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Home/ Questions/Q 587843
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T15:15:19+00:00 2026-05-13T15:15:19+00:00

I am attempting to implement a very simple Trie in Java that supports 3

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I am attempting to implement a very simple Trie in Java that supports 3 operations. I’d like it to have an insert method, a has method (ie is a certain word in the trie), and a toString method to return the trie in string form. I believe I have insertion working properly, but has and toString are proving to be difficult. Here’s what I have so far.

The trie class.


public class CaseInsensitiveTrie implements SimpleTrie {

    //root node
    private TrieNode r;

    public CaseInsensitiveTrie() {
        r = new TrieNode();
    }

    public boolean has(String word) throws InvalidArgumentUosException {
        return r.has(word);
    }

    public void insert(String word) throws InvalidArgumentUosException {
        r.insert(word);
    }

    public String toString() {
        return r.toString();
    }

    public static void main(String[] args) {

        CaseInsensitiveTrie t = new CaseInsensitiveTrie();

        System.out.println("Testing some strings");
        t.insert("TEST");
        t.insert("TATTER");
        System.out.println(t.has("TEST"));
    }
}

And the node class


public class TrieNode {

    //make child nodes
    private TrieNode[] c;
    //flag for end of word
    private boolean flag = false;

    public TrieNode() {
        c = new TrieNode[26]; //1 for each letter in alphabet
    }

    protected void insert(String word) {
        int val = word.charAt(0) - 64;

        //if the value of the child node at val is null, make a new node
                //there to represent the letter
        if (c[val] == null) {
            c[val] = new TrieNode();
        }

        //if word length > 1, then word is not finished being added.
        //otherwise, set the flag to true so we know a word ends there.
        if (word.length() > 1) {
            c[val].insert(word.substring(1));
        } else {
            c[val].flag = true;
        }
    }

    public boolean has(String word) {
        int val = word.charAt(0) - 64;
        if (c[val]!=null && word.length()>1) {
            c[val].has(word.substring(1));
        } else if (c[val].flag==true && word.length()==1) {
            return true;
        }

        return false;
    }

    public String toString() { 
        return "";
    }
}

So basically, when creating a Trie, a TrieNode is created as the root with 26 children. When an insert is attempted, insert is called on that root node, which recursively creates a new node at the correct position, and continues until the word is complete. I believe that method is working properly.

My has function is very broken, because I have to have that return statement outside of the brackets for some reason. I cannot contain it within an else clause or the compiler complains. Other than that, I am thinking that method should work with some tweaks, but I cannot figure it out for the life of me.

toString is a beast I have tried to tackle, but nothing I throw at it works, so I will leave that be until I solve the has problem. If I get has working I may be able to figure a way to reformat it into a toString function.

The purpose of the int val = word.charAt(0) – 64; is because each string entered must be all caps (I will create a string formatting function to ensure this afterwards) so the first letter’s int value – 64 will be it’s position in the array. ie array index 0 is A, so A = 64, A – 64 = 0. B = 65, B – 64 = 1, and so on.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T15:15:20+00:00Added an answer on May 13, 2026 at 3:15 pm

    Your has function should probably look like this:

    if (c[val]!=null && word.length()>1) {
        return c[val].has(word.substring(1)); //<-- Change is on this line
    } else if (c[val].flag==true && word.length()==1) {
        ...etc
    

    You perform the recursive call, but you really need to let that value propagate back out to the original caller.

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