• 2022-06-19 问题

    The Chinese government has been providing training to migrant workers inorder to help them have better job opportunities.

    The Chinese government has been providing training to migrant workers inorder to help them have better job opportunities.

  • 2022-06-16 问题

    void PreOrder(BinTree bt)//递归先序遍历算法{ if(bt==NULL) return; //递归出口visit(bt); //访问根结点 InOrder (leftchild(bt)); //中序遍历左子树 InOrder (rightchild(bt)); //中序遍历右子树 }void InOrder(BinTree bt)//递归中序遍历算法{ if(bt==NULL) return; //递归出口 PreOrder (leftchild(bt)); //先序遍历左子树 visit(bt); //访问根结点 PreOrder (rightchild(bt)); //先序遍历右子树 }void main(){ bt = CreateBinTree();//创建一棵二叉树 Preorder(bt); //入口}对下面二叉树执行以上程序,则输出序列是()[img=94x192]1803078d93c9821.png[/img] A: 1,2,3,4,5 B: 1,3,5,4,2 C: 5,4,3,2,1 D: 1,3,4,5,2

    void PreOrder(BinTree bt)//递归先序遍历算法{ if(bt==NULL) return; //递归出口visit(bt); //访问根结点 InOrder (leftchild(bt)); //中序遍历左子树 InOrder (rightchild(bt)); //中序遍历右子树 }void InOrder(BinTree bt)//递归中序遍历算法{ if(bt==NULL) return; //递归出口 PreOrder (leftchild(bt)); //先序遍历左子树 visit(bt); //访问根结点 PreOrder (rightchild(bt)); //先序遍历右子树 }void main(){ bt = CreateBinTree();//创建一棵二叉树 Preorder(bt); //入口}对下面二叉树执行以上程序,则输出序列是()[img=94x192]1803078d93c9821.png[/img] A: 1,2,3,4,5 B: 1,3,5,4,2 C: 5,4,3,2,1 D: 1,3,4,5,2

  • 2022-06-16 问题

    ​void PreOrder(BinTree bt)//递归先序遍历算法‌​{‌​ if(bt==NULL) return; //递归出口‌​ visit(bt); //访问根结点 ‌​ InOrder (leftchild(bt)); //中序遍历左子树‌​ InOrder (rightchild(bt)); //中序遍历右子树 ‌​}‌​void InOrder(BinTree bt)//递归中序遍历算法‌​{‌​ if(bt==NULL) return; //递归出口‌​ PreOrder (leftchild(bt)); //先序遍历左子树‌​ visit(bt); //访问根结点 ‌​ PreOrder (rightchild(bt)); //先序遍历右子树 ‌​}‌​void main()‌​{‌​ bt = CreateBinTree(); //创建一棵二叉树‌​ Preorder(bt); //入口‌​}‌​对下面二叉树执行以上程序,则输出序列是()‌​[img=94x192]18031cb3c2815d5.png[/img]‌ A: 1,3,5,4,2 B: 1,2,3,4,5 C: 5,4,3,2,1 D: 1,3,4,5,2

    ​void PreOrder(BinTree bt)//递归先序遍历算法‌​{‌​ if(bt==NULL) return; //递归出口‌​ visit(bt); //访问根结点 ‌​ InOrder (leftchild(bt)); //中序遍历左子树‌​ InOrder (rightchild(bt)); //中序遍历右子树 ‌​}‌​void InOrder(BinTree bt)//递归中序遍历算法‌​{‌​ if(bt==NULL) return; //递归出口‌​ PreOrder (leftchild(bt)); //先序遍历左子树‌​ visit(bt); //访问根结点 ‌​ PreOrder (rightchild(bt)); //先序遍历右子树 ‌​}‌​void main()‌​{‌​ bt = CreateBinTree(); //创建一棵二叉树‌​ Preorder(bt); //入口‌​}‌​对下面二叉树执行以上程序,则输出序列是()‌​[img=94x192]18031cb3c2815d5.png[/img]‌ A: 1,3,5,4,2 B: 1,2,3,4,5 C: 5,4,3,2,1 D: 1,3,4,5,2

  • 2022-06-05 问题

    Inorder to take steps to fulfill our ultimate, splendid dreams, we have to_______. A: take action without planning B: start with goals C: make a thorough plan before taking actions D: believe in our specific objectives

    Inorder to take steps to fulfill our ultimate, splendid dreams, we have to_______. A: take action without planning B: start with goals C: make a thorough plan before taking actions D: believe in our specific objectives

  • 2022-06-06 问题

    二叉树中序遍历的非递归算法如下所示。请填写算法中下划线的空白处。[br][/br]Status Inorder(BiTree T){[br][/br]             InitStack(s);push(s,T);            While ( (1)){[br][/br]                            While (gettop(s,p)&&p) push (s, (2));              pop(s,p);              if(!StackEmpty(s)){                        pop(s,p);printf( (3));                        push(s, (4));[br][/br]              }//if[br][/br]        }//while[br][/br]         return ok;[br][/br]}//Inorder[br][/br]注:[br][/br]InitStack(s);初始化一个栈s;[br][/br]push(s,p);将所指向的结点进s栈[br][/br];pop(s,p);s栈顶元素出栈;[br][/br]gettop(s,p);取s栈顶元素;[br][/br]StackEmpty(s);判栈s是否为空。

    二叉树中序遍历的非递归算法如下所示。请填写算法中下划线的空白处。[br][/br]Status Inorder(BiTree T){[br][/br]             InitStack(s);push(s,T);            While ( (1)){[br][/br]                            While (gettop(s,p)&&p) push (s, (2));              pop(s,p);              if(!StackEmpty(s)){                        pop(s,p);printf( (3));                        push(s, (4));[br][/br]              }//if[br][/br]        }//while[br][/br]         return ok;[br][/br]}//Inorder[br][/br]注:[br][/br]InitStack(s);初始化一个栈s;[br][/br]push(s,p);将所指向的结点进s栈[br][/br];pop(s,p);s栈顶元素出栈;[br][/br]gettop(s,p);取s栈顶元素;[br][/br]StackEmpty(s);判栈s是否为空。

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