<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://assets.leetcode.com/uploads/2019/06/24/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>

class Solution:
    def pathInZigZagTree(self, label: int) -> List[int]:
        cur_row = []
        tree = []
        #create tree of all rows
        for i in range(21):
            if pow(2, i) > label:
                break
            row = list(range(pow(2, i), pow(2, i+1)))
            if i % 2 == 0:
                cur_row = row
            else:
                cur_row = row[::-1]
            tree.append(cur_row)
        #go through binary tree to find root
        cur_index = tree[-1].index(label)
        path = []
        for row in tree[::-1]:
            path.insert(0, row[cur_index])
            cur_index //= 2
        return(path)

            
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