Path In Zigzag Labelled Binary Tree
<p>In an infinite binary tree where every node has two children, the nodes are labelled in row order.</p>

<p>In the odd numbered rows (ie., the first, third, fifth,...), the labelling is left to right, while in the even numbered rows (second, fourth, sixth,...), the labelling is right to left.</p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1100-1199/1104.Path%20In%20Zigzag%20Labelled%20Binary%20Tree/images/tree.png" style="width: 300px; height: 138px;" /></p>

<p>Given the <code>label</code> of a node in this tree, return the labels in the path from the root of the tree to the&nbsp;node with that <code>label</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> label = 14
<strong>Output:</strong> [1,3,4,14]
</pre>

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

<pre>
<strong>Input:</strong> label = 26
<strong>Output:</strong> [1,2,6,10,26]
</pre>

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

<ul>
	<li><code>1 &lt;= label &lt;= 10^6</code></li>
</ul>
Tree; Math; Binary Tree
class Solution:
    def pathInZigZagTree(self, label: int) -> List[int]:
        x = i = 1
        while (x << 1) <= label:
            x <<= 1
            i += 1
        ans = [0] * i
        while i:
            ans[i - 1] = label
            label = ((1 << (i - 1)) + (1 << i) - 1 - label) >> 1
            i -= 1
        return ans
