Let's dive into a fascinating problem: finding the right side view of a binary tree using Depth-First Search (DFS). Guys, this isn't just another algorithm; it's a journey into the heart of tree traversal. We'll break down the problem, explore the DFS approach, and equip you with the knowledge to conquer this challenge. So, buckle up and get ready to explore the right side of binary trees!

Understanding the Problem

Before we jump into the code, let's make sure we're all on the same page. Imagine you're standing to the right of a binary tree. What nodes would you see? The right side view consists of the rightmost node at each level of the tree. Our mission is to write an algorithm that returns these nodes in order from top to bottom. For example, consider the following binary tree:

 1
 / \
 2 3
 / \ \
4 5 6
 / \
7 8 9

The right side view would be [1, 3, 6, 9]. Notice how at each level, we only pick the rightmost node that is visible from the right side. This seemingly simple problem opens the door to understanding tree traversal and level-based processing.

Depth-First Search (DFS) to the Rescue

So, how can we use DFS to solve this? DFS is a powerful technique for exploring trees (and graphs) by going as deep as possible along each branch before backtracking. In our case, we can adapt DFS to keep track of the rightmost node seen at each level. Here's the core idea:

1. Maintain a level variable: This tells us which level of the tree we're currently exploring.
2. Keep track of the rightmost nodes: We'll use a list (or array) to store the rightmost nodes we've encountered at each level.
3. Prioritize the right subtree: When traversing, we'll always visit the right subtree before the left subtree. This ensures that the first node we see at each level is the rightmost one.
4. Update the rightmost nodes list: If we're visiting a level for the first time (i.e., the level is greater than the current size of our rightmost nodes list), we add the current node's value to the list.

By following these steps, we can efficiently traverse the tree and identify the nodes that form the right side view. The beauty of DFS lies in its ability to naturally explore the tree's structure, making it a great fit for this problem. Let's translate this idea into code!

Implementing the DFS Approach

Here's how you might implement the DFS approach in Python:

class TreeNode:
 def __init__(self, val=0, left=None, right=None):
 self.val = val
 self.left = left
 self.right = right


def rightSideView(root):
 if not root:
 return []

 rightmost_values = []

 def dfs(node, level):
 if not node:
 return

 if level == len(rightmost_values):
 rightmost_values.append(node.val)

 dfs(node.right, level + 1)
 dfs(node.left, level + 1)

 dfs(root, 0)
 return rightmost_values

Let's break down this code:

• TreeNode class: This defines the structure of a node in our binary tree.
• rightSideView(root) function: This is the main function that takes the root of the tree as input and returns the right side view as a list.
• rightmost_values list: This list stores the rightmost node values for each level.
• dfs(node, level) function: This is the recursive DFS function. It takes the current node and its level as input.
  • Base case: If the node is None, we simply return.
  • Rightmost node check: If the current level is equal to the length of rightmost_values, it means we haven't seen a node at this level yet. So, we append the current node's value to rightmost_values.
  • Recursive calls: We recursively call dfs on the right subtree first, then on the left subtree. This ensures that the rightmost node is visited first.
• Initial call to dfs: We start the DFS traversal from the root node at level 0.

This code efficiently finds the right side view of the binary tree using DFS. The key is the order in which we traverse the subtrees – right first, then left – and the clever use of the rightmost_values list to keep track of the visible nodes.

Example Walkthrough

Let's walk through an example to solidify our understanding. Consider the tree we discussed earlier:

 1
 / \
 2 3
 / \ \
4 5 6
 / \
7 8 9

1. rightSideView(1): We start at the root node (1) and call dfs(1, 0). rightmost_values is initially empty [].
2. dfs(1, 0): Since level (0) equals len(rightmost_values) (0), we append 1 to rightmost_values. rightmost_values becomes [1]. We then call dfs(3, 1) (right subtree) and dfs(2, 1) (left subtree).
3. dfs(3, 1): Since level (1) equals len(rightmost_values) (1), we append 3 to rightmost_values. rightmost_values becomes [1, 3]. We then call dfs(6, 2) (right subtree) and dfs(None, 2) (left subtree, which returns immediately).
4. dfs(6, 2): Since level (2) equals len(rightmost_values) (2), we append 6 to rightmost_values. rightmost_values becomes [1, 3, 6]. We then call dfs(9, 3) (right subtree) and dfs(None, 3) (left subtree, which returns immediately).
5. dfs(9, 3): Since level (3) equals len(rightmost_values) (3), we append 9 to rightmost_values. rightmost_values becomes [1, 3, 6, 9]. We then call dfs(None, 4) (right subtree) and dfs(None, 4) (left subtree, both of which return immediately).
6. Backtracking: The calls to dfs(6, 2), dfs(3, 1), and dfs(1, 0) complete.
7. dfs(2, 1): This call happens after dfs(3, 1). Even though node 2 is at level 1, we've already added a node (3) at that level. Therefore, we don't add 2 to rightmost_values. We continue recursively down the left and right subtrees of node 2.
8. The algorithm continues, but no further additions are made to rightmost_values because the rightmost node at each level has already been recorded.
9. Finally, rightSideView(1) returns [1, 3, 6, 9], which is the correct right side view of the tree.

This walkthrough illustrates how the DFS algorithm, combined with the level tracking and right-subtree-first traversal, effectively identifies the rightmost nodes at each level.

Complexity Analysis

Let's analyze the time and space complexity of our DFS solution.

• Time Complexity: The algorithm visits each node in the tree exactly once. Therefore, the time complexity is O(N), where N is the number of nodes in the tree.
• Space Complexity: In the worst case, the recursion stack can grow to the height of the tree. In a skewed tree (where all nodes are on one side), the height can be equal to N. Therefore, the worst-case space complexity is O(N). However, in a balanced tree, the height is log₂(N), so the space complexity would be O(log N). Additionally, we use a list rightmost_values to store the result. In the worst case, the size of this list will be equal to the height of the tree, contributing another O(N) in the worst-case (skewed tree) or O(log N) in the average case (balanced tree).

Therefore, while the time complexity is always linear, the space complexity can vary depending on the shape of the tree.

Advantages of DFS

While other approaches like Breadth-First Search (BFS) can also solve this problem, DFS offers certain advantages in this specific scenario:

• Simplicity: The DFS approach is often more concise and easier to implement compared to BFS, especially when dealing with tree traversal problems.
• Natural fit: DFS naturally explores the tree in a depth-first manner, making it easier to keep track of the rightmost node at each level.
• Less memory overhead (potentially): In some cases, DFS can have a lower memory footprint than BFS, especially for very deep trees. BFS typically requires storing all nodes at a given level in a queue, which can consume more memory.

Conclusion

Guys, the binary tree right side view problem is a great example of how DFS can be used to solve tree traversal problems efficiently. By understanding the core principles of DFS and adapting them to the specific requirements of the problem, we can develop elegant and effective solutions. Remember to prioritize the right subtree, keep track of the level, and update the rightmost nodes list accordingly. With this knowledge in hand, you're well-equipped to tackle similar tree-related challenges! Keep coding, and keep exploring the fascinating world of algorithms!