaysOfCode