将二叉树的所有路径打印为0和1的字符串

Question

I am trying to print all the paths of a binary tree. If the method goes left, it should append 0 to the return value. If the method goes right it should append 1 to the return value. The final product should look something like this:

A
B 0
C 00
D 01
E 1

Unfortunately my code is only printing out zeros. I would assume that my method is not going right, but I am unable to determine why. Any help or suggestions would be greatly appreciated. Thanks!

private static String getAllPaths(final BinaryNodeInterface<Character> root)
{   
     String returnVal = "";
     returnVal = privateGetAllPaths(root, returnVal);
     return returnVal;

}
private static String privateGetAllPaths(final BinaryNodeInterface<Character> root, String numbers){
    String returnVal = "";
    String tempVal;
     if (root == null)
         return null;
     if (root != null)
         returnVal +=  root.getData() + numbers + '\n';
    tempVal = privateGetAllPaths(root.getLeftChild(), numbers += 0);
        if(tempVal != null)
          {
           returnVal += tempVal +"0";
            return returnVal;
          }
         tempVal  =  privateGetAllPaths(root.getRightChild(), numbers +=1);
          if(tempVal != null)
           {
            return returnVal;
           }


    return returnVal;
}

Answer 1

You have way to many return statements in your code. And the complete code is a single mess in every way (layout, style, etc.).

public static void listPaths(BinaryNodeInterface<Character> node , StringBuilder builder , String path){
     builder.append('\n');
     builder.append(node.getData());
     builder.append(" " + path);

     if(node.getLeftChild() != null)
          listPaths(node.getLeftChild() , builder , path + "0");

     if(node.getRightChild() != null)
          listPaths(node.getRightChild() , builder , path + "1");
}

public String listPaths(BinaryNodeInterface<Character> node){
     StringBuilder builder = new StringBuilder();
     listPaths(node , builder , "");
     builder.deleteChar(0);//delete first char (useless '\n')
     return builder.toString();
}

Should work like a charm (not tested).

Answer 2

As fabian said you used too many return -s. If your code follows the left link then you return and skip processing a right subtree of the current node.

Additionaly you use += where should be + , thus when you finally go to the right child, the numbers will have the "01" suffix instead of "1".

Test also if numbers + 0 does what you expect (I'm not sure if it is equivalent to numbers + '0' )...