Binary Tree Preorder Traversal in Java on LeetCode

Day 33 of #100DaysOfLeetCode 💻✅ Solved #145. Binary Tree Postorder Traversal on LeetCode using Java. Approach: • Used recursion to perform postorder traversal • Followed Left → Right → Root order strictly • Traversed left subtree first • Then traversed right subtree • Added node value after visiting both subtrees Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) ✓ Memory: 43.12 MB (Beats 71.37% submissions) Key Learning: ✓ Clearly understood difference between preorder, inorder, and postorder ✓ Strengthened recursive tree traversal concepts ✓ Improved confidence in handling binary tree problems Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinaryTree #Recursion #TreeTraversal #ProblemSolving #CodingJourney #100DaysOfCode

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 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 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

Today I solved “Fibonacci Number” on LeetCode using Java. 💡 Problem Summary The Fibonacci sequence is defined as: F(0) = 0 F(1) = 1 F(n) = F(n-1) + F(n-2) We need to compute F(n). 🧠 Approach I Used (Iterative Optimization) Instead of recursion (which is slow and redundant), I used an iterative approach: Maintain two variables → previous two values Keep updating them in a loop Build the answer step by step ⚙️ Why Not Recursion? Recursion leads to repeated calculations Time complexity becomes O(2ⁿ) ❌ Iterative solution reduces it to O(n) ✅ 📊 Complexity Analysis ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) ⚡ Result ✅ Accepted ⚡ Runtime: 0 ms (Beats 100%) 📚 Key Learning Always try to optimize recursion → iteration Use variables smartly to reduce space Even simple problems teach important optimization patterns Slowly building strong fundamentals, one problem at a time 💯 Day 15 done. Let’s keep the streak alive 🔥 #DSA #LeetCode #Java #100DaysOfCode #Algorithms #ProblemSolving #CodingJourney #Consistency #LearningInPublic

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

Day 46 of #100DaysOfLeetCode 💻✅ Solved #374. Guess Number Higher or Lower problem in Java. Approach: • Used Binary Search to efficiently guess the number. • Calculated mid using low + (high - low) / 2 to avoid overflow. • Used the guess(mid) API to check if the number is higher, lower, or correct. • Updated the search range accordingly until the number was found. Performance: ✓ Runtime: 0 ms (Beats 100% submissions) 🚀 ✓ Memory: 41.68 MB (Beats 97.84% submissions) Key Learning: ✓ Practiced implementing binary search on a real problem ✓ Learned how to safely calculate mid to prevent integer overflow ✓ Strengthened problem-solving skills with search algorithms Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinarySearch #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