Help with Huffman Binary Tree JAVA coding with strings

public String codificar(String text) {
    String res="";
    if (!text.isEmpty()){
        for (int i = 0; i<text.length(); i++){
            res+=busquedaDelCaracter(text.charAt(i), this.huffmanTree, this.huffmanTree.root(), res);
        }
    }
    return res;
}


public String busquedaDelCaracter(Character ch, BinaryTree<Character> tree, Position<Character> pos, String path) {
    if(tree.isExternal(pos)){
        return path+busquedaDelCaracter(ch,tree,pos,path)==null?"":busquedaDelCaracter(ch,tree,pos,path);
    }
    if(tree.hasLeft(pos)){
        busquedaDelCaracter(ch,tree,tree.left(pos),"0"+path);
    }
    return path+busquedaDelCaracter(ch,tree,pos,path) == null?"":busquedaDelCaracter(ch,tree,tree.right(pos),"1"+path);     
}

I put you in situation:

This is an exercise based on Huffman coding, only work with simple data structures.

Example of Huffman coding:

            .
           / \
           a  .
               \
                .
               / \
              c  d

codificar("acd") ---> "0110111"
A la derecha = 1
A la izquierda = 0

Currently I have the code that I have shown you to code and the recursive method busquedaDelCaracter . I do not know where the fault may be, could you guide me a bit?

Note, if you can not find a character, it's null but when you return it you can not concatenate nulls.

Code already done and with 0 problems:

public String decode(String textocifrado) {

    Position <Character> cursor = this.huffmanTree.root();
    String decodeText="";

    for(int i=0; i<textocifrado.length(); i++){
        if(textocifrado.charAt(i) == '1'){
            cursor = this.huffmanTree.right(cursor);
        }
        if(textocifrado.charAt(i) == '0'){
            cursor = this.huffmanTree.left(cursor);
        }
        if(this.huffmanTree.isExternal(cursor)){
            decodeText+=cursor.element();
            cursor=this.huffmanTree.root();
        }           
    }   
    return decodeText;
}

And the constructor only works with a char[]paths with a LinkedBinaryTree called HuffmanTree and what it does is form the tree (you do not need to call the constructor) in the exercise I'm asking for help.