初始化二叉树的两个子树而不出现“二元运算符的错误操作数类型”错误？

Question

I'm having trouble understanding why I can't initialize both sides of a tree in the same statement. The task I have is to recursively return a list of all the leaves of a binary tree (and return null if the tree is empty), but all I get is

"error: bad operand types for binary operator '&&'
    return nbrLeaves(root.left, pong) && nbrLeaves(root.right, pong);"

I am to assume that the binary tree class with nodes is already implemented.

My code is as follows:

public List<E> leaves(){
    List<E> pong = new ArrayList<E>();
     if (root == null){
        return pong;
    }
    nbrLeaves(root, pong);
    return pong;
    }


    public List<E> nbrLeaves(Node<E> root, List<E> pong){
    
    if (root.left == null && root.right == null){
        pong.add(root.element);
    }
    if (root.left != null && root.right == null){
        return nbrLeaves(root.left, pong);
    } 
    if (root.left == null && root.right != null){
        return nbrLeaves(root.right, pong);
    }
    return nbrLeaves(root.left, pong) && nbrLeaves(root.right, pong);
}

Answer 1

&& is the binary AND operator. It only accepts boolean arguments, so you can't pass List s to it.

Since you are adding the output to the ArrayList passed to your method, it doesn't require a return type, and you can eliminate all the return statements.

You can write it as follows:

public void nbrLeaves(Node<E> root, List<E> pong) {
    if (root.left == null && root.right == null) {
        pong.add(root.element);
    } else if (root.left != null && root.right == null) {
        nbrLeaves(root.left, pong);
    } else if (root.left == null && root.right != null) {
        nbrLeaves(root.right, pong);
    } else {
        nbrLeaves(root.left, pong);
        nbrLeaves(root.right, pong);
    }
}

If you wish the output List to be created by the recursive method instead of being passed to it, you can write it as follows:

public List<E> nbrLeaves(Node<E> root) {
    if (root.left == null && root.right == null) {
        List<E> pong = new ArrayList<>;
        pong.add(root.element);
        return pong;
    } else if (root.left != null && root.right == null) {
        return nbrLeaves(root.left);
    } else if (root.left == null && root.right != null) {
        return nbrLeaves(root.right);
    } else {
        List<E> left = nbrLeaves(root.left);
        List<E> right = nbrLeaves(root.right);
        left.addAll(right);
        return left;
    }
}