花了大把的時間研究裡面二叉樹應用解決計算表達式的問題
#include <iostream>
using namespace std;
class calc
{
enum Type {DATA, ADD, SUB, MULTI, DIV, OPAREN, CPAREN, EOL};
struct node
{
Type type;
int data;
node *lchild, *rchild;
node(Type t, int d=0, node *lc=NULL, node *rc=NULL)
{
type=t; data=d; lchild=lc; rchild=rc;
}
};
node *root;
node *create(char * &s);
Type getToken (char * &s, int &value);
int result (node *t);
public:
calc (char *s) {root=create(s);}
int result() {if (root==NULL) return 0;
return result(root);}
};
calc::node *calc::create(char * &s)
{
node *p, *root=NULL;
Type returnType,flag=DATA;
int value;
while (*s)
{
flag=returnType;
returnType=getToken(s,value);
switch (returnType)
{
case DATA:
case OPAREN:
if (returnType == DATA) p=new node(DATA,value);
else p=create(s);
if (root==NULL) root=p;
else if (root->rchild==NULL) root->rchild=p;
else root->rchild->rchild=p;
break;
case CPAREN:
case EOL: return root;
case ADD:
case SUB:
root=new node(returnType,0,root);
break;
case MULTI:
case DIV:
if (root->type==DATA || root->type==MULTI || root->type==DIV || flag==OPAREN)
root=new node(returnType,0,root);
else root->rchild=new node(returnType,0,root->rchild);
}
}
return root;
}
calc::Type calc::getToken(char *&s, int &data)
{
char type;
while (*s==' ') ++s;
if (*s>='0' && *s<='9')
{
data=0;
while (*s>='0' && *s<='9') {data=data*10+ *s-'0'; ++s;}
return DATA;
}
if (*s == '\0') return EOL;
type =*s; ++s;
switch(type)
{
case '+':return ADD;
case '-':return SUB;
case '*':return MULTI;
case '/':return DIV;
case '(':return OPAREN;
case ')':return CPAREN;
default: return EOL;
}
}
int calc::result(node *t)
{
int num1,num2;
if (t->type == DATA) return t->data;
num1=result(t->lchild);
num2=result(t->rchild);
switch(t->type)
{
case ADD:t->data=num1+num2;break;
case SUB:t->data=num1-num2;break;
case MULTI: t->data=num1*num2;break;
case DIV:t->data=num1/num2;break;
}
return t->data;
}
int main()
{
char equation[256];
cin>>equation;
//calc exp("3*(2+(6/2))");
calc exp(equation);
cout<<exp.result()<<endl;
return 0;
}
*
