【发布时间】：2014-11-24 12:08:10
【问题描述】：
我不确定我将如何攻击我的霍夫曼树的遍历。这棵树是正确的，我只是很难弄清楚如何以一种好的方式遍历它。由于某种原因，我的遍历方法没有结果......
更新：清理代码，使其更加面向对象
节点类：
public class Node
{
public int frekvens; //Frequency
public char tegn; //Symbol
public Node venstre; //Left child
public Node høyre; //Right child
public string s; //result string
public string resultat;
public Node (char c) // Node constructor containing symbol.
{
frekvens = 1;
tegn = c;
}
public Node (int f, Node venstre, Node høyre) // Node Constructor containing frequency and children
{
frekvens = f;
this.venstre = venstre;
this.høyre = høyre;
}
public Node (Node node) // Node constructor containing a node
{
frekvens = node.frekvens;
tegn = node.tegn;
this.venstre = venstre;
this.høyre = høyre;
}
public void ØkMed1() // Inkrement frequency by one
{
frekvens = frekvens + 1;
}
public char getVenstreTegn ()
{
return venstre.tegn;
}
public char getHøyreTegn ()
{
return venstre.tegn;
}
public int getVenstreFrekvens ()
{
return venstre.frekvens;
}
public int getHøyreFrekvens ()
{
return høyre.frekvens;
}
public int getFrekvens()
{
return frekvens;
}
public bool ErTegn(char c)
{
if ( c == tegn)
{
return false;
}
else
{
return true;
}
}
//Pretty sure this does not work as intended
public string traverser (Node n) //Traverse the tree
{
if (n.tegn != '\0') //If the node containes a symbol --> a leaf
{
resultat += s;
}
else
{
if (n.getVenstreTegn() == '\0') //If left child does not have a symbol
{
s += "0";
traverser(n.venstre);
}
if (n.getHøyreTegn() == '\0') //If right child does not have a symbol
{
s += "1";
traverser(n.høyre);
}
}
return resultat;
}
public string Resultat() //Used priviously to check if i got the correct huffman tree
{
string resultat;
resultat = "Tegn: " + Convert.ToString(tegn) +" frekvens: " + Convert.ToString(frekvens) + "\n";
return resultat;
}
}
Huffman_Tree 类：
public class Huffman_Tre
{
string treString;
List<Node> noder = new List<Node>();
public Node rot;
public void bygg (string input)
{
bool funnet; //Found
char karakter; //character
for (int i = 0; i < input.Length;i++) //Loops through string and sets character
//with coresponding freqeuncy in the node list
{
karakter = input[i];
funnet = false; //default
for (int j = 0; j< noder.Count; j++)
{
if (noder[j].ErTegn(karakter) == false) //if the character already exists
{
noder[j].ØkMed1(); //inkrement frequency by one
funnet = true;
break;
}
}
if (!funnet) //if the character does not exist
{
noder.Add(new Node(karakter)); //add the character to list
}
}
//Sorting node list acending by frequency
var sortertListe = noder.OrderBy(c => c.frekvens).ToList();
noder = sortertListe;
do
{
noder.Add(new Node((noder[0].frekvens + noder[1].frekvens), noder[0],noder[1]));
//Remove the leaf nodes
noder.RemoveAt(0);
noder.RemoveAt(0);
} while(noder.Count >= 2);
}
public Node getRot()
{
return rot;
}
public string visTre()
{
foreach (Node node in noder)
{
treString += node.Resultat();
}
return treString;
}
public bool erNull()
{
if (noder[0].tegn == '\0')
{
return true;
}
else
return false;
}
}
主程序：
private void btnKomprimer_Click(object sender, System.Windows.RoutedEventArgs e)
{
string input; //The string input I want to compress
input = txtInput.Text; //initialize input to text input
input = input.ToLower();
txtOutput.Text = "";
Huffman_Tre tre = new Huffman_Tre();
tre.bygg(input);
Node rot = new Node(tre.getRot());
txtOutput.Text += rot.traverser(rot);
}
}
【问题讨论】：
-
Sindre，也许先使用二叉树会更方便。由于霍夫曼算法将根据字符的频率或形成根的频率总和来确定二进制数字（1 或 0）。如果您还需要更多提示，请大喊
-
我忽略了您已经在使用二叉树的事实。我很抱歉。尽管如此，它还是促使我构建了一个带有遍历的霍夫曼算法，我已经在我的答案中发布了它。
-
顺便说一句，由于您实际上没有遍历树的左右节点，因此您的遍历不起作用。
-
感谢您的回复，在我昨天收到 gmail 之前，我以为这个线程已经死了 :) 我写的代码是大约写的。在高中学习c#一个月后，当我现在看它时，它并不好。我将很快重新制作此应用程序并使其完全正常工作。这个哈夫曼代码是一个非常简洁的算法
标签： c# recursion traversal huffman-code