When I use the stack to deposit the root node，only A B two root nodes can enter the stack, and then the program ends
- input:
AB@@C@D@@# - output:
BA
Stack definition, initialization, judgment whether it is empty, into the stack, out of the stack.
//定义栈
typedef struct
{
char *base;//栈底指针
char *top;//栈顶指针
int sta
SqStack lnitStack(SqStack & S)//返回结为构体类型的顺序栈的地址
{
S.base = (char*)malloc(MAXSIZE * sizeof(char));
if(NULL == S.base)
exit(-1);
else
{
S.top = S.base;
S.stacksize = MAXSIZE;
}
return S;
}
void Push(SqStack & S,BiTree *p)
{
*(S.top)= (*p)->data;
//printf("入栈：%c",*(S.top));
S.top++;
}
void Pop(SqStack & S,BiTree* q)
{
--S.top;
//printf("出栈：%c",*(S.top));
(*q)->data = *(S.top);
}
bool empty(SqStack S)//判断空栈
{
if(S.base == S.top)
return true;
else
return false;
}
Non-recursive mid-order traversal binary trees:
void InOrderTraverse(BiTree T)
{
BiTree p,q;
q = (BiTree)malloc(sizeof(BiNode));
SqStack S;
lnitStack(S);
p = T;//因为原本是指针T指向二叉树的根节点，将T赋值给p，则使指针p也指向根节点
while(p || !empty(S))
{
if(p)//如果是节点不为空（节点不为空就是把该节点当作根）
{
printf("p = %c",p->data);
Push(S,&p);//则根节点其入栈
p = p->lchild;//访问它的左子树
}
else//当根节点为空栈为空时
{
Pop(S,&q);//弹出栈顶元素--根节点
printf("%3c", q->data);
p = q->rchild;//访问根节点的右子树
}
}
return ;
}
I want to know which step wrong, why program can't output complete results?
