<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 {
public:
    int get(int label, int row){
        return (1<<row) + (1<<(row-1)) - 1 - label;
    }

    vector<int> pathInZigZagTree(int label) {
        vector<int> res;
        int row = 1, count =1;
        while(count*2 <= label){
            row++;
            count*=2;
        }
        if(row%2 == 0) label = get(label,row);
        while(row){
            if(row%2==0){
                res.push_back(get(label,row));
            }else{
                res.push_back(label);
            }

            row--;
            label/=2;
        }
        reverse(res.begin(),res.end());
        return res;    
    }
};
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