< - pr8ae7 - Online C Compiler & Debugging Tool
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#include&stdio.h&#include&malloc.h& #include &stdlib.h& #include&string.h&
#define P 100; typedef struct{ int *base; int *top; int stacksize;}Sqstack; int InitStack(Sqstack &s){
s.base=(int*)(100*sizeof(int)); if(!s.base) (0); s.top=s.base; s.stacksize=100; return 1;}int Push(Sqstack &s,int e){
if(s.top-s.base&=s.stacksize){ s.base=(int*)(s.base,(s.stacksize+100)*sizeof(int)); if(!s.base){
(&OVERFLOE!&); } s.top=s.base+s.stacksize; s.stacksize+=100; } *s.top++=e; return 1;}int Pop(Sqstack &s,int &e){ if(s.top==s.base) return 0;
e=*(--s.top); return 1;}void Stack(char a[],int m){
//应用栈进行数值转换
int n,e,x,i=0;
int t=0; Sqstack s;
InitStack(s);
//构造数组
x = (a); for(i;i&x;i++){
if(a[i]-'0'&=m&&a[i]&'A'||(a[i]-'A'+10&=m)){
//判断输入数据是否否和数制
(&数据错误!!请重新输入\n&);
(0);
if(a[i]&='0'&&a[i]&='9'){
//判断是否为数字
Push(s,a[i]-'0');
if(a[i]&='A'&&(a[i]-'A'+10&m)){
//判断是否为字母
Push(s,a[i]-'A'+10);
while(s.top!=s.base){
Pop(s,n);
t=t*m+n; } }
(&转化为十进制为：\n%d\n&,t);
//转化为十进制数 (&要转化成进制为：\n&); (&%d&,&n); while(t){
Push(s,t%n);
t=t/n; } (&转化后为：\n&);
while(s.top!=s.base){
Pop(s,e);
(&%d&,e); }}
void arry(int T){
int y,i=0;
int a[100]; (&要转化成进制为：\n&); (&%d&,&y); while(T){
a[i]=T%y;
i++; } (&转化后为:&); i=i-1; for(i;i&=0;i--) (&%d&,a[i]);}void Arry(char a[],int m){
//应用数组进行数值转换
int t1,x,i=0; int t=0; x = (a); for(i;i&x;i++){
if(a[i]-'0'&=m&&a[i]&'A'||(a[i]-'A'+10&=m)){
(&数据错误!!\n&);
(0);
if(a[i]&='0'&&a[i]&='9'){
//判断是否为数字
t1=a[i]-'0';
if(a[i]&='A'&&(a[i]-'A'+10&m)){
//判断是否为字母
t1=a[i]-'A'+10;
t=t*m+t1; }
(&转化为十进制为：\n%d\n&,t);
arry(t);}int main(){ (&*************************************\n&); (&**************数值转换***************\n&);
(&**************应用说明***************\n&); (&**************1:运用数组转换*********\n&); (&**************2:运用栈转换***********\n&); (&*************************************\n&); int n,m; char a[100]; (&输入非十进制数:\n&); (a); strupr(a);
(&输入它的进制:\n&); (&%d&,&m);
n:(&输入指令:&); (&%d&,&n); switch(n){
Arry(a,m);
Stack(a,m);
(&输入错误！！请重新选择\n&);
goto n; } return 0;}&
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Standard input is empty
compilation info
prog.c:11:23: error: expected ';', ',' or ')' before '&' token
int InitStack(Sqstack &s){
prog.c:19:18: error: expected ';', ',' or ')' before '&' token
int Push(Sqstack &s,int e){
prog.c:31:17: error: expected ';', ',' or ')' before '&' token
int Pop(Sqstack &s,int &e){
prog.c: In function 'Stack':
prog.c:41:5: warning: implicit declaration of function 'InitStack' [-Wimplicit-function-declaration]
InitStack(s);
//构造数组
prog.c:43:2: warning: statement with no effect [-Wunused-value]
for(i;i&x;i++){
prog.c:44:17: warning: suggest parentheses around '&&' within '||' [-Wparentheses]
if(a[i]-'0'&=m&&a[i]&'A'||(a[i]-'A'+10&=m)){
//判断输入数据是否否和数制
