Create 2471. Minimum Number of Operations to Sort a Binary Tree by Level

2471. Minimum Number of Operations to Sort a Binary Tree by Level

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+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    int minimumOperations(TreeNode* root) {
+        int ans = 0;
+        queue<TreeNode*> q{{root}};
+        while (!q.empty()){
+            vector<int> vals;
+            vector<int> child(q.size());
+            for (int size = q.size(); size > 0; size--){
+                TreeNode* node = q.front();
+                q.pop();
+                vals.push_back(node->val);
+                if (node->left != nullptr)
+                    q.push(node->left);
+                if (node->right != nullptr)
+                    q.push(node->right);
+            }
+            iota(child.begin(), child.end(), 0);
+            ranges::sort(child, [&vals](int i, int j){ return vals[i] < vals[j]; });
+            for (int i = 0; i < child.size(); i++)
+                for ( ;child[i] != i; ans++)
+                    swap(child[i], child[child[i]]);
+        }
+        return ans;
+    }
+};