为二叉树实现节点类

Question

我必须实现一种binary tree形式，其中显示了一个就业层次结构，其中老板是树的根，而较小的雇员是左节点，而右节点是同一级别的雇员。 相同级别的员工将根据其员工编号进行放置。

public class Node {
private Person data;
private Node seniorEmployee;
private Node nextEmployee;
private Node lesserEmployee;

public Node(Member data, Node seniorEmployee) {
    this.data = data;
    this.seniorEmployee = seniorEmployee;
    this.nextEmployee = null;
    this.lesserColleague = null;
}

public Member getPerson() {
    return this.data;
}

public Node getSeniorEmployee() {
    return this.seniorEmployee;
}

public Node getNextEmployee() {
    return this.nextEmployee;
}

public Node getLesserEmployee() {
    return this.lesserEmployee;
}

public void setEmployee(Person e1) {
    if (nextEmployee == null) {
        nextEmployee = new Node(e1, this.seniorEmployee);
    } else {
        if (e1.compareTo(nextEmployee.getPerson()) > 0) {

    } else {

    }
}

public void setLesserEmployee(Member p1) {
    if (lesserEmployee == null) {
        lesserEmployee = new Node(e1, this.seniorEmployee);
    } else {
        if (e1.compareTo(lesserEmployee.getPerson()) > 0) {

    } else {

    }
}

到目前为止，这是我设法做到的，但是我不知道如何实现set方法。

Answer 1

我认为对于相同级别的员工来说也可以：

public void setEmployee(Person e1) {
if (nextEmployee == null) {
    nextEmployee = new Node(e1, this.seniorEmployee);
} else {
    nextEmployee.setEmployee(e1);
}

}