prog.c:49:4: warning: implicit declaration of function 'Push' [-Wimplicit-function-declaration]
Push(s,a[i]-'0');
prog.c:55:3: warning: implicit declaration of function 'Pop' [-Wimplicit-function-declaration]
prog.c: In function 'arry':
prog.c:84:2: warning: statement with no effect [-Wunused-value]
for(i;i&=0;i--)
prog.c: In function 'Arry':
prog.c:91:2: warning: statement with no effect [-Wunused-value]
for(i;i&x;i++){
prog.c:92:17: warning: suggest parentheses around '&&' within '||' [-Wparentheses]
if(a[i]-'0'&=m&&a[i]&'A'||(a[i]-'A'+10&=m)){
prog.c: In function 'main':
prog.c:117:2: warning: implicit declaration of function 'gets' [-Wimplicit-function-declaration]
prog.c:118:2: warning: implicit declaration of function 'strupr' [-Wimplicit-function-declaration]
strupr(a);
Standard output is empty
visibility:
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三、运算题（每题 6 分，共24）1.写出下列中缀表达式的后缀形式：（1）3X/(Y-2)+1（2）2+X*(Y+3)&图22.试对图2中的二叉树画出：（1）顺序存储表示的示意图；（2）二叉链表存储表示的示意图。 4.已知一个图的顶点集V和边集E分别为：V={1,2,3,4,5,6,7};E={(1,2)3,(1,3)5,(1,4)8,(2,5)10,(2,3)6,(3,4)15,(3,5)12,(3,6)9,(4,6)4,(4,7)20,(5,6)18,(6,7)25};按照普里姆算法从顶点1出发得到最小生成树，试写出在最小生成树中依次得到的各条边。四、阅读算法（每题7分，共14分）1.void AE(Stack& S){&&&&&& &&&&&& InitStack(S);&&&&&& &&&&&& Push(S,3);&&&&& &&&&&& Push(S,4);&&&&&& &&&&&& int x=Pop(S)+2*Pop(S);&&&&&& &&&&&& Push(S,x);&&&&&& &&&&&& int i,a[5]={1,5,8,12,15};&&&&&& &&&&&& for(i=0;i&5;i++) Push(S,2*a[i]);&&&&&& &&&&&& while(!StackEmpty(S)) cout&&Pop(S)&&' ';&&&&&& }该算法被调用后得到的输出结果为：_____________。2.void ABC (BTNode *BT,int &c1,int &c2) {if (BT !=NULL ) {&&&& ABC(BT-&left,c1,c2);&&&& c1++;&&&& if (BT-&left==NULL&&BT-&right==NULL)& c2++;&&&&& ABC(BT-&right,c1,c2);& &&& }//if& }该函数执行的功能是什么？五、算法填空（共8分）向单链表的末尾添加一个元素的算法。Void InsertRear(LNode*& HL,const ElemType& item){LNode*newptr=new LNIf (______________________){cerr&&"Memory allocation failare!"&&exit(1);}________________________=newptr-&next=NULL;if (HL==NULL)& HL=__________________________;else{LNode* P=HL;While (P-&next!=NULL)&& ____________________;p-&next=}&&&&&&& }六、编写算法（共8分）编写从类型为List的线性表L中将第i个元素删除的算法，（假定不需要对i的值进行有效性检查,也不用判别L是否为空表。）void Delete(List& L, int i)三、运算题1.（1）3& X& * Y& 2& - / 1& +&&&（2）2& X& Y& 3& + ...
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品评校花校草，体验校园广场关于数据结构中二叉树的中序遍历非递归算法最近做一个二叉树的中序遍历的非递归算法，但是编译通过就是运行出问题，哪位高手帮我改改错啊~~我通过debug知道问题一定出在入栈了，因为执行完GoFarLeft(T,s)这个函数后栈的top指针并没有向下移动一个单位，所以再次入栈的时候就把原来栈中的元素覆盖了……但是实在不知道该怎么改，请帮帮忙~~~：） C/C++ code #include &stdafx.h& #include &stdio.h& #include &malloc.h& #include &stdlib.h&
#define OVERFLOW -2 #define TRUE 1 #define FALSE 0 #define ERROR 0 #define STACK_INIT_SEZE 100 #define STACKINCREMENT 10
typedef struct BiTNode{
struct BiTNode* lchild,* }BiTNode,*BiT typedef struct Stack{
BiTree* }S
void InitStack(Stack &s){
s.base = (BiTree*)malloc(STACK_INIT_SEZE*sizeof(BiTNode));
if(!s.base) exit(OVERFLOW);
s.top = s.
