With pattern: AI检测代码解析 if(node == NULL){ // do something } left = do_something(root->left) right = do_something(root->right) do_something with root, left, right 1. 2. 3. 4. 5. 6. we can get that, left and right return its depth and the deepest node’s parents Error...
A0-indexedstringnumof lengthn +1is created using the following conditions: numconsists of the digits'1'to'9', where each digit is usedat mostonce. Ifpattern[i] =='I', thennum[i] < num[i +1]. Ifpattern[i] =='D', thennum[i] > num[i +1]. Return the lexicographicallysmallestp...