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+---
+layout: post
+title: 逐层排序二叉树所需的最少操作数目
+date: 2024-12-23 17:30:00 +0800
+categories: algorithm
+tags: algorithm
+published: true
+---
+
+* content
+{:toc}
+
+## Minimum Number of Operations to Sort a Binary Tree by Level
+
+[LeetCode 2471](https://leetcode.com/problems/minimum-number-of-operations-to-sort-a-binary-tree-by-level/description/){:target="_blank"}
+
+You are given the root of a binary tree with unique values.
+
+In one operation, you can choose any two nodes at the same level and swap their values.
+
+Return the minimum number of operations needed to make the values at each level sorted in a strictly increasing order.
+
+The level of a node is the number of edges along the path between it and the root node.
+
+**Example 1:**
+
+![example-1]({{ '/styles/images/leetcode-2471/example-1.png' | prepend: site.baseurl }}){:width="404" height="auto"} 
+
+```
+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.
+```
+
+**Example 2:**
+
+![example-1]({{ '/styles/images/leetcode-2471/example-2.png' | prepend: site.baseurl }}){:width="293" height="auto"} 
+
+```
+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.
+```
+
+**Example 3:**
+
+![example-1]({{ '/styles/images/leetcode-2471/example-3.png' | prepend: site.baseurl }}){:width="354" height="auto"} 
+
+```
+Input: root = [1,2,3,4,5,6]
+Output: 0
+Explanation: Each level is already sorted in increasing order so return 0.
+```
+
+**Constraints:**
+
+* The number of nodes in the tree is in the range [1, $$10^5$$].
+* 1 <= Node.val <= $$10^5$$
+* All the values of the tree are unique.
+
+## Hash Map
+
+```kotlin
+class Solution {
+    fun minimumOperations(root: TreeNode?): Int {
+        if (root == null) return 0
+        var operators = 0
+        val queue = LinkedList<TreeNode>()
+        queue.offer(root)
+
+        while (queue.isNotEmpty()) {
+            val size = queue.size
+            val list = MutableList(size) { 0 }
+            repeat(size) {
+                val node = queue.pop()
+                list[it] = node.`val`
+                node.left?.apply(queue::offer)
+                node.right?.apply(queue::offer)
+            }
+            operators += minSwaps(list)
+        }
+        return operators
+    }
+
+    fun minSwaps(original: MutableList<Int>): Int {
+        var swaps = 0
+        val target = original.sorted()
+        val value2Index = original.withIndex().associate { Pair(it.value, it.index) }.toMutableMap()
+        repeat(original.size) { i ->
+            if (original[i] != target[i]) {
+                swaps++
+                val index = value2Index[target[i]]!!
+                value2Index[original[i]] = index
+                original[index] = original[i]
+            }
+        }
+        return swaps
+    }
+}
+```
+
+## Bit Manipulation
+
+```kotlin
+class Solution {
+    val shift = 20
+    val mask = 1L.shl(shift) - 1
+
+    fun minimumOperations(root: TreeNode?): Int {
+        if (root == null) return 0
+        var operators = 0
+        val queue = LinkedList<TreeNode>()
+        queue.offer(root)
+        while (queue.isNotEmpty()) {
+            val size = queue.size
+            val list = MutableList(size) { 0L }
+            repeat(size) {
+                val node = queue.pop()
+                list[it] = node.`val`.toLong().shl(shift) + it
+                node.left?.apply(queue::offer)
+                node.right?.apply(queue::offer)
+            }
+            list.sort()
+            var i = 0
+            while (i < list.size) {
+                val index = list[i].and(mask).toInt()
+                if (index != i) {
+                    list[i] = list[index].also { list[index] = list[i] }
+                    operators++
+                } else i++
+            }
+        }
+        return operators
+    }
+}
+```
+
+## Cycle Sort
+
+```kotlin
+class Solution {
+    fun minimumOperations(root: TreeNode?): Int {
+        if (root == null) return 0
+        var operators = 0
+        val queue = LinkedList<TreeNode>()
+        queue.offer(root)
+        while (queue.isNotEmpty()) {
+            val size = queue.size
+            val list = MutableList(size) { 0 }
+            repeat(size) {
+                val node = queue.pop()
+                list[it] = node.`val`
+                node.left?.apply(queue::offer)
+                node.right?.apply(queue::offer)
+            }
+            operators += cycleSort(list)
+        }
+        return operators
+    }
+
+    fun cycleSort(list: MutableList<Int>): Int {
+        val value2Index = list.withIndex().associate { Pair(it.value, it.index) }
+        list.sort()
+        val visited = BooleanArray(list.size) { false }
+        var ans = 0
+        repeat(list.size) { i ->
+            if (visited[i] || value2Index[list[i]] == i) return@repeat
+            var j = i
+            var cycleSize = 0
+            while (!visited[j]) {
+                visited[j] = true
+                j = value2Index[list[j]]!!
+                cycleSize++
+            }
+            ans += cycleSize - 1
+        }
+        return ans
+    }
+}
+```
+
+## Bit Manipulation + Cycle Sort
+
+```kotlin
+class Solution {
+    val shift = 20
+    val mask = 1L.shl(shift) - 1
+
+    fun minimumOperations(root: TreeNode?): Int {
+        if (root == null) return 0
+        var operators = 0
+        val queue = LinkedList<TreeNode>()
+        queue.offer(root)
+        while (queue.isNotEmpty()) {
+            val size = queue.size
+            val list = MutableList(size) { 0L }
+            repeat(size) {
+                val node = queue.pop()
+                list[it] = node.`val`.toLong().shl(shift) + it
+                node.left?.apply(queue::offer)
+                node.right?.apply(queue::offer)
+            }
+            operators += cycleSort(list)
+        }
+        return operators
+    }
+
+    fun cycleSort(list: MutableList<Long>): Int {
+        list.sort()
+        val visited = BooleanArray(list.size) { false }
+        var ans = 0
+        repeat(list.size) { i ->
+            if (visited[i] || list[i].and(mask).toInt() == i) return@repeat
+            var j = i
+            var cycleSize = 0
+            while (!visited[j]) {
+                visited[j] = true
+                j = list[j].and(mask).toInt()
+                cycleSize++
+            }
+            ans += cycleSize - 1
+        }
+        return ans
+    }
+}
+```
+
+<!-- https://leetcode.com/problems/minimum-number-of-operations-to-sort-a-binary-tree-by-level/editorial/ -->
+<!-- https://leetcode.com/problems/minimum-number-of-operations-to-sort-a-binary-tree-by-level/solutions/6175754/graph-based-approach-cycle-count-by-sume-rfxl/ -->
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