特里(删除)

在上一则有关trie的文章中，我们描述了如何在trie中插入和搜索节点。这是一种如何从特里删除节点的算法。
在删除操作过程中，我们使用递归从下往上的方式删除密钥。从trie删除密钥时，以下是可能的条件，

1. 可能没有密码。删除操作不应修改特里。
2. 密钥作为唯一密钥存在(密钥的任何部分都不包含另一个密钥(前缀)，密钥本身也不是Trie中另一个密钥的前缀)。删除所有节点。
3. 键是trie中另一个长键的前缀键。取消标记叶子节点。
4. 出现在trie中的密钥，具有至少一个其他密钥作为前缀密钥。从密钥结尾删除节点，直到删除最长前缀密钥的第一个叶子节点。

下面的代码介绍了实现以上条件的算法。

C++

// C++ implementation of delete
// operations on Trie
#include 
using namespace std;
 
const int ALPHABET_SIZE = 26;
 
// trie node
struct TrieNode {
    struct TrieNode* children[ALPHABET_SIZE];
 
    // isEndOfWord is true if the node represents
    // end of a word
    bool isEndOfWord;
};
 
// Returns new trie node (initialized to NULLs)
struct TrieNode* getNode(void)
{
    struct TrieNode* pNode = new TrieNode;
 
    pNode->isEndOfWord = false;
 
    for (int i = 0; i < ALPHABET_SIZE; i++)
        pNode->children[i] = NULL;
 
    return pNode;
}
 
// If not present, inserts key into trie
// If the key is prefix of trie node, just
// marks leaf node
void insert(struct TrieNode* root, string key)
{
    struct TrieNode* pCrawl = root;
 
    for (int i = 0; i < key.length(); i++) {
        int index = key[i] - 'a';
        if (!pCrawl->children[index])
            pCrawl->children[index] = getNode();
 
        pCrawl = pCrawl->children[index];
    }
 
    // mark last node as leaf
    pCrawl->isEndOfWord = true;
}
 
// Returns true if key presents in trie, else
// false
bool search(struct TrieNode* root, string key)
{
    struct TrieNode* pCrawl = root;
 
    for (int i = 0; i < key.length(); i++) {
        int index = key[i] - 'a';
        if (!pCrawl->children[index])
            return false;
 
        pCrawl = pCrawl->children[index];
    }
 
    return (pCrawl != NULL && pCrawl->isEndOfWord);
}
 
// Returns true if root has no children, else false
bool isEmpty(TrieNode* root)
{
    for (int i = 0; i < ALPHABET_SIZE; i++)
        if (root->children[i])
            return false;
    return true;
}
 
// Recursive function to delete a key from given Trie
TrieNode* remove(TrieNode* root, string key, int depth = 0)
{
    // If tree is empty
    if (!root)
        return NULL;
 
    // If last character of key is being processed
    if (depth == key.size()) {
 
        // This node is no more end of word after
        // removal of given key
        if (root->isEndOfWord)
            root->isEndOfWord = false;
 
        // If given is not prefix of any other word
        if (isEmpty(root)) {
            delete (root);
            root = NULL;
        }
 
        return root;
    }
 
    // If not last character, recur for the child
    // obtained using ASCII value
    int index = key[depth] - 'a';
    root->children[index] =
          remove(root->children[index], key, depth + 1);
 
    // If root does not have any child (its only child got
    // deleted), and it is not end of another word.
    if (isEmpty(root) && root->isEndOfWord == false) {
        delete (root);
        root = NULL;
    }
 
    return root;
}
 
// Driver
int main()
{
    // Input keys (use only 'a' through 'z'
    // and lower case)
    string keys[] = { "the", "a", "there",
                      "answer", "any", "by",
                      "bye", "their", "hero", "heroplane" };
    int n = sizeof(keys) / sizeof(keys[0]);
 
    struct TrieNode* root = getNode();
 
    // Construct trie
    for (int i = 0; i < n; i++)
        insert(root, keys[i]);
 
    // Search for different keys
    search(root, "the") ? cout << "Yes\n" : cout << "No\n";
    search(root, "these") ? cout << "Yes\n" : cout << "No\n";
 
    remove(root, "heroplane");
    search(root, "hero") ? cout << "Yes\n" : cout << "No\n";
    return 0;
}

输出：

Yes
No
Yes

时间复杂度：删除操作的时间复杂度为O(n)，其中n是密钥长度。