Check Completeness of a Binary Tree
<p>Given the <code>root</code> of a binary tree, determine if it is a <em>complete binary tree</em>.</p>

<p>In a <strong><a href="http://en.wikipedia.org/wiki/Binary_tree#Types_of_binary_trees" target="_blank">complete binary tree</a></strong>, every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between <code>1</code> and <code>2<sup>h</sup></code> nodes inclusive at the last level <code>h</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0900-0999/0958.Check%20Completeness%20of%20a%20Binary%20Tree/images/complete-binary-tree-1.png" style="width: 180px; height: 145px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,5,6]
<strong>Output:</strong> true
<strong>Explanation:</strong> Every level before the last is full (ie. levels with node-values {1} and {2, 3}), and all nodes in the last level ({4, 5, 6}) are as far left as possible.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0900-0999/0958.Check%20Completeness%20of%20a%20Binary%20Tree/images/complete-binary-tree-2.png" style="width: 200px; height: 145px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,5,null,7]
<strong>Output:</strong> false
<strong>Explanation:</strong> The node with value 7 isn&#39;t as far left as possible.
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 100]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 1000</code></li>
</ul>
Tree; Breadth-First Search; Binary Tree
# 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 isCompleteTree(self, root: TreeNode) -> bool:
        q = deque([root])
        while q:
            node = q.popleft()
            if node is None:
                break
            q.append(node.left)
            q.append(node.right)
        return all(node is None for node in q)
