Hey guys! Ever wondered how to peek at the rightmost nodes of a binary tree? Today, we're diving into a cool algorithm that lets us do just that using Depth-First Search (DFS). It's like taking a snapshot of the tree from the right, capturing only the nodes you can see from that vantage point. Trust me; it's simpler than it sounds, and by the end of this article, you'll be a pro at it!

Understanding the Problem

Before we get our hands dirty with the code, let's make sure we're all on the same page. The right side view of a binary tree is the set of nodes visible when you look at the tree from the right side. Imagine you're standing to the right of the tree; you'd only see the rightmost node at each level. Our mission is to write a function that returns these nodes in the order they appear from top to bottom.

For example, consider this binary tree:

        1
       /  \
      2    3
       \    \
        5     4

The right side view would be [1, 3, 4]. Node 1 is at the top, then 3 is the rightmost node on the second level, and finally, 4 is the rightmost on the third level. Notice that 5 is hidden behind 4, so it's not included in our view.

Diving into Depth-First Search (DFS)

So, how do we use DFS to solve this? DFS is an algorithm for traversing or searching tree or graph data structures. It starts at the root node and explores as far as possible along each branch before backtracking. In our case, we can modify DFS to keep track of the rightmost node we encounter at each level.

The key idea is to visit the right child before the left child. By doing so, the first node we encounter at each level will be the rightmost node. We use a list to store the values of these rightmost nodes as we traverse the tree. We also need to keep track of the current depth (or level) we are at. This can be achieved by passing level number in dfs recursive call.

The Algorithm Explained Step-by-Step

1. Initialization: Start with an empty list called result to store the right side view nodes.
2. DFS Function: Define a recursive function, let's call it dfs(node, level), that takes the current node and its level as input.
3. Base Case: If the current node is null, return (do nothing).
4. Check Level: Before processing the current node, check if the current level is already present in the result list. This check is simply, if the current level number is equal to the size of the result list. If the level is not yet in result, it means we haven't found the rightmost node for this level yet. Add the current node's value to the result list.
5. Recursive Calls: Recursively call dfs on the right child first (dfs(node.right, level + 1)), then on the left child (dfs(node.left, level + 1)). This ensures that we always prioritize the right side.
6. Starting the Traversal: Call the dfs function with the root node and an initial level of 0.
7. Return the Result: After the dfs function completes, return the result list.

Let's walk through the example tree again to see how this works:

        1 (Level 0)
       /  \
      2 (Level 1) 3 (Level 1)
       \    \
        5 (Level 2) 4 (Level 2)

1. We start at node 1 (level 0). Since result is empty, we add 1 to result. result is now [1]. Then call dfs on right node.
2. Next, we move to node 3 (level 1). Since result only has one element (size is 1), we add 3 to result. result is now [1, 3]. Then call dfs on right node.
3. Then, we visit node 4 (level 2). Since result has only two elements (size is 2), we add 4 to result. result is now [1, 3, 4]. Since the right node is null, return. Then call dfs on left node.
4. Now, we backtrack to node 3 and call dfs on the left child (which is null).
5. Back at node 1, we call dfs on the left child, node 2 (level 1). However, result already has an element at level 1 (the size of result is 2), so we don't add 2 to result. We continue the process for node 5 (level 2), but again, result already has an element at level 2 (the size of result is 3), so we don't add 5.
6. Finally, we return result, which is [1, 3, 4]. Voila!

Code Implementation (Java)

Alright, let's translate this algorithm into Java code. This should make it crystal clear how everything fits together.

import java.util.ArrayList;
import java.util.List;

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int val) {
        this.val = val;
    }
}

public class BinaryTreeRightSideView {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        dfs(root, 0, result);
        return result;
    }

    private void dfs(TreeNode node, int level, List<Integer> result) {
        if (node == null) {
            return;
        }

        if (level == result.size()) {
            result.add(node.val);
        }

        dfs(node.right, level + 1, result);
        dfs(node.left, level + 1, result);
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
        root.left.right = new TreeNode(5);
        root.right.right = new TreeNode(4);

        BinaryTreeRightSideView solution = new BinaryTreeRightSideView();
        List<Integer> rightView = solution.rightSideView(root);
        System.out.println("Right side view: " + rightView);
    }
}

Code Breakdown

• TreeNode Class: A basic class to represent a node in a binary tree.
• rightSideView Method: This is the main method that takes the root node as input and returns the list of nodes in the right side view.
• dfs Method: A recursive method that performs the Depth-First Search. It takes the current node, the current level, and the result list as parameters. It adds the node's value to the result list if it's the first node encountered at that level.

Why DFS Works So Well

You might be wondering, why does DFS work so effectively for this problem? The beauty of DFS lies in its ability to explore each branch as deeply as possible before backtracking. By prioritizing the right child in our recursive calls, we ensure that the first node we encounter at each level is the rightmost node.

This approach is also quite efficient. In the worst-case scenario, where the tree is skewed to the left, we would still visit each node once, resulting in a time complexity of O(N), where N is the number of nodes in the tree. The space complexity is also O(H), where H is the height of the tree, due to the recursive call stack.

Other Approaches

While DFS is a great way to solve this problem, there are other approaches you might want to consider. One popular alternative is using Breadth-First Search (BFS).

Breadth-First Search (BFS)

BFS involves traversing the tree level by level. For each level, the rightmost node is simply the last node we visit. This approach also has a time complexity of O(N), but it uses a queue to store nodes at each level, which can potentially require more space than DFS in certain scenarios.

Tips and Tricks

• Understand Recursion: Make sure you have a solid understanding of recursion. It's crucial for implementing DFS effectively.
• Visualize: Draw out the tree and trace the algorithm's execution. This will help you understand how the right side view is constructed.
• Practice: Solve similar tree traversal problems to strengthen your understanding of DFS and BFS.

Conclusion

So there you have it! We've explored how to find the right side view of a binary tree using DFS. It's a simple yet powerful algorithm that leverages the strengths of depth-first traversal to efficiently capture the nodes visible from the right. Whether you're prepping for a coding interview or just expanding your knowledge of tree algorithms, this technique is a valuable addition to your toolkit. Keep coding, keep exploring, and you'll be mastering these algorithms in no time!