完整二叉树的动态编程

Question

我正在尝试在具有15个节点的完整二叉树上应用动态编程，其中每个节点都是Employee类型，并且具有double类型和ID的值（Evalution_Score）。

我创建了一个double类型的数组来存储最大值。方法find返回ID = j或I

的节点

public class BT<T> {
    BTNode<T> root, current;

    public BT() {
        root = current = null;
    }

    public void max(){
       BTNode<Employee> tmproot = (BTNode<Employee>)root;
       BTNode<Employee> tmp=null;
       double x[] = new double[33];
       Employee p[] = new Employee[33];

       for(int i=14;i>=0;i--) {
           tmp = search(i+1, tmproot);
           x[i] = Math.max(x[i+1], (tmp.data.Evaluation_Score + x[i+2+2]));
       }

       for(int j=14;j>=0;j--) {
           tmp = search(j+1, tmproot);
       if(x[j+1]<x[j])
           p[j] = tmp.data;
       }

       for(Employee e: p)
       System.out.println(e);
    }

    public BTNode<Employee> search(int id, BTNode<Employee> node){
        if(node != null){
            if(node.data.ID ==id){
               return node;
            } else {
                BTNode<Employee> foundNode = search(id, node.left);
                if(foundNode == null) {
                    foundNode = search(id, node.right);
                }
                return foundNode;
             }
        } else {
            return null;
        }
    }
}

class BTNode <T> {
    public T data;
    public BTNode<T> left, right;

    /** Creates a new instance of BTNode */
    public BTNode(T val) {
        data = val;
        left = right = null;
    }

    public BTNode(T val, BTNode<T> l, BTNode<T> r){
        data = val;
        left = l;
        right = r;
    }
}

class Employee {
    int ID_of_parent;
    String Name;
    int ID;
    double Evaluation_Score;

    public Employee(int ID_of_parent , String Name , int ID , double Evaluation_Score) {
        this.ID_of_parent = ID_of_parent;
        this.Name = Name;
        this.ID = ID;
        this.Evaluation_Score = Evaluation_Score;
    }

    public String toString() {
        return ID_of_parent + ":" + Name + ":" + ID + ":" + Evaluation_Score + "\n";
    }
}

但由于某种原因，它不能完全打印所需的输出。