Efficient & Clear Solution !
Steps
Performing a level-order traversal of the binary tree.
Extracting the values at each level into an array.
Calculating the minimum number of swaps required

# Steps

1. Performing a level-order traversal of the binary tree.
2. Extracting the values at each level into an array.
3. Calculating the minimum number of swaps required to sort the array.

The `minSwaps` function computes the minimum number of swaps using a cycle detection method.

---



# Code Implementation 💻

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int minimumOperations(TreeNode root) {
        Queue<TreeNode> currentLevel = new LinkedList<TreeNode>();
        currentLevel.add(root);
        int totalOperations = 0;

        while (!currentLevel.isEmpty()) {
            Queue<TreeNode> nextLevel = new LinkedList<TreeNode>();
            TreeNode[] currentNodes = currentLevel.toArray(new TreeNode[0]);
            int[] currentValues = new int[currentNodes.length];

            for (int i = 0; i < currentNodes.length; i++) { // O(n)
                TreeNode node = currentNodes[i];
                currentValues[i] = node.val;

                if (node.left != null) 
                    nextLevel.add(node.left);
                if (node.right != null) 
                    nextLevel.add(node.right);
            }

            totalOperations += minSwaps(currentValues);
            currentLevel = nextLevel;
        }
        return totalOperations;
    }    

    public int minSwaps(int[] values) {
        int[] sortedValues = values.clone();
        Arrays.sort(sortedValues);

        // Map each value to its correct index
        HashMap<Integer, Integer> sortedIndex = new HashMap<Integer, Integer>();
        for (int i = 0; i < sortedValues.length; i++) {
            sortedIndex.put(sortedValues[i], i); // each value with its correct index
        }

        int min = 0;
        HashSet<Integer> visited = new HashSet<Integer>();

        // Process each cycle
        for (int i = 0; i < values.length; i++) {
            if (visited.contains(i) || i == sortedIndex.get(values[i])) {
                continue;
            }

            int path = 0, currIndex = i;

            while (!visited.contains(currIndex)) {
                path++;
                visited.add(currIndex);
                currIndex = sortedIndex.get(values[currIndex]); // Traverse the cycle
            }

            min += (path > 0) ? path - 1 : 0;
        }

        return min;
    }
}

```

---

### Explanation

#### **Key Components:**
1. **Level Order Traversal:**
   - Use a queue to traverse the tree level by level.
   - Extract the values at each level into an array.

2. **Minimum Swaps Calculation:**
   - The `minSwaps` function determines the minimum swaps needed to sort an array using cycle detection.
   - It uses a hash map to map values to their correct indices in the sorted array.
   - A visited set ensures that each index is processed only once.

#### **How `minSwaps` Works:**
The `minSwaps` function is based on the idea of detecting cycles in a graph representation of the array. Each value in the array points to its correct index in the sorted array. This can be visualized as a directed graph:

1. **Sorted Indices Mapping:**
   - A hash map stores the mapping of values to their correct positions in the sorted array.

2. **Cycle Detection:**
   - Starting from each unvisited index, traverse the graph by jumping to the correct index for each value.
   - Count the number of nodes (indices) in this cycle.

3. **Calculating Swaps:**
   - For a cycle of length \(k\), the minimum number of swaps needed to sort the cycle is \(k-1\).
   - This is because in each swap, one value reaches its correct position, reducing the cycle size by 1.

4. **Efficiency:**
   - By marking indices as visited, each index is processed only once, ensuring optimal performance.

Example:
- Input array: [7, 6, 5, 4]
- Sorted array: [4, 5, 6, 7]
- Graph representation:
  - 7 -> 4 (index 0 -> index 3)
  - 4 -> 5 (index 3 -> index 2)
  - 5 -> 6 (index 2 -> index 1)
  - 6 -> 7 (index 1 -> index 0)
- This forms a cycle of length 4. Minimum swaps = 4 - 1 = 3.

---

### Time Complexity
- **Level Order Traversal:** \(O(N)\), where \(N\) is the number of nodes in the tree.
- **Min Swaps Calculation:** \(O(M log M + M)\), where \(M\) is the number of nodes at a level.

### Space Complexity
- \(O(N)\) for storing nodes in the queue and arrays.
- \(O(M)\) for the hash map and visited set during `minSwaps` calculation.

---

### Example
#### Input:
```
      1
     / \
    3   2
   / \
  7   6
```
#### Output:
```
2
```
#### Explanation:
1. Level 1: [1] - No swaps needed.
2. Level 2: [3, 2] - Swap to get [2, 3] (1 swap).
3. Level 3: [7, 6] - Swap to get [6, 7] (1 swap).
4. Total swaps = 1 + 1 = 2.

---

### Notes
- This solution handles both balanced and unbalanced trees.
- It ensures minimal operations by processing each level independently.

---

['Tree', 'Breadth-First Search', 'Java']