s.stacksize = STACK_INIT_SEZE; } int StackEmpty(Stack s){
if(s.top == s.base)
return FALSE;
else return TRUE; } void Push(Stack &s,BiTree T){
*s.top = T;
s.top++; } BiTree Pop(Stack &s){
BiTree e = NULL;
if(s.base == s.top) return ERROR;
BiTree CreaBiTree(BiTree &T){
scanf(&%c&,&ch);
if(ch == 'x')
if(!(T = (BiTNode*)malloc(sizeof(BiTNode))))
exit(OVERFLOW);
CreaBiTree(T-&lchild);
CreaBiTree(T-&rchild);
return T; } void visit(char& e){
printf(&%c&,e); } BiTNode* GoFarLeft(BiTree T,Stack s){
if(!T) return NULL;
while(T-&lchild){
Push(s,T);
return T; } void Inorder_Mid(BiTree T,Stack &s,void(*visit)(char &e)){
BiTree t = GoFarLeft(T,s);
visit(t-&data);
if(t-&rchild)
t = GoFarLeft(t-&rchild,s);
if(!StackEmpty(s))
t = Pop(s);
else t = NULL;
int main(int argc, char* argv[]) {
BiTree tree = CreaBiTree(tree);
InitStack(s);
Inorder_Mid(tree,s,visit);
printf(&Hello World!\n&);
return 0; }
回答1： C/C++ code 1.先序遍历非递归算法
#define maxsize 100
typedef struct
Bitree Elem[maxsize];
void PreOrderUnrec(Bitree t)
StackInit(s);
while (p!=null || !StackEmpty(s))
while (p!=null)
//遍历左子树
visite(p-&data);
push(s,p);
}//endwhile
if (!StackEmpty(s))
//通过下一次循环中的内嵌while实现右子树遍历
}//endwhile
}//PreOrderUnrec
2.中序遍历非递归算法
#define maxsize 100
typedef struct
Bitree Elem[maxsize];
void InOrderUnrec(Bitree t)
StackInit(s);
while (p!=null || !StackEmpty(s))
while (p!=null)
//遍历左子树
push(s,p);
}//endwhile
if (!StackEmpty(s))
visite(p-&data);
//访问根结点
//通过下一次循环实现右子树遍历
}//endwhile
}//InOrderUnrec
3.后序遍历非递归算法
#define maxsize 100
typedef struct
typedef struct
stacknode Elem[maxsize];
void PostOrderUnrec(Bitree t)
StackInit(s);
while (p!=null)
//遍历左子树
x.tag = L;
//标记为左子树
push(s,x);
while (!StackEmpty(s) && s.Elem[s.top].tag==R)
x = pop(s);
visite(p-&data);
//tag为R，表示右子树访问完毕，故访问根结点
if (!StackEmpty(s))
s.Elem[s.top].tag =R;
//遍历右子树
p=s.Elem[s.top].ptr-&
}while (!StackEmpty(s));
}//PostOrderUnrec
=============================================================
Status InOrderTraverse(BiTree T, Status(*Visit)(TElemType e))
//采用二叉链表存储结构，Visit是对数据元素操作的应用函数。
//中序遍历二叉树T的非递归算法，对每个数据元素调用函数Visit。
InitStack(S);
Push(S,T); //根指针进栈
while(!StackEmpty(s))
while(GetTop(S,p)&&p)
Push(S,p-&lchild); //向左走到尽头
Pop(S,p); //空指针退栈
if(!StackEmpty(S)) //访问结点，向右一步
if(!Visit(p-&data))
return ERROR;
Push(S,p-&rchild);
return OK;
}//InOrderTraverse 算法一
============================================================
============================================================
Status InOrderTraverse(BiTree T, Status (*Visit)(TelemType e))
//采用二叉链表存储结构，Visit是对数据元素操作的应用函数。
//中序遍历二叉树T的非递归算法，对每个数据元素调用函数Visit。
InitStack(S);
while(p||!StackEmpty(S))
Push(S,p);
}//if 根指针进栈，遍历左子树
if(!Visit(p-&data))
return ERROR;
}//else 根指针退栈，访问根结点，遍历右子树
}InOrderTraverse() 算法二
============================================================
