<p>Given the <code>root</code> of a binary tree, return <em>an array of the largest value in each row</em> of the tree <strong>(0-indexed)</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/21/largest_e1.jpg" style="width: 300px; height: 172px;" />
<pre>
<strong>Input:</strong> root = [1,3,2,5,3,null,9]
<strong>Output:</strong> [1,3,9]
</pre>

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

<pre>
<strong>Input:</strong> root = [1,2,3]
<strong>Output:</strong> [1,3]
</pre>

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

<ul>
	<li>The number of nodes in the tree will be in the range <code>[0, 10<sup>4</sup>]</code>.</li>
	<li><code>-2<sup>31</sup> &lt;= Node.val &lt;= 2<sup>31</sup> - 1</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 largestValues(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []
        return self.bfs(root)
        
        
    def bfs(self, root):
        res = []
        nodes = [root]
        
        while nodes:
            temp = []
            item_max = float('-inf')
            for node in nodes:
                item_max = max(item_max, node.val)
                if node.left:
                    temp.append(node.left)
                if node.right:
                    temp.append(node.right)
            res.append(item_max)
            nodes = temp
        return res
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