Binary Tree Right Side View

Background Knowledge

The problem involves a binary tree, which is a data structure where each node has at most two children (i.e., left child and right child). The tree is given as a level-order array, meaning that the nodes are arranged in a way that all nodes at a given depth are visited before moving on to the next depth level. To solve this problem, it's essential to understand the concepts of tree traversal and level-order traversal. Level-order traversal visits all nodes at a given depth before moving on to the next depth level, which can be achieved using a queue data structure.

The problem requires finding the values of the nodes that can be seen when looking at the tree from the right side. This means we need to focus on the rightmost node at each depth level. Understanding the structure of a binary tree and how to traverse it level by level is crucial to solving this problem. Additionally, familiarity with Breadth-First Search (BFS) algorithms will be helpful, as it is a common approach used for level-order traversals.

The problem also involves understanding the concept of node values and how to access them. In a binary tree, each node has a value associated with it, and we need to find the values of the rightmost nodes at each depth level. This requires understanding how to access and manipulate node values during the traversal process.

Algorithm/Approach

The general approach to solve this type of problem involves using a Breadth-First Search (BFS) algorithm, which is suitable for level-order traversals. The BFS algorithm uses a queue data structure to keep track of nodes to visit at each depth level. By using a queue, we can efficiently visit all nodes at a given depth level before moving on to the next level.

Step-by-Step Strategy

To solve this problem, follow these steps:

• Create a queue to store nodes to visit at each depth level
• Enqueue the root node as the starting point
• While the queue is not empty, dequeue a node and process it
• For each node, enqueue its right child and left child (if they exist)
• Keep track of the rightmost node at each depth level
• Once all nodes have been visited, return the values of the rightmost nodes at each depth level

Common Pitfalls

When implementing the solution, watch out for the following:

• Forgetting to handle the case where a node has no children
• Not properly keeping track of the rightmost node at each depth level
• Incorrectly implementing the queue operations (enqueue and dequeue)

Time & Space Complexity

The expected time complexity for this problem is $O(n)$, where $n$ is the number of nodes in the tree, since we visit each node once. The expected space complexity is $O(n)$ as well, as in the worst case, the queue will store all nodes at the last depth level.