Binary Tree Right Side View Leetcode Problem 199 [Python Solution]
Welcome back, coding enthusiasts!
In this article, we will dive into solving a LeetCode problem called Binary Tree Right Side View This problem falls under the category of Trees and is rated as a medium difficulty challenge.
We'll provide you with a detailed Python solution to tackle this problem efficiently.
So, let's get started!
Problem Overview
The problem statement is relatively straightforward.
Given the root of a binary tree, imagine yourself standing on the right side of it.
Your task is to return the values of the nodes you can see when standing on the right side of the tree, ordered from top to bottom.
To give you a clearer picture, let's take a look at some examples:
Example 1:
Input: root = [1,2,3,null,5,null,4]
Output: [1,3,4]
Example 2:
Input: root = [1,null,3]
Output: [1,3]
Example 3:
Input: root = []
Output: []
Constraints:
- The number of nodes in the tree is in the range [0, 100].
- Node values are in the range of -100 to 100. Now that we understand the problem, let's explore how to solve it efficiently.
Understanding the Constraints
Before we dive into the solution, it's essential to understand the problem's constraints.
These constraints help us tailor our approach and ensure that our solution will work efficiently even for the worst-case scenarios.
-
The number of nodes in the tree is limited to a range of 0 to 100. This means that we need to develop a solution that can handle trees of various sizes, including very small or very large ones.
-
The values of the tree nodes can be any integer in the range of -100 to 100. This range is not exceptionally large, so we don't need to worry about integer overflow or extreme values.
Binary Tree Right Side View LeetCode Problem Solution
1. Bruteforce Approach
One approach to solving this problem is to perform a simple breadth-first traversal of the binary tree.
We can keep track of the rightmost node at each level and add it to our result.
Here's how it works:
def rightSideView(self, root: TreeNode) -> List[int]:
res = []
q = collections.deque([root])
while q:
rightSide = None
qLen = len(q)
for i in range(qLen):
node = q.popleft()
if node:
rightSide = node
q.append(node.left)
q.append(node.right)
if rightSide:
res.append(rightSide.val)
return res
Now, let's break down this solution step by step.
We start by initializing two important data structures:
– res: This is our result array where we'll store the values of the rightmost nodes.
– q: This is a double-ended queue (deque) used for performing the breadth-first traversal.
We start with the root node in the queue.
The main loop continues as long as there are elements in the queue (while q).
Within this loop, we keep track of the rightmost node for the current level using the rightSide variable.
Initially, it's set to None.
We also store the length of the queue (qLen) at the beginning of each iteration.
This is crucial because the number of elements in the queue at the start of each loop represents the nodes at the current level.
Now, we enter another loop to process the nodes at the current level.
We run this loop for the number of nodes in the current level (qLen times).
For each node in the queue, we check if it's not None.
If the node is not None, it means it's a valid node, so we update the rightSide variable to this node, as this is the rightmost node for the current level.
Next, we add the left and right children of the current node to the queue.
This ensures that we explore the tree level by level from left to right.
Once we've processed all nodes in the current level, we add the value of the rightSide node to our result array res if it's not None.
This accounts for cases where the rightmost node might be None (for example, at the last level of the tree).
Finally, we return the res array, which contains the values of the rightmost nodes in the tree.
This solution is both intuitive and efficient.
It effectively traverses the tree, ensuring that the rightmost nodes at each level are recorded.
Additionally, it gracefully handles the given constraints.
Time and Space Complexity
Now, let's analyze the time and space complexity of our solution.
Time Complexity: We perform a breadth-first traversal of the binary tree.
In the worst case, we visit all the nodes once, leading to a time complexity of O(N), where N is the number of nodes in the tree.
Space Complexity: The space complexity is determined by the space required for the queue and the result array.
In the worst case, when the tree is perfectly balanced, the maximum number of nodes at any level is N/2. Therefore, the space complexity is O(N) for both the queue and the result array.
Reasoning Behind Our Approach
The reasoning behind the chosen approach is to leverage a breadth-first traversal of the binary tree to ensure that we process nodes level by level, from left to right.
This approach guarantees that the rightmost node at each level is encountered first and added to our result.
The use of a double-ended queue (deque) allows for efficient pop and append operations at both ends, making it suitable for the breadth-first traversal.
Additionally, we keep track of the rightmost node at each level by updating the rightSide variable, ensuring that we capture the correct values.
The if node is not None check ensures that we do not consider null nodes when updating the rightSide.
This is essential for accurately identifying the rightmost node.
In terms of time and space complexity, the solution is efficient and well-suited to handle the constraints presented in the problem statement.
It achieves a time complexity of O(N) and a space complexity of O(N), where N represents the number of nodes in the tree.
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Conclusion
In this article, we tackled the LeetCode problem Binary Tree Right Side View We provided a detailed Python solution that efficiently finds and orders the values of the rightmost nodes in a binary tree.
We explored the problem's constraints, which informed our approach to developing a solution suitable for various tree sizes and node value ranges.
By utilizing breadth-first traversal and carefully maintaining the rightmost node at each level, we achieved an effective solution.
If you found this article helpful or have any questions, feel free to comment, ask questions, make suggestions, and share the content.
Your feedback and engagement are highly encouraged!
Thank you for reading, and happy coding!
