Hey guys! Let's dive into a super interesting problem today: finding the right side view of a binary tree using Depth-First Search (DFS). This is a classic tree traversal question that not only tests your understanding of tree data structures but also your ability to think recursively. So, grab your favorite coding beverage, and let’s get started!

What is the Right Side View of a Binary Tree?

Before we jump into the code, let's clarify what we mean by the “right side view.” Imagine you are standing to the right of a binary tree. The right side view is the set of nodes you can see from that vantage point, ordered from top to bottom. In other words, for each level of the tree, you pick the rightmost node.

For example, consider the following binary tree:

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

The right side view would be [1, 3, 6, 8]. Node 1 is the rightmost node at the root level, node 3 is the rightmost at the second level, node 6 at the third level, and node 8 at the fourth level. Notice that nodes 2, 4, 5, and 7 are not visible from the right side.

Understanding Depth-First Search (DFS)

Okay, so how do we use DFS to solve this? If you're unfamiliar, DFS is a traversal algorithm that explores as far as possible along each branch before backtracking. Think of it as going deep into the tree before exploring siblings.

There are three main types of DFS: In-order, Pre-order, and Post-order. For this particular problem, we can use a modified version of Pre-order traversal. Here’s why:

1. Pre-order Traversal: We visit the current node first, then the left child, and finally the right child (Node -> Left -> Right).
2. Modified Pre-order: We will prioritize visiting the right child before the left child (Node -> Right -> Left). This ensures that the first node we encounter at each level is the rightmost node.

The key idea is to keep track of the current level during the traversal. We maintain a list to store the right side view nodes. When we visit a node at a new level (i.e., a level we haven't seen before), we add it to our list. Because we are visiting the right child first, we ensure that we are adding the rightmost node at each level.

DFS Implementation for the Right Side View

Let’s break down the code step-by-step. We’ll use Python for this example, but the logic can be easily translated to other languages like Java or C++.

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

def rightSideView(root):
    result = []

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

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

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

    dfs(root, 0)
    return result

Code Explanation

1. TreeNode Class:

   • We define a TreeNode class to represent each node in the binary tree. Each node has a value (val), a left child (left), and a right child (right).

2. rightSideView Function:

   • This is the main function that takes the root of the binary tree as input and returns a list representing the right side view.
   • It initializes an empty list called result to store the nodes visible from the right side.

3. dfs Function:

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   • This is a recursive helper function that performs the Depth-First Search.
   • node: The current node being visited.
   • level: The level of the current node in the tree (the root is at level 0).

4. Base Case:

   • if not node: return
   • If the current node is None (i.e., we've reached the end of a branch), we simply return.

5. Adding to the Result:

   • if level == len(result): result.append(node.val)
   • This is a crucial step. We check if the current level is equal to the length of the result list. If it is, it means we are visiting this level for the first time. Because we are visiting the right child first, the current node must be the rightmost node at this level. So, we append its value to the result list.

6. Recursive Calls:

   • dfs(node.right, level + 1)
   • We recursively call the dfs function on the right child, incrementing the level by 1.
   • dfs(node.left, level + 1)
   • We then recursively call the dfs function on the left child, also incrementing the level by 1.

7. Calling DFS:

   • dfs(root, 0)
   • We start the DFS traversal from the root node at level 0.

8. Returning the Result:

   • return result
   • Finally, we return the result list, which contains the right side view of the binary tree.

Example Walkthrough

Let’s walk through our earlier example tree to see how the code works:

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

1. We start at the root node 1 at level 0. Since result is empty, we add 1 to result. result is now [1]. We then call dfs on the right child 3 and then on the left child 2.
2. At node 3 (level 1), len(result) is 1, so we add 3 to result. result is now [1, 3]. We call dfs on the right child 6 of node 3.
3. At node 6 (level 2), len(result) is 2, so we add 6 to result. result is now [1, 3, 6]. We call dfs on the right child 8 of node 6.
4. At node 8 (level 3), len(result) is 3, so we add 8 to result. result is now [1, 3, 6, 8]. There are no more children for node 8, so we return.
5. Back at node 2 (level 1), len(result) is 2, so we don't add 2 to result because we have already added a node at level 1 (node 3). We call dfs on the right child 5 and then on the left child 4.
6. The algorithm continues similarly, but no other nodes are added to result because we have already seen all the levels.

Finally, we return result, which is [1, 3, 6, 8]. This is the right side view of the binary tree.

Why DFS Works

The magic of DFS lies in its ability to explore each branch deeply before moving to the next. By prioritizing the right child, we ensure that the first node we encounter at each level is the rightmost one. The check level == len(result) ensures that we only add one node per level, thus giving us the right side view.

Complexity Analysis

Time Complexity

The time complexity of this algorithm is O(N), where N is the number of nodes in the binary tree. This is because we visit each node exactly once.

Space Complexity

The space complexity is O(H), where H is the height of the binary tree. In the worst case (a skewed tree), H can be equal to N, so the space complexity can be O(N). This space is used by the call stack during the recursive DFS traversal.

Conclusion

And there you have it! We've successfully found the right side view of a binary tree using Depth-First Search. This problem is a great example of how understanding tree traversals and recursion can help you solve complex problems in a clean and efficient manner. Keep practicing, and you'll become a tree traversal master in no time! Happy coding, guys!