从有序列表构建霍夫曼编码树

Building Huffman encoding tree from ordered list

本文关键字:霍夫曼 编码 构建 列表      更新时间:2023-10-16

我正在从以最低频率开始的有序链表(按字母频率排序)构建一个霍夫曼编码树。创建树后,我遍历了它,似乎树实现不正确。当我遍历树时,有序链表中的一些节点似乎被遗漏了。(我不认为这是因为我的遍历是错误的。这是我的树代码:

//My class for the nodes in the ordered linked list that will be converted to a tree
class fList{
public:
  fList();
  int frequency;
  char letter;
  fList* next;
  fList* left;
  fList* right;
};
fList::fList(){
  frequency = 0;
  letter = NULL;
  next = NULL;
  left = NULL;
  right = NULL;
}
fList* head = NULL;
    .
    .
    .
    .
    .
//Create the huffman encoding tree from the linked list stored in head
while(head->next != NULL){
    fList *tree = new fList();
    fList *temp = new fList();
    fList *trail = new fList();
    /* Take the first two frequency numbers, add them, create a new node                             
     * with the total frequency number and have new node point to the first                          
     * two nodes (right child and left child) 
     */                                                       
    total = (head->frequency + head->next->frequency);
    tree->frequency = total;
    //Set a new head node
    tree->left = head;
    tree->right = head->next;
    head = head->next->next;
    tree->left->next = NULL;
    tree->right->next = NULL;
    //place tree node in its correct place in sorted list                                           
    temp = head;
    trail = temp;
    if(head->frequency >= tree->frequency){
      tree->next = head;
      head = tree;
    }
    else if(temp->next != NULL){
      while(temp != NULL){
        if(temp->frequency >= tree->frequency){
          tree->next = temp;
          trail->next = tree;
          break;
        }
        else{
          trail = temp;
          temp = temp->next;
        }
      }//while                 
    //insert at the end of list                                                                   
    if(temp == NULL){
      temp = tree->next;
      trail->next = tree;
    }
  }//else if !=NULL 
  else if(head == NULL || head->next == NULL) head = tree;
}

在您发布的代码段的末尾,在行中

else if(temp->next = NULL && head != NULL) head = tree;

您无意中截断了树,方法是将temp->next = NULL设置在您可能打算询问是否temp->next == NULL 的位置。这可能就是为什么某些条目(由temp链接到的条目)被排除在最终结果之外的原因。