<p>Given the <code>root</code> of a <strong>perfect</strong> binary tree, reverse the node values at each <strong>odd</strong> level of the tree.</p>

<ul>
	<li>For example, suppose the node values at level 3 are <code>[2,1,3,4,7,11,29,18]</code>, then it should become <code>[18,29,11,7,4,3,1,2]</code>.</li>
</ul>

<p>Return <em>the root of the reversed tree</em>.</p>

<p>A binary tree is <strong>perfect</strong> if all parent nodes have two children and all leaves are on the same level.</p>

<p>The <strong>level</strong> of a node is the number of edges along the path between it and the root node.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/07/28/first_case1.png" style="width: 626px; height: 191px;" />
<pre>
<strong>Input:</strong> root = [2,3,5,8,13,21,34]
<strong>Output:</strong> [2,5,3,8,13,21,34]
<strong>Explanation:</strong> 
The tree has only one odd level.
The nodes at level 1 are 3, 5 respectively, which are reversed and become 5, 3.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/07/28/second_case3.png" style="width: 591px; height: 111px;" />
<pre>
<strong>Input:</strong> root = [7,13,11]
<strong>Output:</strong> [7,11,13]
<strong>Explanation:</strong> 
The nodes at level 1 are 13, 11, which are reversed and become 11, 13.
</pre>

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

<pre>
<strong>Input:</strong> root = [0,1,2,0,0,0,0,1,1,1,1,2,2,2,2]
<strong>Output:</strong> [0,2,1,0,0,0,0,2,2,2,2,1,1,1,1]
<strong>Explanation:</strong> 
The odd levels have non-zero values.
The nodes at level 1 were 1, 2, and are 2, 1 after the reversal.
The nodes at level 3 were 1, 1, 1, 1, 2, 2, 2, 2, and are 2, 2, 2, 2, 1, 1, 1, 1 after the reversal.
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 2<sup>14</sup>]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
	<li><code>root</code> is a <strong>perfect</strong> binary tree.</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

from collections import deque

class Solution:
    def reverseOddLevels(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return None
        level = 0
        revrese_list_utils = []
        queue = deque()
        queue.append(root)

        while len(queue):
            for _ in range(len(queue)):
                node = queue.popleft()
                if level % 2 != 0:
                    revrese_list_utils.append(node)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            if revrese_list_utils:
                self.reverse_order(revrese_list_utils)
            revrese_list_utils = []
            level += 1
        return root

    def reverse_order(self, list_of_nodes):
        left = 0
        right = len(list_of_nodes) - 1

        while left < right:
            temp = list_of_nodes[left].val
            list_of_nodes[left].val = list_of_nodes[right].val
            list_of_nodes[right].val = temp
            left += 1
            right -= 1  


        
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