Huffman tree decoding

Description:


Huffman coding assigns variable length codewords to fixed length input
characters based on their frequencies. More frequent characters are
assigned shorter codewords and less frequent characters are assigned
longer codewords. A Huffman tree is made for the input string and
characters are decoded based on their position in the tree. We add a
'0' to the codeword when we move left in the binary tree and a '1'
when we move right in the binary tree. We assign codes to the leaf
nodes which represent the input characters.


You have to decode an encoded string using the Huffman tree.


You are given pointer to the root of the Huffman tree and a binary
coded string. You need to print the actual string.

Code:

void decode(String S ,Node root)
    {
        if (root == null) return;
        StringBuilder sb = new StringBuilder();
        int pos = 0;
        Node current = root;
        char[] chars = S.toCharArray();
        while (pos < chars.length) {
            char c = chars[pos];
            if (c == '0' && current.left != null) {
                current = current.left;
            }
            else if (c == '1' && current.right != null) {
                current = current.right;
            }
            if (current.left == null && current.right == null) {
                sb.append(current.data);
                current = root;
            }
            pos++;
        }
        System.out.print(sb.toString());
    }

You could replace the while with a for-each loop. That will get rid of the otherwise unused pos variable, and make the code slightly more compact and idiomatic.

for (char c : S.toCharArray()) {
    // ...
}

Instead of printing the output in the same function that does the main computation, you could return the string result, to minimize the function's responsibilities.

Other than these minor points, the code is fine.

StackExchange Code Review Q#150466, answer score: 2