<p>Given the <code>root</code> of a binary tree, imagine yourself standing on the <strong>right side</strong> of it, return <em>the values of the nodes you can see ordered from top to bottom</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/14/tree.jpg" style="width: 401px; height: 301px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,null,5,null,4]
<strong>Output:</strong> [1,3,4]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [1,null,3]
<strong>Output:</strong> [1,3]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = []
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 100]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

# 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 []


        queue = deque()
        queue.append(root)
        res = [root.val]


        while queue:
            tmp_queue = deque()
            while queue:
                node = queue.popleft()
                if node.left:
                    tmp_queue.append(node.left)
                if node.right:
                    tmp_queue.append(node.right)

            if tmp_queue:
                res.append(tmp_queue[-1].val)
            queue = tmp_queue


        return res
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