<p>Given the <code>root</code> of a <strong>complete</strong> binary tree, return the number of the nodes in the tree.</p>

<p>According to <strong><a href="http://en.wikipedia.org/wiki/Binary_tree#Types_of_binary_trees" target="_blank">Wikipedia</a></strong>, every level, except possibly the last, is completely filled in a complete binary tree, 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>Design an algorithm that runs in less than&nbsp;<code data-stringify-type="code">O(n)</code>&nbsp;time complexity.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/14/complete.jpg" style="width: 372px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,5,6]
<strong>Output:</strong> 6
</pre>

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

<pre>
<strong>Input:</strong> root = []
<strong>Output:</strong> 0
</pre>

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

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

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 5 * 10<sup>4</sup>]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 5 * 10<sup>4</sup></code></li>
	<li>The tree is guaranteed to be <strong>complete</strong>.</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 countNodes(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0
        elif not root.left and not root.right:
            return 1
        elif root.left and not root.right:
            return 2
        def level(node):
            height = 0
            while node:
               height += 1
               node = node.left
            return height
        left = level(root.left)
        right = level(root.right)
        if left == right:
            return 2**left + self.countNodes(root.right)
        else:
            return self.countNodes(root.left) + 2**right
        

        
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