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

/**
 * 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;
    }
};