Minimum Number of Operations to Sort a Binary Tree by Level
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2400-2499/2471.Minimum%20Number%20of%20Operations%20to%20Sort%20a%20Binary%20Tree%20by%20Level/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2400-2499/2471.Minimum%20Number%20of%20Operations%20to%20Sort%20a%20Binary%20Tree%20by%20Level/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2400-2499/2471.Minimum%20Number%20of%20Operations%20to%20Sort%20a%20Binary%20Tree%20by%20Level/images/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>
Tree; Breadth-First Search; Binary Tree
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function minimumOperations(root: TreeNode | null): number {
    const queue = [root];
    let ans = 0;
    while (queue.length !== 0) {
        const n = queue.length;
        const row: number[] = [];
        for (let i = 0; i < n; i++) {
            const { val, left, right } = queue.shift();
            row.push(val);
            left && queue.push(left);
            right && queue.push(right);
        }
        for (let i = 0; i < n - 1; i++) {
            let minIdx = i;
            for (let j = i + 1; j < n; j++) {
                if (row[j] < row[minIdx]) {
                    minIdx = j;
                }
            }
            if (i !== minIdx) {
                [row[i], row[minIdx]] = [row[minIdx], row[i]];
                ans++;
            }
        }
    }
    return ans;
}
