Minimum Number of Operations to Sort a Binary

lcpy gen -n 2471   # by problem number
lcpy gen -s minimum_operations_to_sort_a_binary_tree_by_level   # by problem name

Input: root = [1,4,3,7,6,8,5,null,null,null,null,9,null,10]
Output: 3
Explanation:
- Swap 4 and 3. The 2nd level becomes [3,4].
- Swap 7 and 5. The 3rd level becomes [5,6,8,7].
- Swap 8 and 7. The 3rd 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.

Input: root = [1,3,2,7,6,5,4]
Output: 3
Explanation:
- Swap 3 and 2. The 2nd level becomes [2,3].
- Swap 7 and 4. The 3rd level becomes [4,6,5,7].
- Swap 6 and 5. The 3rd 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.

Input: root = [1,2,3,4,5,6]
Output: 0
Explanation: Each level is already sorted in increasing order so return 0.

from leetcode_py import TreeNode


class Solution:
    # Time: O(n log n)
    # Space: O(n)
    def minimum_operations(self, root: TreeNode[int] | None) -> int:
        if root is None:
            return 0

        total = 0
        level = [root]
        while level:
            total += self._min_swaps([node.val for node in level])
            level = [child for node in level for child in (node.left, node.right) if child]
        return total

    def _min_swaps(self, vals: list[int]) -> int:
        order = sorted(range(len(vals)), key=lambda i: vals[i])
        seen = [False] * len(vals)
        swaps = 0
        for i in range(len(vals)):
            if seen[i] or order[i] == i:
                continue
            cycle = 0
            j = i
            while not seen[j]:
                seen[j] = True
                j = order[j]
                cycle += 1
            swaps += cycle - 1
        return swaps