Introduction

In this blog post, we will explore how to solve the LeetCode problem "199" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. Example 1: Input: root = [1,2,3,null,5,null,4] Output: [1,3,4] Explanation: Example 2: Input: root = [1,2,3,4,null,null,null,5] Output: [1,3,4,5] Explanation: Example 3: Input: root = [1,null,3] Output: [1,3] Example 4: Input: root = [] Output: [] Constraints: The number of nodes in the tree is in the range [0, 100]. -100 <= Node.val <= 100

Explanation

Here's a solution to the problem of finding the right side view of a binary tree:

• Level-Order Traversal: Perform a level-order (breadth-first) traversal of the tree.
• Rightmost Node: For each level, the rightmost node encountered during the traversal is the one visible from the right side. Store the value of this node.

• Result: Collect the values of the rightmost nodes from each level in a list.

• Time Complexity: O(N), where N is the number of nodes in the tree.

• Space Complexity: O(W), where W is the maximum width of the tree (worst case O(N) if the tree is skewed).

Code

    from collections import deque

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

def rightSideView(root):
    """
    Given the root of a binary tree, imagine yourself standing on the right side of it,
    return the values of the nodes you can see ordered from top to bottom.
    """
    if not root:
        return []

    result = []
    queue = deque([root])

    while queue:
        level_size = len(queue)

        for i in range(level_size):
            node = queue.popleft()

            if i == level_size - 1:
                result.append(node.val)

            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)

    return result