Q: In my implementation of a binary tree below, why does the compiler choke at
if (data.compareTo(this.data) <= 0),
producing
Error: incompatible types: java.lang.Comparable<T> cannot be converted to T?
Both data and this.data are of type Comparable<T> and should be able to use or be an argument to the compareTo() method...right? Well, clearly not. But I really don't understand why. Generics are still baffling me.
public class MyBinaryTreeNodeG<T>{
Comparable<T> data;
MyBinaryTreeNodeG<T> parent;
MyBinaryTreeNodeG<T> left;
MyBinaryTreeNodeG<T> right;
public MyBinaryTreeNodeG(Comparable<T> data){
this.data = data;
}
public MyBinaryTreeNodeG<T> addChild(Comparable<T> data){
if (data.compareTo(this.data) <= 0) { //this is the line on which the compiler chockes
//check if left tree node is null. If so, add. Otherwise, recurse.
} else {
//same for the right tree node
return null;
}
The following is a clip from a more standard implementation of a binary tree. This compiles fine. But I still fail to see why this is a "better" (according to compiler) implementation than mine above.
public class MyBinaryTreeNodeG<T extends Comparable<T>>{
T data;
MyBinaryTreeNodeG<T> parent;
MyBinaryTreeNodeG<T> left;
MyBinaryTreeNodeG<T> right;
public MyBinaryTreeNodeG(T data){
this.data = data;
}
public MyBinaryTreeNodeG<T> addChild(T data){
if (data.compareTo(this.data) <= 0) {
//left node stuff
} else {
//right node stuff
return null;
}
Look at the javadoc of Comparable :
compareTo(T)is provided by theComparableinterface. This method allows to compare a object instance ofTclass implementing this interface with anotherTinstance. In your first example, you're comparing aComparable<T>with aComparable<T>(and not withT)This should work :