199. Binary Tree Right Side View

A simple question? We can find the length of an amoeba or a brick? BRICK!!(So calculate size Between while and for)
This simple fact made my head spin today when i misplaced the size of the queue in the for loop! To tackle this problem I put root in queue, thereon adding the left and right child for every element of queue.Standard BST. Is it!! NO!!!

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I calculated the size inside the for loop, not paying attention that this will recalculate the size of queue for every element in queue.
Once that was spotted (AFTER 2 HOURS and using AI tool) I corrected it and got the solution

# 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]:
        if not root:
            return []
        q, res = [],[]
        q.append(root)
        while q:
            # tmp = []
            size = len(q)
            for index in range(size):
                print(index, len(q))
                
                node = q.pop(0)
                if index ==  size-1:
                    print(f'here{index} -> {node.val}')
                    res.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            # res.append(tmp)
        print(res)
        return res

Back to fight leetcode harder than ever! More solutions to follow:)