Trie | (Insert and Search) | GeeksforGeeks

http://hehejun.blogspot.com/2015/02/data-structuretrie.html
Trie比哈希和BST更快，尤其是在string很多的情况下
public class Trie<Value> {

    private class Node {
        private Value val;
        private TreeMap<Character, Node> children; // use hashmap or char[256]
       
        public Node() {
            val = null;
            children = new TreeMap<Character, Node>();
        }
    }
   
    private Node root;
   
    public Trie() {
        root = null;
    }
   
    public boolean containsKey(String key) {
        Node node = getNode(key);
        if (node == null)
            return false;
        return node.val != null;
    }
   
    public void put(String key, Value val) {
        root = put(root, key, val, 0);
    }
   
    private Node put(Node node, String key, Value val, int index) {
        if (node == null)
            node = new Node();
        if (index == key.length()) {
            node.val = val;
            return node;
        }
        Node next = put(node.children.get(key.charAt(index)), key, val, index + 1);
        node.children.put(key.charAt(index), next);
        return node;
    }
   
    public Value get(String key) {
        Node node = getNode(key);
        return node == null? null: node.val;
    }
   
    private Node getNode(String key) {
        int len = key.length();
        Node node = root;
        for (int i = 0; i < len; i++) {
            if (node == null)
                return null;
            node = node.children.get(key.charAt(i));
        }
        return node;
    }
   
    public void delete(String key) {
        root = delete(root, key, 0);
    }
   
    private Node delete(Node node, String key, int len) {
        if (node == null)
            return null;
        if(key.length() == len) {
            //it key is not a word in trie
            if (node.val == null)
                return node;
            else {
                //if the node has children
                if (node.children.size() > 0) {
                    node.val = null;
                    return node;
                } else
                    return null;
            }
        }
        Node next = delete(node.children.get(key.charAt(len)), key, len + 1);
        if (next == null) {
            //if word is not in the trie
            if (!node.children.containsKey(key.charAt(len)))
                return node;
            else {
                //if has more the one children or this node has value(it is a word in trie)
                if (node.children.size() > 1 || node.val != null) {
                    node.children.remove(key.charAt(len));
                    return node;
                } else
                    return null;
            }
        } else
            return node;
    }
   
    public Iterable<String> keys() {
        LinkedList<String> list = new LinkedList<String>();
        collect(root, list, "");
        return list;
    }
   
    private void collect(Node node, LinkedList<String> list, String str) {
        if (node == null)
            return;
        if (node.val != null)
            list.add(new String(str));
        Iterator<Entry<Character, Node>> iter = node.children.entrySet().iterator();
        while (iter.hasNext()) {
            Entry<Character, Node> entry = iter.next();
            collect(entry.getValue(), list, str + entry.getKey());
        }
    }
   
    public Iterable<String> keysWithPrefix(String prefix) {
        LinkedList<String> list = new LinkedList<String>();
        Node node = getNode(prefix);
        collect(node, list, prefix);
        return list;
    }
   
    public String longestPrefixOf(String key) {
        int length = search(root, key, 0);
        return key.substring(0, length);
    }
   
    private int search(Node node, String key, int len) {
        if (node == null)
            return len;
        if (key.length() == len)
            return len;
        Node next = node.children.get(key.charAt(len));
        if (next == null)
            return len;
        return search(next, key, len + 1);
           
    }
   
    public String longestKeyAsPrefixOf(String key) {
        int length = search(root, key, 0, 0);
        return key.substring(0, length);
    }
   
    private int search(Node node, String key, int longestLen, int currLen) {
        if (node == null)
            return longestLen;
        if (node.val != null)
            longestLen = currLen;
        if (key.length() == currLen)
            return longestLen;
        Node next = node.children.get(key.charAt(currLen));
        return search(next, key, longestLen, currLen + 1);
    }
}
Trie | (Insert and Search) | GeeksforGeeks

Trie is an efficient information retrieval data structure. Using trie, search complexities can be brought to optimal limit (key length).  Using trie, we can search the key in O(M) time. However the penalty is on trie storage requirements.

struct trie_node

{

    int value; /* Used to mark leaf nodes */

    trie_node_t *children[ALPHABET_SIZE];

};

void insert(trie_t *pTrie, char key[])

{

    int level;

    int length = strlen(key);

    int index;

    trie_node_t *pCrawl;

    pTrie->count++;

    pCrawl = pTrie->root;

    for( level = 0; level < length; level++ )

    {

        index = CHAR_TO_INDEX(key[level]);

        if( !pCrawl->children[index] )

        {

            pCrawl->children[index] = getNode();

        }

        pCrawl = pCrawl->children[index];

    }

    // mark last node as leaf

    pCrawl->value = pTrie->count;

}

// Returns non zero, if key presents in trie

int search(trie_t *pTrie, char key[])

{

    int level;

    int length = strlen(key);

    int index;

    trie_node_t *pCrawl;

    pCrawl = pTrie->root;

    for( level = 0; level < length; level++ )

    {

        index = CHAR_TO_INDEX(key[level]);

        if( !pCrawl->children[index] )

        {

            return 0;

        }

        pCrawl = pCrawl->children[index];

    }

    return (0 != pCrawl && pCrawl->value);

}

Trie | (Delete)

During delete operation we delete the key in bottom up manner using recursion. The following are possible conditions when deleting key from trie,

1. Key may not be there in trie. Delete operation should not modify trie.
2. Key present as unique key (no part of key contains another key (prefix), nor the key itself is prefix of another key in trie). Delete all the nodes.
3. Key is prefix key of another long key in trie. Unmark the leaf node.
4. Key present in trie, having atleast one other key as prefix key. Delete nodes from end of key until first leaf node of longest prefix key.

bool deleteHelper(trie_node_t *pNode, char key[], int level, int len)

{

    if( pNode )

    {

        // Base case

        if( level == len )

        {

            if( pNode->value )

            {

                // Unmark leaf node

                pNode->value = 0;

                // If empty, node to be deleted

                if( isItFreeNode(pNode) )

                {

                    return true;

                }

                return false;

            }

        }

        else // Recursive case

        {

            int index = INDEX(key[level]);

            if( deleteHelper(pNode->children[index], key, level+1, len) )

            {

                // last node marked, delete it

                FREE(pNode->children[index]);

                // recursively climb up, and delete eligible nodes

                return ( !leafNode(pNode) && isItFreeNode(pNode) );

            }

        }

    }

    return false;

}

int isItFreeNode(trie_node_t *pNode)

{

    int i;

    for(i = 0; i < ALPHABET_SIZE; i++)

    {

        if( pNode->children[i] )

            return 0;

    }

    return 1;

}

int leafNode(trie_node_t *pNode)

{

    return (pNode->value != 0);

}

void deleteKey(trie_t *pTrie, char key[])

{

    int len = strlen(key);

    if( len > 0 )

    {

        deleteHelper(pTrie->root, key, 0, len);

    }

}

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