Java程序在trie中查找特定单词的出现次数

Question

我正在制作一个trie数据结构。我想找到该特里的特定单词的出现次数，例如如果“类似”这个词在一个三重奏中出现四次，程序应该将该单词的输出指定为4.有人可以引导我朝正确的方向发展吗？这是我的类代码，它告诉该特定单词的索引。

public class TST<Value> {
    private int N;       // size
    private Node root;   // root of TST

    private class Node {
        private char c;                 // character
        private Node left, mid, right;  // left, middle, and right subtries
        private Value val;              // value associated with string
    }

    // return number of key-value pairs
    public int size() {
        return N;
    }
    public boolean contains(String key) {
        return get(key) != null;
    }

    public Value get(String key) {
        if (key == null) throw new NullPointerException();
        if (key.length() == 0) throw new IllegalArgumentException("key must have length >= 1");
        Node x = get(root, key, 0);
        if (x == null) return null;
        return x.val;
    }

    // return subtrie corresponding to given key
    private Node get(Node x, String key, int d) {
        if (key == null) throw new NullPointerException();
        if (key.length() == 0) throw new IllegalArgumentException("key must have length >= 1");
        if (x == null) return null;
        char c = key.charAt(d);
        if      (c < x.c)              return get(x.left,  key, d);
        else if (c > x.c)              return get(x.right, key, d);
        else if (d < key.length() - 1) return get(x.mid,   key, d+1);
        else                           return x;
    }
public void put(String s, Value val) {
        if (!contains(s)) N++;
        root = put(root, s, val, 0);
    }

    private Node put(Node x, String s, Value val, int d) {
        char c = s.charAt(d);
        if (x == null) {
            x = new Node();
            x.c = c;
        }
        if      (c < x.c)             x.left  = put(x.left,  s, val, d);
        else if (c > x.c)             x.right = put(x.right, s, val, d);
        else if (d < s.length() - 1)  x.mid   = put(x.mid,   s, val, d+1);
        else                          x.val   = val;
        return x;
    }