LeetCode Problem #199 — Binary Tree Right Side View Solve LeetCode Binary Tree Right Side View using JavaScript and Binary Search. This solution finds the optimal result using Modified Binary Search in O(n) time.
| Property | Value |
|---|---|
| Problem | Binary Tree Right Side View |
| LeetCode | #199 |
| Difficulty | Medium |
| Language | JavaScript |
| Algorithm | Modified Binary Search |
| Data Structure | Sorted Array |
| Pattern | Binary Search |
| Time Complexity | O(n) |
| Space Complexity | O(1) |
Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom .
Exploit sorted ordering or monotonic properties to eliminate half of the search space at each step in
We iterate through the input using Depth-First Search (DFS) / Breadth-First Search (BFS). By maintaining state efficiently in a Tree / Graph / Grid, we eliminate redundant operations and process each element in optimal time.
- Initialize state variables / data structure (Sorted Array).
- Process elements sequentially using Modified Binary Search.
- Validate boundary conditions and return optimal result.
Consider the standard input for Binary Tree Right Side View. Applying Modified Binary Search yields the target result step by step.
- Time Complexity: O(n)
- Space Complexity: O(1)
Binary Search
- Binary Search
- Divide and Conquer
- Search Space
JavaScript
By utilizing Binary Search, each element is processed efficiently, ensuring optimal performance while avoiding unnecessary re-computations.
- Applying standard binary search without accounting for array rotation or duplicates.
- Off-by-one errors when updating boundary pointers (
left = mid + 1vsright = mid - 1). - Integer overflow during midpoint calculation (use
mid = left + (right - left) // 2).
- Tests: Logarithmic search space reduction, boundary handling, and invariant preservation.
- Follow-up: How does performance change if the array contains duplicate elements?