<p>You are given the <code>root</code> of a binary tree with <strong>unique values</strong>.</p>

<p>In one operation, you can choose any two nodes <strong>at the same level</strong> and swap their values.</p>

<p>Return <em>the minimum number of operations needed to make the values at each level sorted in a <strong>strictly increasing order</strong></em>.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img src="https://assets.leetcode.com/uploads/2022/09/18/image-20220918174006-2.png" style="width: 500px; height: 324px;" />
<pre>
<strong>Input:</strong> root = [1,4,3,7,6,8,5,null,null,null,null,9,null,10]
<strong>Output:</strong> 3
<strong>Explanation:</strong>
- Swap 4 and 3. The 2<sup>nd</sup> level becomes [3,4].
- Swap 7 and 5. The 3<sup>rd</sup> level becomes [5,6,8,7].
- Swap 8 and 7. The 3<sup>rd</sup> level becomes [5,6,7,8].
We used 3 operations so return 3.
It can be proven that 3 is the minimum number of operations needed.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img src="https://assets.leetcode.com/uploads/2022/09/18/image-20220918174026-3.png" style="width: 400px; height: 303px;" />
<pre>
<strong>Input:</strong> root = [1,3,2,7,6,5,4]
<strong>Output:</strong> 3
<strong>Explanation:</strong>
- Swap 3 and 2. The 2<sup>nd</sup> level becomes [2,3].
- Swap 7 and 4. The 3<sup>rd</sup> level becomes [4,6,5,7].
- Swap 6 and 5. The 3<sup>rd</sup> level becomes [4,5,6,7].
We used 3 operations so return 3.
It can be proven that 3 is the minimum number of operations needed.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img src="https://assets.leetcode.com/uploads/2022/09/18/image-20220918174052-4.png" style="width: 400px; height: 274px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,5,6]
<strong>Output:</strong> 0
<strong>Explanation:</strong> Each level is already sorted in increasing order so return 0.
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>5</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
	<li>All the values of the tree are <strong>unique</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 minimumOperations(self, root: Optional[TreeNode]) -> int:
        def minSwaps(arr):
            n = len(arr)
            arrpos = [*enumerate(arr)]
            arrpos.sort(key=lambda it: it[1])
            vis = {k: False for k in range(n)}
            ans = 0
            for i in range(n):
                if vis[i] or arrpos[i][0] == i:
                    continue
                cycle_size = 0
                x = i
                while not vis[x]:
                    vis[x] = True
                    x = arrpos[x][0]
                    cycle_size += 1
                if cycle_size > 0:
                    ans += (cycle_size - 1)
            return ans

        if not root:
            return 0

        queue = [root]
        operations = 0

        while queue:
            level_size = len(queue)
            level_values = []
            for _ in range(level_size):
                node = queue.pop(0)
                level_values.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            operations += minSwaps(level_values)

        return operations
750