此函數(shù)驗(yàn)證式子是否正確

1、#include<iostream> #include<string> #include<stack> #include"Tree.h" /Checks if expression if ok/ bool isok(string exp) /此函數(shù)驗(yàn)證式子是否正確，即是否符合運(yùn)算規(guī)則。 char check; int error=0; int lb=0; int rb=0; if(exp.size()=1)return false; else if(IsOperator(exp0)|IsOperator(expexp.size()-1)&

2、amp;&exp0!='('&&expexp.size()-1!=')') /此處若不加，在遇到某些式子時(shí)，會(huì)出現(xiàn)非法操作。 return false; for(int m=0;m<exp.size();m+) check=expm; if(IsOperand(check); /如果是數(shù)字，跳過，不管。 else if(IsOperator(check) if(check=')') rb+; if(IsOperator(expm+1)&&(expm+1='+'|expm+1='

3、-'|expm+1='*'|expm+1='/'|expm+1=''|expm+1=')') m+; if(expm=')') rb+; else if(IsOperator(expm+1) error+; else if(check='(') lb+; if(expm+1='(') m+; lb+; else if(IsOperator(expm+1) error+; else if(IsOperator(expm+1)&&expm+1='('

4、;) m+; lb+; else if(IsOperator(expm+1) error+; else error+; if(error=0&&lb=rb) return(true); else return(false); / int main() /主函數(shù)開始 binary_tree etree; stack<binary_tree>NodeStack; stack<char>OpStack; string infix; char choice='y' char c; while(choice='y'|choice=&

5、#39;Y') cout<<"nn請(qǐng)輸入表達(dá)式，不要帶空格;n" cin>>infix; cout<<"-"<<'n' cout<<"表達(dá)式為: "<<infix<<'n' if(isok(infix) for(int i=0;i<infix.size();i+) c=infixi; if(IsOperand(c) string tempstring; tempstring=tempstring+c; wh

6、ile(i+1<infix.size()&&IsOperand(infixi+1) tempstring=tempstring+infix+i; binary_tree temp; temp.root=build_node(tempstring); NodeStack.push(temp); / else if(c='+'|c='-'|c='*'|c='/'|c='') if(OpStack.empty() OpStack.push(c); else if(OpStack.top()='

7、;(') OpStack.push(c); else if(TakesPrecedence(c,OpStack.top() OpStack.push(c); else while(!OpStack.empty()&&(TakesPrecedence(OpStack.top(),c)|addition(OpStack.top(),c) binary_tree temp_tree; string thisstring="" thisstring=thisstring+OpStack.top(); OpStack.pop(); etree.root=bui

8、ld_node(thisstring); copy(temp_tree.root,NodeStack.top().root); NodeStack.pop(); etree.root->right_child=temp_tree.root; temp_tree.root=NULL; copy(temp_tree.root,NodeStack.top().root); etree.root->left_child=temp_tree.root; NodeStack.pop(); temp_tree.root=NULL; copy(temp_tree.root,etree.root);

9、 NodeStack.push(temp_tree); etree.root=NULL; OpStack.push(c); else if(c='(') OpStack.push(c); else if(c=')') while(OpStack.top()!='(') binary_tree temp_tree; string thisstring="" thisstring=thisstring+OpStack.top(); OpStack.pop(); etree.root=build_node(thisstring);

10、copy(temp_tree.root,NodeStack.top().root); NodeStack.pop(); etree.root->right_child=temp_tree.root; temp_tree.root=NULL; copy(temp_tree.root,NodeStack.top().root); etree.root->left_child=temp_tree.root; NodeStack.pop(); temp_tree.root=NULL; copy(temp_tree.root,etree.root); NodeStack.push(temp_

11、tree); etree.root=NULL; OpStack.pop(); / while(!OpStack.empty() binary_tree temp_tree; string thisstring="" thisstring=thisstring+OpStack.top(); OpStack.pop(); etree.root=build_node(thisstring); copy(temp_tree.root,NodeStack.top().root); NodeStack.pop(); etree.root->right_child=temp_tre

12、e.root; temp_tree.root=NULL; copy(temp_tree.root,NodeStack.top().root); etree.root->left_child=temp_tree.root; NodeStack.pop(); temp_tree.root=NULL; copy(temp_tree.root,etree.root); NodeStack.push(temp_tree); if(!OpStack.empty() etree.root=NULL; cout<<"打印結(jié)點(diǎn)如下: " etree.print(); bin

13、ary_tree temp; copy(temp.root,etree.root); cout<<'n' cout<<"結(jié)點(diǎn)個(gè)數(shù)為："<<etree.counter()<<'n' cout<<"以下是，中間的計(jì)算結(jié)果："<<'n' etree.evaluate(); cout<<'n' cout<<"結(jié)果是: " cout<<etree.root->data&

14、lt;<'n' cout<<"-"<<'n' cout<<"是不是要進(jìn)行二叉樹排序，并顯示？若是升序點(diǎn)A，若是降序點(diǎn)B,若不是點(diǎn)C。"<<'n' char c1; cin>>c1; if(c1='A'|c1='a') binary_tree temp1; copy(temp1.root,temp.order().root); /此處代碼無需再改 temp1.inprint(temp1.root);/前邊程序有錯(cuò),此處TEMP1為空.所以沒有輸出. else if(c1='B'|c1='b') binary_tree temp1; copy(temp1.root,temp.order().root); temp1.deprint(temp1.root); cout<<"nnRun Program again?Enter<y/n>:" cin>>choice; else cout<<&quo