NullPointerException in binary search tree

Question

Each time I run it, I get a NullPointerException.

public StrongBuilder inorder() {
    return inorder(root);

}

private StringBuilder inorder(Node root) {
    StringBuilder val;
    if (root != null) {
        inorder(root.getLeft())

    }
    val = new StringBuilder().append(root.getValue().getAccountNumber()).append(root.getValue().getName());
    inorder(root.getRight());
    return val;
}

Can anyone help spot the error for me? Thanks.

Probably `root.getValue()` is `null`. But how would we know, you don't post the relevant code or stack trace. — lexicore, Dec 11 '16 at 17:05

score 0 · Answer 1 · answered Dec 11 '16 at 17:08

Your problem is here:

val = new StringBuilder().append(root.getValue().getAccountNumber()).append(root.getValue().getName());

Eather this code returns null: root.getValue().getAccountNumber() Or this is the problem: root.getValue().getName() - root.getValue() returns null.

NullPointerException in binary search tree

1 Answers1