We delve into a popular algorithmic problem from LeetCode: the "Binary Tree Right Side View". This problem is an excellent exercise for understanding tree-based data structures and breadth-first traversal techniques.

Problem Statement

The "Binary Tree Right Side View" problem asks us to find the values of the nodes you can see when you look at a binary tree from the right side. This perspective means we only observe the rightmost node at each level of the tree.

example

Given a binary tree like this:

1
/ \
2 3
\ \
5 4

The right side view will be [1, 3, 4].

Solution

To solve this problem, we employ a breadth-first search (BFS) approach using a queue. BFS allows us to traverse the tree level by level, ensuring we can access the rightmost element of each level.

Method Signature: The method rightSideView takes the root of a binary tree as its input and returns a list of integers representing the tree's right side view.

Base Case Check: We first check if the root is null. If it is, we return an empty list, as there's no tree to traverse.

Queue Initialization: We use a Deque to store nodes at each level.

BFS Loop: We iterate through the tree using a while loop, processing nodes level by level.

Level Processing: For each level, we iterate through the nodes at that level. The key is to add the value of the last node of each level to our output list.

Child Nodes: We add the left and right child nodes of each processed node to the queue for subsequent processing.

Time Complexity

The time complexity of this solution is O(N), where N is the number of nodes in the tree. This is because each node is visited exactly once during the BFS traversal.

Queue Operations: Each node is added and removed from the queue exactly once.

Level-wise Traversal: Since we process nodes level by level, we ensure that each node is considered only once.