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c++ - 使用访问者设计模式重复访问 N 叉树中的子节点值

转载 作者:行者123 更新时间:2023-11-30 03:16:41 25 4
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这棵树长什么样:

          A
/ \ \
AA AB AC
/ \ \
AAA ABA ABB

目前,使用数组我能够得到这个输出:

A AA AAA AAA AAA AA AA A AB ABA ABA ABA AB ABB ABB ABB AB A AC AC AC

输出应该是这样的:

A AA AAA AB ABA AC ABB

使用硬编码,我只能得到这个输出:

A AA AAA AB ABA ABB

这个问题有什么建议吗?如何停止重复已读取的子节点?任何帮助将非常感激!

这里是代码文件,如果你想运行和测试的话。

主要.cpp

#include "pch.h"
#include <iostream>
#include "NTree.h"
#include <string>
using namespace std;
int main()
{
string A("A");
string A1("AA");
string A2("AB");
string A3("AC");
string AA1("AAA");
string AB1("ABA");
string AB2("ABB");
typedef NTree<string, 3> NS3Tree;

NS3Tree root(A);

NS3Tree nodeA1(A1);
NS3Tree nodeA2(A2);
NS3Tree nodeA3(A3);
NS3Tree nodeAA1(AA1);
NS3Tree nodeAB1(AB1);
NS3Tree nodeAB2(AB2);
root.attachNTree(0, &nodeA1);
root.attachNTree(1, &nodeA2);
root.attachNTree(2, &nodeA3);
root[0].attachNTree(0, &nodeAA1);
root[1].attachNTree(0, &nodeAB1);
root[1].attachNTree(1, &nodeAB2);

cout << "root: " << root.key() << endl;
cout << "root[0]: " << root[0].key() << endl;
cout << "root[1]: " << root[1].key() << endl;
cout << "root[2]: " << root[2].key() << endl;
cout << "root[0][0]: " << root[0][0].key() << endl;
cout << "root[1][0]: " << root[1][0].key() << endl;
cout << "root[1][1]: " << root[1][1].key() << endl;


//test traversal
PreOrderVisitor<string> v1;
PostOrderVisitor<string> v2;
cout << "Pre-order traversal:" << endl;
root.transverseDepthFirst(v1);
cout << endl;
cout << "Post-order traversal:" << endl;
root.transverseDepthFirst(v2);
return 0;
}

树.h

#pragma once
#include <stdexcept>
#include "TreeVisitor.h"
template<class T,int N>
class NTree {
private:
const T* fKey;
NTree<T, N>* fNodes[N];
NTree() :fKey((T*)0) {
for (int i = 0; i < N; i++) {
fNodes[i] = &NIL;
}
};
public:
static NTree<T, N> NIL;
NTree(const T& aKey);
~NTree();

bool isEmpty() const;
const T& key() const;

NTree& operator[](int aIndex) const;
void attachNTree(int aIndex, NTree<T, N>* aNTree);
NTree* detachNTree(int aIndex);
void transverseDepthFirst(const TreeVisitor<T>& aVisitor)const;
};

template<class T,int N>
void NTree<T, N>::transverseDepthFirst(const TreeVisitor<T>& aVisitor)const {
/*for (int i = 0; i < N ; i++) {
if (!isEmpty()) {
aVisitor.preVisit(key());
fNodes[i]->transverseDepthFirst(aVisitor);
aVisitor.postVisit(key());
}
}*/
if (!isEmpty()) {
aVisitor.preVisit(key());
fNodes[0]->transverseDepthFirst(aVisitor);
aVisitor.postVisit(key());
fNodes[1]->transverseDepthFirst(aVisitor);
}

}
template<class T,int N>
NTree<T, N>::~NTree() {
for (int i = 0; i < N; i++) {
if (fNodes[i] != &NIL) {
delete fNodes[i];
}
}
}
template<class T,int N>
NTree<T, N>::NTree(const T& aKey) :fKey(&aKey){
for (int i = 0; i < N; i++) {
fNodes[i] = &NIL;
}
}
template<class T,int N>
void NTree<T, N>::attachNTree(int aIndex, NTree<T, N>* aNTree) {
if (isEmpty()) {
throw std::domain_error("Empty NTree");
}
if (fNodes[aIndex] != &NIL) {
throw std::domain_error("Non-empty sub tree");
}
fNodes[aIndex] = new NTree<T, N>(*aNTree);
}

template<class T,int N>
NTree<T, N>* NTree<T, N>::detachNTree(int aIndex) {
if (isEmpty()) {
throw std::domain_error("Empty NTree");
}
NTree<T, N>& Result = *fNodes[aIndex];
fNodes[aIndex] = &NIL;
return &Result;
}

template<class T,int N>
NTree<T, N>& NTree<T, N>::operator[](int aIndex) const {
return *fNodes[aIndex];
}


template<class T,int N>
bool NTree<T,N>::isEmpty() const {
return this == &NIL;
}

template<class T,int N>
const T& NTree<T, N>::key() const {
if (isEmpty()) {
throw std::domain_error("Empty NTree");
}
return *fKey;
}

template<class T,int N>
NTree<T,N> NTree<T,N>::NIL;

TreeVisitor.h

#pragma once
#include<iostream>

template <class T>
class TreeVisitor {
public:
virtual ~TreeVisitor(){}
virtual void preVisit(const T& aKey) const{}
virtual void postVisit(const T& aKey) const{}
virtual void inVisit(const T& aKey) const{}

virtual void visit(const T& aKey)const
{
std::cout << aKey << " ";
}
};

template<class T>
class PostOrderVisitor :public TreeVisitor<T> {
public:
virtual void postVisit(const T& aKey) const {
this->visit(aKey);
}
};

template<class T>
class PreOrderVisitor :public TreeVisitor<T> {
public:
virtual void preVisit(const T& aKey) const {
this->visit(aKey);
}
};

template<class T>
class InOrderVisitor :public TreeVisitor<T> {
public:
virtual void inVisit(const T& aKey) const {
this->visit(aKey);
}
};

最佳答案

template<class T,int N> void NTree<T, N>::transverseDepthFirst(const TreeVisitor<T>& aVisitor)

必须是:

template<class T,int N>
void NTree<T, N>::transverseDepthFirst(const TreeVisitor<T>& aVisitor)const {
if (!isEmpty()) {
aVisitor.preVisit(key());
for (int i = 0; i < N ; i++) {
fNodes[i]->transverseDepthFirst(aVisitor);
}
aVisitor.postVisit(key());
}
}

结果如预期:

root: A
root[0]: AA
root[1]: AB
root[2]: AC
root[0][0]: AAA
root[1][0]: ABA
root[1][1]: ABB
Pre-order traversal:
A AA AAA AB ABA ABB AC
Post-order traversal:
AAA AA ABA ABB AB AC A

警告你的树不是

          A
/ \ \
AA AB AC
/ \ \
AAA ABA ABB

但是

          A
/ | \
AA AB AC
/ / \
AAA ABA ABB

因为 ABB 附加到 AB 而不是 AC:root[1][1]: ABB而不是 root[2][0]: ABB

关于c++ - 使用访问者设计模式重复访问 N 叉树中的子节点值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56091527/

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