我有一个简单的n-ary（最多3个子节点），其中插入的第一个节点将是根。以前，我添加任何其他节点，我必须搜索树并插入作为子节点从先前插入的节点，如果条件满足。 我的插入方法因首次插入和后续插入而过载。
我能够使用此方法插入第一个节点：
void Tree::AddSkill(char* name, char* desc, int level)
{
Skill s(name, desc, level);
Node * newNode = new Node(s);
//newNode->aSkill = Skill(name, desc, level);
newNode->parent = NULL;
for (int i = 0; i<CHILD_MAX; i++)
{
newNode->children[i] = NULL;
}
if (this->root == NULL)
{
this->root = newNode;
}
else
{
this->root->parent = newNode;
newNode->children[0] = this->root;
this->root = newNode;
}
}
我在后续插入树时遇到一些问题， 这是我到目前为止的代码：
void Tree::AddSkill(char* name, char* desc, int level, char* parentName)
{
if (this->root == NULL)
{
cout << "Error: no nodes in tree.\n";
return;
}
Node* node = NULL;
Skill s(name, desc, level);
Node * child = new Node(s);
while (root != NULL)
{
if (strcmp(child->aSkill.GetName(), parentName) == 0)
{
for (int i = 0; i < CHILD_MAX; i++)
{
if (node->children[i] == NULL)
{
child->aSkill = s;
child->parent = node;
node->children[i] = child;
return;
}
}
}
}
}
当我通过VS Debugger运行代码时，第二个AddSkill方法中的while循环会无休止地重复。
我不太确定我做错了什么或我需要实施什么概念，我们将不胜感激。
P.S。这是一个家庭作业（不确定适当的标签是什么）。
更新：
我尝试使用Queue实现重载AddSkill()。
这就是我尝试过的。
void SkillTree::AddSkill(char* name, char* desc, int level, char* parentName)
{
if (this->root == NULL)
{
cout << "Error: no nodes in tree.\n";
return;
}
queue<Node*> q;
q.push(this->root);
while (!q.empty())
{
Node * n = q.front();
q.pop();
if (strcmp(n->aSkill.GetName(), parentName) == 0)
{
for (int i = 0; i<CHILD_MAX; i++)
{
if (n->children[i] == NULL)
{
Skill s(name, desc, level);
Node * child = new Node(s);
//When I comment out the next 3 lines, program does not crash. Not sure what the problem is here.
child->aSkill = s;
child->parent = n;
n->children[i] = child;
return;
}
}
return;
}
for (int i = 0; i<CHILD_MAX; i++)
{
if (n->children[i] != NULL)
{
q.push(n->children[i]);
}
}
}
}
技能等级
#include <iostream>
#include "Skill.h"
Skill::Skill()
{
name = NULL;
desc = NULL;
level = 0;
}
Skill::Skill(char* name, char* desc, int level) : level(level), name(new char[strlen(name) + 1]), desc(new char[strlen(desc) + 1])
{
strcpy_s(this->name, (strlen(name) + 1), name);
strcpy_s(this->desc, (strlen(desc) + 1), desc);
}
Skill::Skill(const Skill& aSkill)
{
this->name = new char[strlen(aSkill.name) + 1];
strcpy_s(this->name, (strlen(aSkill.name) + 1), aSkill.name);
this->level = aSkill.level;
this->desc = new char[strlen(aSkill.desc) + 1];
strcpy_s(this->desc, (strlen(aSkill.desc) + 1), aSkill.desc);
}
Skill& Skill::operator=(const Skill& aSkill)
{
if (this == &aSkill)
return *this;
else
{
delete[] name;
delete[] desc;
name = new char[strlen(aSkill.name) + 1];
strcpy_s(name, (strlen(aSkill.name) + 1), aSkill.name);
desc = new char[strlen(aSkill.desc) + 1];
strcpy_s(name, (strlen(aSkill.desc) + 1), aSkill.desc);
level = aSkill.level;
return *this;
}
}
Skill::~Skill()
{
delete[] name;
delete[] desc;
}
char* Skill::GetName() const
{
return name;
}
char* Skill::GetDesc() const
{
return desc;
}
int Skill::GetLevel() const
{
return level;
