我有两个单独的程序,都处理与中缀和后缀评估表达式树。一个基于结构,而另一个基于类。现在我被困在上面写着我的任务的一部分(类版本):类树vs结构TreeNode
“说完createExpressionTree()方法的实现”
和
“它与您之前的实现非常相似,只不过您将使用”类树“的实例而不是”结构树节点“的实例。”
这两者比以下有更多的内容,但我认为你可以得到事物的要点,所以我问的是:类实现与结构类实现的相似程度如何?我是否可以复制并粘贴以下所示的旧代码并进行微调?我一直在尝试,但在访问私人会员等方面遇到问题。
因此,这里的分类和我createExpressionTree的新版本是应该去用它
#ifndef TREE_H
#define TREE_H
#include <vector>
#include <stack>
#include <sstream>
#include <map>
# define TYPE_NUMBER 0
# define TYPE_VARIABLE 1
# define TYPE_OPERATOR 2
class Tree
{
public:
Tree(std::string input,Tree *leftSubTree=NULL,Tree *rightSubTree=NULL);
Tree(const Tree &inTree); //COPY CONSTRUCTOR
~Tree(); //DESTRUCTOR
int evaluate(std::map< std::string, int > ipMap); //EVALUATE THE EXPRESSION
void postOrderPrint();
void inOrderPrint();
private:
Tree *leftPtr;
std::string Op;
Tree *rightPtr;
int NodeType;
};
与树类
Tree::Tree(std::string input,Tree *leftSubTree,Tree *rightSubTree){
Op = input;
leftPtr = leftSubTree;
rightPtr = rightSubTree;
int num;
if (input == "+"|input == "-"|input == "*"|input == "/")
NodeType = TYPE_OPERATOR;
else if(std::istringstream(Op)>>num)
NodeType = TYPE_NUMBER;
else
NodeType = TYPE_VARIABLE;
}
// copy constructor
Tree::Tree(const Tree &inTree){
Op = inTree.Op;
NodeType = inTree.NodeType;
if (inTree.leftPtr == NULL){
leftPtr = NULL;
}
else {
leftPtr = new Tree(*(inTree.leftPtr));
}
if (inTree.rightPtr == NULL){
rightPtr = NULL;
}
else {
rightPtr = new Tree(*(inTree.rightPtr));
}
}
// tree destructor
Tree::~Tree(){
std::cout << "Tree destructor called" << std::endl;
if (leftPtr != NULL) {
delete(leftPtr);
leftPtr = NULL;
}
if (rightPtr != NULL) {
delete(rightPtr);
rightPtr = NULL;
}
}
#endif
新createExpressionTree相关的代码,我会喜欢一些帮助与:
void arithmetic_expression::createExpressionTree(std::vector<std::string> expression)
{
std::stack <Tree> localStack;
std::string Op;
//Very similar to old implementation
}
这里是以前的实施th Ë结构treeNode的,并且在完成前面的createExpressionTree
struct treeNode {
treeNode *leftPtr; /* pointer to left subtree */
std::string Op; /* integer data value */
treeNode *rightPtr; /* pointer to right subtree */
};
typedef struct treeNode TreeNode;
typedef TreeNode * TreeNodePtr;
以前createExpressionTree
void arithmetic_expression::createExpressionTree(std::vector<std::string> expression)
{
std::stack <TreeNodePtr> localStack;
std::string Op;
TreeNodePtr ptr;
for(int i=0; i<expression.size();i++)
{
Op = expression[i];
ptr = createNewTreeNode(Op);
if(char_is_operator(Op))
{
// adding element to right tree
if (localStack.empty())
{
std::cout<< "Invalid expression: tree not created " << std::endl;
topPtr = NULL;
return;
}
else
{
ptr->rightPtr = localStack.top();
localStack.pop();
}
// adding element to left tree
if (localStack.empty()) {
std::cout<< "Invalid expression: tree not created " << std::endl;
topPtr = NULL;
return;
}
else
{
ptr->leftPtr = localStack.top();
localStack.pop();
}
}
// pushing element to stack
localStack.push(ptr);
}
if (localStack.empty()) {
std::cout<< "Invalid expression: tree not created " << std::endl;
topPtr = NULL;
}
else
{
topPtr = localStack.top();
localStack.pop();
if (!localStack.empty()) {
std::cout<< "Invalid expression: tree not created " << std::endl;
topPtr = NULL;
}
}
}
我没有访问leftPtr/rightPtr就像我在结构版本中所做的那样。那么我该如何为班级版本做些什么呢?也许这是一个更具体的问题,我应该问 – user1325578
如果你想访问私人会员字段有两种方法去实现它。第一种方法是在Tree类中创建一个返回私有变量的访问器方法。第二种方法是在树类中声明朋友void arithmetic_expression :: createExpressionTree(std :: vector表达式),以便它可以访问私有字段。我相信,一般来说,你想要做第一种方法,因为使用朋友被认为是骇人听闻。两者都将允许您访问私有字段,并允许您重用您的createExpressionTree。 –