C# Level-Wise Transformation with BFS and Minimum Swaps Solution
Intuition

Intuition About Tree Structure: When reading the problem, it becomes clear that levels of the tree are independent in terms of sorting. Therefore, th
# Intuition
- <!-- Describe your first thoughts on how to solve this problem. -->Intuition About Tree Structure: When reading the problem, it becomes clear that levels of the tree are independent in terms of sorting. Therefore, the solution must treat each level separately while ensuring the tree traversal captures these levels in order.
- Immediate Thoughts About BFS: BFS is the natural choice for grouping nodes level by level because it processes all nodes at the same depth before moving to the next depth.
- Observation of Minimum Swaps: Sorting a level in-place using swaps immediately brings to mind problems like cycle detection in a permutation. By calculating cycles in the array formed by the current positions of the level's values, we can determine the minimum swaps required.
- Step-by-Step Plan: First, traverse the tree level by level (using BFS).
- Then, for each level, extract the node values and calculate the minimum swaps needed to sort them using cycle detection.
- Sum up the swaps for all levels to find the total minimum operations.
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# Approach
- <!-- Describe your approach to solving the problem. -->BFS to Traverse Levels:
- *We use BFS to traverse the binary tree level by level.
- *For each level, we collect the values of the nodes into a list.
- Sort Each Level: For the collected values at each level, we calculate the minimum number of swaps needed to sort them in strictly increasing order. This is done using a helper method MinimumSwapsToSort.
- Minimum Swaps to Sort:
- *For sorting the array, we create a list of tuples where each tuple contains the value and its original index.
- *Sort the list based on the values.
- *Use a visited array to determine cycles in the permutation of the array indices. Each cycle contributes (cycle size - 1) to the number of swaps.
- Sum up Swaps: Add the swaps needed for all levels to get the total minimum number of operations.
___
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->O(n log k), where 'k' is the average number of nodes per level.
___
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->O(k), where 'k' is the number of nodes at the largest level.
___
# Code
```csharp []
using System;
using System.Collections.Generic;

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public int MinimumOperations(TreeNode root) {
        if (root == null) return 0;
        
        // Perform BFS to group the tree nodes level by level
        Queue<TreeNode> queue = new Queue<TreeNode>();
        queue.Enqueue(root);
        int totalSwaps = 0;
        
        while (queue.Count > 0) {
            int levelSize = queue.Count;
            List<int> levelValues = new List<int>();
            
            // Collect all node values for the current level
            for (int i = 0; i < levelSize; i++) {
                TreeNode node = queue.Dequeue();
                levelValues.Add(node.val);
                
                // Add child nodes to the queue for the next level
                if (node.left != null) queue.Enqueue(node.left);
                if (node.right != null) queue.Enqueue(node.right);
            }
            
            // Count minimum swaps needed to sort the current level
            totalSwaps += MinimumSwapsToSort(levelValues);
        }
        
        return totalSwaps;
    }

    private int MinimumSwapsToSort(List<int> values) {
        int n = values.Count;
        int swaps = 0;

        // Create a list of value-index pairs and sort it based on the values
        List<(int value, int index)> sorted = new List<(int value, int index)>();
        for (int i = 0; i < n; i++) {
            sorted.Add((values[i], i));
        }
        sorted.Sort((a, b) => a.value.CompareTo(b.value));

        // Create a visited array to track processed elements
        bool[] visited = new bool[n];
        
        for (int i = 0; i < n; i++) {
            // If the element is already visited or is in the correct position, skip it
            if (visited[i] || sorted[i].index == i) continue;
            
            // Calculate the size of the cycle
            int cycleSize = 0;
            int j = i;
            
            while (!visited[j]) {
                visited[j] = true;
                j = sorted[j].index;
                cycleSize++;
            }
            
            // If there is a cycle of size k, we need (k - 1) swaps
            if (cycleSize > 1) {
                swaps += (cycleSize - 1);
            }
        }
        
        return swaps;
    }
}
```
['C#']