I wrote a simple huffman coding algorithm for learning and practice. I just used the technique given on the wikipedia page.

Could you tell me my missing points and mistakes?


package ged.gont.bst.huffmancode;

public class Node implements Comparable<Node> {

    private char letter;
    private int freq;
    private Node leftChild;
    private Node rightChild;

    public Node(char letter, int freq, Node leftChild, Node rightChild) {
        this.letter = letter;
        this.freq = freq;
        this.leftChild = leftChild;
        this.rightChild = rightChild;

    public char getLetter() {
        return letter;

    public int getFreq() {
        return freq;

    public Node getLeftChild() {
        return leftChild;

    public Node getRightChild() {
        return rightChild;

    public boolean isLeaf() {
        return this.leftChild == null && this.rightChild == null;

    public int compareTo(Node arg0) {
        return this.freq - arg0.freq;


package ged.gont.bst.huffmancode;

import java.util.LinkedHashMap;
import java.util.Map;
import java.util.PriorityQueue;

public class HuffmanCode {

    private Node root;
    Map<Character, String> charMap = new LinkedHashMap<>();

     * Encodes string in which most used characters have min codeword length
     * @param inputString compressed string
     * @return encoded string
     * @throws IllegelArgumentException if inputString contains invalid ASCII character 
    public String encode(String inputString) {

        char[] letters = inputString.toCharArray();
        Map<Character, Integer> charFreq = new LinkedHashMap<>();
        PriorityQueue<Node> priorityQueue = new PriorityQueue<>();
        String encodedString = "";

        for (char c : letters) {
            if ((int) c > 255) {
                throw new IllegalArgumentException("Input contains invalid ASCII character");

            if (charFreq.containsKey(c)) {
                charFreq.put(c, charFreq.get(c) + 1);
            } else {
                charFreq.put(c, 1);

        for (Character c : charFreq.keySet()) {
            priorityQueue.offer(new Node(c, charFreq.get(c), null, null));

        while (priorityQueue.size() > 1) {
            Node leftChild = priorityQueue.remove();
            Node rightChild = priorityQueue.remove();
            priorityQueue.offer(new Node(Character.MIN_VALUE, leftChild.getFreq() + rightChild.getFreq(), leftChild, rightChild));

        root = priorityQueue.remove();
        generatePrefix(root, "");

        for (int i = 0; i < inputString.length(); i++) {
            encodedString += (charMap.get(inputString.charAt(i)));
        return encodedString;


     * Generates prefix code in bit string format
     * @param root
     * @param code
    private void generatePrefix(Node root, String prefix) {
        if (!root.isLeaf()) {
            generatePrefix(root.getLeftChild(), prefix.concat("0"));
            generatePrefix(root.getRightChild(), prefix.concat("1"));
        } else {
            charMap.put(root.getLetter(), prefix);

     * Decodes the given encoded string
     * @param encodedString
     * @return decoded string
    public String decode(String encodedString) {
        String decodedString = "";
        Node currentNode = root;

        for (int i = 0; i < encodedString.length(); i++) {

            if (encodedString.charAt(i) == '0') {
                currentNode = currentNode.getLeftChild();
            } else if (encodedString.charAt(i) == '1') {
                currentNode = currentNode.getRightChild();
            if (currentNode.isLeaf()) {
                decodedString += currentNode.getLetter();
                currentNode = root;

        return decodedString;


package ged.gont.testbst.testhuffmancode;

import static org.junit.jupiter.api.Assertions.assertEquals;
import org.junit.jupiter.api.*;
import ged.gont.bst.huffmancode.*;

public class TestHuffmanCode {

    static HuffmanCode huffmanCode;
    static String inputString;
    static String expectedEncodedString;

    public static void init() {
        huffmanCode = new HuffmanCode();
        expectedEncodedString = "1000011101001000110010011101100111001001000111110010011111011111100010001111110100111001001011111011101000111111001";

    public void testEncode() {
        assertEquals(expectedEncodedString, huffmanCode.encode(inputString));

    public void testDecode() {
        assertEquals(inputString, huffmanCode.decode(huffmanCode.encode(inputString)));
  • 3
    \$\begingroup\$ What's going on in code.prefix("0")? It doesn't look like code is defined (but it looks like it used to be a parameter and isn't anymore) \$\endgroup\$
    – user555045
    Commented Feb 19, 2022 at 12:06
  • 2
    \$\begingroup\$ You represent the encoded text as a string consisting of characters '0' and '1'. This is by the factor of 16 larger space than a java.util.BitSet would normally require. \$\endgroup\$
    – coderodde
    Commented Feb 19, 2022 at 12:45
  • 1
    \$\begingroup\$ Is there a reason why you are not using standard classes for binary tree representation avaible in Java? \$\endgroup\$
    – convert
    Commented Feb 20, 2022 at 14:02
  • \$\begingroup\$ @convert: suggest a JRE class. Then, gedofgont introduced for learning and practice (tagging reinventing-the-wheel is definite where a wheel is well-known). \$\endgroup\$
    – greybeard
    Commented Mar 13, 2022 at 9:42

1 Answer 1



Classes not carefully designed for extension should be marked `final`.

root is problematic as a class-level variable. The API of HuffmanCode implies that decode() depends only on the encoded string, and that one could call decode more than once with different values. That will fail, though. The code only works if you call encode directly before the corresponding decode.

