Java 二叉搜索树中的前序遍历

我遇到了问题，无法理解为什么我的方法在找到对象时不返回对象？它首先递归遍历树的左侧，然后递归遍历树的右侧。println语句在找到客户但return customer始终为null时打印。我做错了什么

    public Customer lookUpCustomer(String lastName, String firstName) {

    Customer customer;
    Customer foundCustomer = null;

    if (left != null) {

        customer = (Customer) left.getItem();

        if(customer.getLastName().equals(lastName) && customer.getFirstName().equals(firstName)) {
            System.out.println("Found customer: " + customer.toString());
            return customer;
            //foundCustomer = customer;             
        }
        left.lookUpCustomer(lastName, firstName);
    }
    if (right != null) {

        customer = (Customer) right.getItem();

        if(customer.getLastName().equals(lastName) && 
       customer.getFirstName().equals(firstName)) {
            System.out.println("Found customer: " + customer.toString());
            return customer;
            //foundCustomer = customer;         
        }   
        right.lookUpCustomer(lastName, firstName);
    }

    return null;
}

---

因为您总是在第一次调用lookUpCustomer的函数时返回null。另外，您没有保存从lookUpCustomer方法的递归调用返回的值

要解决此问题，请返回找到的节点。您可以按以下方式更改代码：

public Customer lookUpCustomer(String lastName, String firstName) {

    Customer customer;
    Customer foundCustomer = null;

    if (left != null) {

        customer = (Customer) left.getItem();

        if(customer.getLastName().equals(lastName) && customer.getFirstName().equals(firstName)) {
            System.out.println("Found customer: " + customer.toString());
            return customer;
            //foundCustomer = customer;             
        }
        foundCustomer = left.lookUpCustomer(lastName, firstName);
    }
    if (foundCustomer==null && right != null) {

        customer = (Customer) right.getItem();

        if(customer.getLastName().equals(lastName) && 
       customer.getFirstName().equals(firstName)) {
            System.out.println("Found customer: " + customer.toString());
            return customer;
            //foundCustomer = customer;         
        }   
        foundCustomer  = right.lookUpCustomer(lastName, firstName);
    }

    return foundCustomer;
}

---

当递归调用该方法时，不会对任何内容使用返回值。您不认为应该吗？另外，您不认为名称比较应该与当前节点进行比较吗？消除重复代码，并确保根节点中的名称也能进行比较。@Andreas我该怎么做？customer=left.lookUpCustomerlastName，类似的firstName？是，如果不为null，则立即返回。这就是重点，对吗？归还找到的顾客？@Andreas是的，这就是重点。