二叉树表达规定形式的代数表达式并计算表达式结果

#include<stdio.h>

//程序缺陷:4/5这样的浮点结果不能正确计算.待改进.

//仔细分析有代表性的实例,发现规律.

typedef struct node    // 结点类型定义
{
  char data;
        struct node *lchild;
        struct node *rchild;
}BTNode;

BTNode *create(char *str,int n)    // 创建代数表达式对应的二叉树
{
  BTNode *root = NULL;   //始终指向当前已经生成的二叉树的根
  BTNode *p = NULL;       //始终指向当前待处理结点
        char ch;
  int j = 0;
  ch = str[j];
while(ch!='\0' && j < n)
{
    switch(ch)
    {
    case '0':   //数字的处理
    case '1':
         case '2':
         case '3':
         case '4':
         case '5':
         case '6':
         case '7':
         case '8':
         case '9':
     p = new BTNode;
     p->data = ch;
     if(root == NULL)    
        root = p;
     else if(root->rchild == NULL)    
        root->rchild = p;
     else    
        root->rchild->rchild=p;    
     break;
    case'+': //扫描到'+''-',作为当前根结点
    case'-':
     p = new BTNode;
     p->data = ch;
     p->rchild = NULL;
     p->lchild = root;
     root = p;
     root->rchild = NULL;
     break;
    case'*':    //'*'和'/'的处理
    case'/':
     if(root->data >= '0' && root->data <= '9' || root->data == '*' || root->data == '/')
     {    
        p=new BTNode;
                         p->data = ch;
        p->rchild = NULL;
        p->lchild = root;
        root = p;
     }
     else
     {
        p = new BTNode;
             p->data = ch;
             p->lchild = root->rchild;
             root->rchild = p;
             p->rchild = NULL;
     }
     break;
    }
    j++;
    ch=str[j];
}
return root;
}
int result(BTNode *b) //计算表达式
{
  int n1,n2;
        if(b->data>='0'&&b->data<='9')
    return (b->data-'0');
        n1=result(b->lchild);//计算左子表达式
        n2=result(b->rchild);//计算右子表达式
  switch(b->data)
  {
  case'+':
    b->data=n1+n2;break;
  case'-':
         b->data=n1-n2;break;
  case'*':
         b->data=n1*n2;break;    
  case'/':
         b->data=n1/n2;break;
  }
  return b->data;
}
void main()
{
  BTNode *b=NULL,*q=NULL;
        char str[100];
        printf("请输入符合要求的代数表达式：");
        scanf("%s",str);
        int n=0,i=0;
  while(str[i++] != '\0')
    n++;
  b = create(str,n);
  printf("该表达式的值为：%d\n",result(b));
}