  • It would be better to localize root and do whatever computation is necessary for decode in that method. This would mean changing the output of encode so that the tree can be rebuilt`.
  • Another option would be to separate out the building of the tree into its own method, and then have encode and decode take the tree as an argument. The downside here is needing to track both the tree and the encoded string.
  • A third option is to build the HuffmanCode class using a static encode factory method. The class would keep the tree as an instance variable. It could have a public toString method to return the encoded value, and a public decode method to return the decoded value.

Which of these is best depends on your needs. But the API right now will result in cranky users calling encode() multiple times and then trying to decode() multiple times and getting nonsense results.


This class is immutable, so all the variables can be marked as final. This conveys the intent that they cannot be changed, and prevents accidental modification after construction time.

This class is only intended for use inside its package, and should therefore have default (package-private) access, not public.

This class has one constructor, but it should have two, one for leaf node construction and one for internal node construction.

arg0 might be better named as node or otherNode.


`charMap` should be `private`.

Abbreviations are confusing and should be avoided. characterFrequency would be preferable to charFreq. Likewise charMap could be characterMap or characterEncodings.

characters can be inlined and still clear to readers.

letters is not a good variable name, since the values are actually ASCII characters, not letters.

It's not necessary to convert a char to an int for numerical comparisons. The compiler can do it natively.

It would be preferable if the invalid ASCII character exception contained both the invalid character and the input string.

Setting the map value to 1 or adding 1 to the map value is cleaner using Map.merge(). It would look like charFreq.merge(c, 1, Integer::sum);

The code could use an int[] instead of a map, since ASCII is a constrained set of ints from 0-255. Using an int[256] and incrementing the values there should be more time and space efficient. Prefer whichever is easier to read unless you have a documented performance bottleneck.

It's preferable to localize variables so they're initialized as close to where they're first used as is reasonable. The convention of declaring them at the top of a method is a holdover from languages where it's not possible to declare them later.

When modifying Strings, it is preferable to use StringBuilder (or StringBuffer if thread safety is an issue). This is for efficiency reasons and clarity-of-intent reasons. The compiler will generally figure it out, but the hint doesn't hurt.

This block:

    for (int i = 0; i < inputString.length(); i++) {

can be replaced with

    for (char c : inputString.toCharArray()) {

In generatePrefix, it's easier to read if the non-negated case is in the if part and the negation is in the else.

generatePrefix is a confusing name, because it's really generating the character encodings.

In generatePrefix, prefix might be better named encoding. root might be better named node.

It would be preferable if decode() handled invalid input more aggressively, rather than silently continuing if it doesn't see a 0 or 1.


Only having one test is not especially useful. There should be tests for all boundary conditions, and a variety of interesting inputs. For instance, zero characters, one character, non-ASCII characters only, non-ascii characters embedded in ASCII characters, etc.
  • 1
    \$\begingroup\$ class-level variable - what is that? Re. Java, I'd accept that designation for static data members. While I'd prefer HuffmanCoDec, what's problematic about instance data member root in HuffmanCode? I agree that documentation of use and interdependency of encode() & decode(), while present, is lacking. Then again: learning and practice. \$\endgroup\$
    – greybeard
    Commented Mar 13, 2022 at 20:04
  • 1
    \$\begingroup\$ @greybeard HuffmanCode hc = new HuffmanCode(); String s1 = hc.encode("abcde"); String s2 = hc.encode("wxyz"); System.out.println(hc.decode(s1)); I would expect the last statement to actually decode s1. It will not, because root is now built to "wxyz". Either HuffmanCode is restricted to the string it's built from, in which case it needs to be constructed with the instance it's encoding, or it's not, in which case the implementation of encode needs to change to embed the node tree. \$\endgroup\$
    – Eric Stein
    Commented Mar 14, 2022 at 0:27
  • 1
    \$\begingroup\$ And I fixed the typo. You can feel free to edit stuff like that directly in the future. :) \$\endgroup\$
    – Eric Stein
    Commented Mar 14, 2022 at 11:55
  • \$\begingroup\$ thank u so much for review and advise, i gonna try refactoring my code in accordance with your answer. \$\endgroup\$
    – gedofgont
    Commented Mar 21, 2022 at 19:58

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