}
void Skill::Display(ostream& out)
{
out << "- " << GetName() << " -- " << GetDesc() << " [Lvl: " << GetLevel() << "]\n";
}
节点：
Skill aSkill;
Node* parent;
Node* children[CHILD_MAX];
Node() : parent(NULL)
{
for (int i = 0; i < CHILD_MAX; i++)
{
children[i] = NULL;
}
};
Node(const Skill& n) : aSkill(n), parent(NULL)
{
for (int i = 0; i < CHILD_MAX; i++)
{
children[i] = NULL;
}
};
以下是main()
SkillTree student("Student");
student.Display(cout);
student.AddSkill("Alphabet","Mastery of letters and sounds",0);
student.Display(cout);
student.AddSkill("Reading","The ability to read all manner of written material",1,"Alphabet");
student.AddSkill("Writing","The ability to put your thoughts on paper",1,"Alphabet");
student.Display(cout);
student.AddSkill("Speed Reading Level 1","Read any text twice as fast as normal",5,"Reading");
student.AddSkill("Speed Reading Level 2","Read any text four times as fast as normal",10,"Speed Reading Level 1");
student.AddSkill("Memorization","Memorize average sized texts",10,"Reading");
student.AddSkill("Massive Memorization","Memorize large sized texts",20,"Memorization");
student.AddSkill("Spell Writing","The ability to write spells",5,"Writing");
student.AddSkill("History","The ability to write (and rewrite) history",10,"Writing");
student.AddSkill("Written Creation","The ability to write things into reality",20,"History");
student.Display(cout);
student.Display(cout);调用的两个函数如下
void Tree::Display(ostream& out)
{
out << "Skill Tree: " << title << "\n";
if (this->root == NULL)
{
cout << "Empty\n";
return;
}
else
Display_r(out, this->root, 1);
}
void Tree::Display_r(ostream& out, Node* n, int depth)
{
for (int i = 0; i<depth; i++)
{
out << " ";
}
n->aSkill.Display(out);
for (int i = 0; i<CHILD_MAX; i++)
{
if (n->children[i] != NULL)
{
Display_r(out, n->children[i], depth + 1);
}
}
}
如果我在AddSkill()的队列实现中注释掉一段代码，我就不会收到任何错误。
答案 0 :(得分：1)
在第一个AddSkill()中，您将新节点插入树的顶部，使其成为新的根。
在第二个AddSkill()中，您打算将新节点作为父技能的子节点插入。方法似乎是：
while循环）for循环）有什么问题？
您的算法存在以下几个缺陷：
root上循环不为空。由于树在这里不是空的，并且因为你没有删除任何节点，所以这种情况将保持为真，允许无限循环。 node是当前节点，并将新child插入node个孩子。此代码未被解释。幸运的是：它将是未定义的行为，因为您已将node设置为NULL并且从未更改此值。如何解决？
要做到这一点，您必须从node root开始，然后检查节点名称是否与parentname匹配，如果是，像你一样插入孩子。
然而，这是最后一个问题。一个相当重要的一个。算法的结构适用于链表遍历，但不适用于tree traversal。树遍历算法需要堆栈/列表来跟踪要探索的所有分支，或者需要递归方法。
这里有一些代码（抱歉，我已将char*替换为string并使用vector<Node*>代替Node*[] ）， AddSkill()的辅助重载，用于执行递归搜索：
// replaces the former one that you had
void Tree::AddSkill(string name, string desc, int level, string parentName)
{
if (root == NULL)
{
cout << "Error: no nodes in tree.\n";
return;
}
Skill s(name, desc, level);
AddSkill(root, s, parentName);
}
// auxiliary helper
void Tree::AddSkill(Node*node, Skill& s, string& parentName)
{
if (node->sk.name == parentName) { // if found, add the new node as childen
Node * child = new Node(s);
child->parent = node;
node->children.push_back(child);
}
else {
for (auto &x : node->children) // for all the children
AddSkill(x, s, parentName); // search recursively
}
}
这里有online demo使用共享指针而不是原始指针。