Question

所以我在网上发现了一些特里树的例子，并试着尝试出来帮助一个正在进行的游戏，它将包含字典中所有单词的特里树。我在网上找到的例子没有任何freeTree的实现，以避免内存泄漏，所以我试图自己做。但是我有一段时间没有和c一起工作，而且遇到了问题。

char keys[][8] = {"the", "a", "there", "answer", "any",
                 "by", "bye", "their"};
char output[][32] = {"Not present in trie", "Present in trie"};
struct TrieNode *root = getNode();

// Construct trie
int i;
for (i = 0; i < ARRAY_SIZE(keys); i++){
    insert(root, keys[i]);
}

// Search for different keys
printf("%s --- %s\n", "the", output[search(root, "the")] );
printf("%s --- %s\n", "these", output[search(root, "these")] );
printf("%s --- %s\n", "their", output[search(root, "their")] );
printf("%s --- %s\n", "thaw", output[search(root, "thaw")] );

freeTree(root);

printf("test after free\n");
printf("%s --- %s\n", "the", output[search(root, "the")] );
printf("%s --- %s\n", "these", output[search(root, "these")] );
printf("%s --- %s\n", "their", output[search(root, "their")] );
printf("%s --- %s\n", "thaw", output[search(root, "thaw")] );

这是一个简单的测试即时运行，免费后的结果与之前相同

the --- Present in trie
these --- Not present in trie
their --- Present in trie
thaw --- Not present in trie
test after free
the --- Present in trie
these --- Not present in trie
their --- Present in trie
thaw --- Not present in trie

这是使用

的结构

struct TrieNode
{
    struct TrieNode *children[ALPHABET_SIZE];
    bool isLeaf;
};

和免费实施

void freeChildren(struct TrieNode *node){
    int i;
    if (node->isLeaf == false){
      for (i=0;i<ALPHABET_SIZE;i++){
         if (node->children[i] != NULL){
            freeChildren(node->children[i]);
         }
      }
    }

   if (node != NULL){
      node = NULL;
      free(node);
   }
}

void freeTree(struct TrieNode *root){
  freeChildren(root);
  free(root);
}

当我创建一个新的树节点时，我为它内存了malloc，所以我知道我需要释放。

Answer 1

我认为你的问题就是这个部分：

if (node != NULL){
  node = NULL;
  free(node);
}

好吧，你需要释放它然后将其设置为NULL。否则你已经失去了地址。

Answer 2

void freeChildren(struct TrieNode *node){
    int i;
    if (node != NULL && node->isLeaf == false){//NULL check important only for the root
      for (i=0;i<ALPHABET_SIZE;i++){
         if (node->children[i] != NULL){
            freeChildren(node->children[i]);
            node->children[i] = NULL]; //you have to remove the address, otherwise it stays, just is no longer allocated (but it is still possible to access it, though it might crash or do whatever else)
         }
      }
    }

    if (node != NULL){ //again only root could be NULL
      free(node);//this has to go first
      //node = NULL; this has no meaning, as this is the local variable, not the place you passed it to
   }
}

void freeTree(struct TrieNode *root){
  freeChildren(root);
  // free(root); this is already done in the freeChildren
  // you might want to realloc the root there though, as otherwise you don't have anything allocated to root
  root = NULL;
}

释放一棵树

2 个答案: