Right Side View Of Binary Tree: LeetCode 199 Solution

by SLV Team 54 views
LeetCode 199: Finding the Right Side View of a Binary Tree

Hey guys! Today, let's dive into a classic tree traversal problem on LeetCode: problem 199, which is all about finding the right side view of a binary tree. This problem is a fantastic way to sharpen your tree traversal skills and understand how to visualize trees in different ways. We'll break down the problem, explore the solution, and I'll share my Python code along with a detailed explanation. So, let's get started!

Understanding the Problem

The problem asks us to imagine standing on the right side of a binary tree and then returning the values of the nodes we can see, ordered from top to bottom. Essentially, we need to find the rightmost node at each level of the tree. This means if there are multiple nodes at the same depth, the one furthest to the right is the one we want to include in our final result.

To make it crystal clear, consider these examples:

  • Example 1: If the tree is [1,2,3,null,5,null,4], the right side view is [1,3,4]. This is because from the right side, you'd first see 1 (the root), then 3 (at the next level), and finally 4 (at the bottom level).
  • Example 2: For a tree like [1,2,3,4,null,null,null,5], the right side view is [1,3,4,5]. Notice how 5 is included because it's the rightmost node at its level.
  • Example 3: A simpler tree [1,null,3] gives a right side view of [1,3]. The null node doesn't count, so we just take the rightmost node at each level.
  • Example 4: An empty tree [] naturally results in an empty list [].

Constraints

Before we jump into the solution, let's quickly look at the constraints. This helps us understand the scale of the problem and think about potential optimizations:

  • The number of nodes in the tree can range from 0 to 100.
  • Node values can be between -100 and 100.

These constraints tell us that we're dealing with a relatively small tree, so efficiency considerations, while important, aren't as critical as correctness.

My Python Solution: Breadth-First Search (BFS)

The approach I used for this problem is Breadth-First Search (BFS). BFS is perfect for traversing trees level by level, which aligns perfectly with our need to find the rightmost node at each level. Here’s the Python code I wrote:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
        result = []
        if root is None:
            return result
        currLvl = [root]
        nextLvl = []
        while len(currLvl) > 0:
            for node in currLvl:
                nextLvl.append(node.left) if node.left else None
                nextLvl.append(node.right) if node.right else None
            result.append(currLvl[-1].val)
            currLvl = nextLvl
            nextLvl = []
        return result

Code Breakdown

Let’s walk through the code step by step to understand what’s happening:

  1. Initialization:
    • We start by initializing an empty list called result, which will hold the values of the nodes in the right side view.
    • We handle the base case where the root is None (empty tree) by returning the empty result list.
  2. Setting up BFS:
    • We use two lists: currLvl to store the nodes at the current level and nextLvl to store the nodes at the next level. We initialize currLvl with the root node.
  3. BFS Traversal:
    • The while loop continues as long as there are nodes in the currLvl list, meaning we haven't traversed all levels of the tree.
    • Inside the loop, we iterate through each node in currLvl.
    • For each node, we append its left and right children to the nextLvl list, but only if they exist (i.e., not None).
    • After processing all nodes at the current level, we append the value of the last node in currLvl (which is the rightmost node at that level) to our result list.
    • We then update currLvl to be nextLvl and reset nextLvl to an empty list to prepare for the next level.
  4. Returning the Result:
    • Finally, after traversing all levels, we return the result list, which contains the right side view of the tree.

Why BFS Works

BFS works perfectly for this problem because it processes nodes level by level. By appending the rightmost node's value from each level to our result, we ensure that we're capturing the nodes visible from the right side.

Alternative Approach: Depth-First Search (DFS)

While I used BFS, it's worth noting that Depth-First Search (DFS) can also solve this problem. The key idea with DFS is to traverse the right subtree before the left subtree. By keeping track of the maximum depth we've seen so far, we can ensure that we only add the first node we encounter at each depth to our result.

I didn’t use DFS in my solution, but it’s a great alternative to consider, especially if you want to explore different tree traversal techniques.

My Thoughts on the Problem

I found this problem to be a great exercise in applying tree traversal algorithms. The concept of visualizing a tree from a specific perspective is quite interesting, and it highlights the importance of choosing the right traversal method for the job. BFS was a natural fit for this problem, but understanding how DFS could also be applied adds another layer of insight.

The problem is relatively straightforward once you grasp the core idea, but it's still a good challenge for solidifying your understanding of tree traversals. If you're preparing for coding interviews, this is definitely a problem worth practicing!

Additional Resources

If you want to dive deeper into tree traversals and explore more related problems, here are a few resources I recommend:

  • LeetCode Tree Problems: LeetCode has a wealth of tree-related problems that you can practice. Start with the easy ones and gradually work your way up to more challenging problems.
  • GeeksforGeeks Tree Data Structure: GeeksforGeeks provides comprehensive articles and tutorials on tree data structures and algorithms.
  • Cracking the Coding Interview by Gayle Laakmann McDowell: This book has an excellent section on trees and graphs, with plenty of practice problems and solutions.

Conclusion

So, there you have it! We’ve explored how to find the right side view of a binary tree using BFS in Python. I hope this explanation was helpful and gave you a better understanding of this problem. Remember, practice is key, so keep coding and keep exploring!

If you have any questions or alternative solutions, feel free to share them in the comments below. Let's learn and grow together!

Until next time, happy coding, and good luck with your job hunt!

baddif@gmail.com

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