Java - 在BT的预先遍历遍历中返回节点x之后访问的节点

Question

在学校的Java中，我被要求“在二进制树的预先遍历遍历中返回节点x之后访问的节点”。我已经创建了一个代码来按顺序列出所有节点，但我不确定如何打印单个节点。

我创建节点的第一个类是：

public class TreeNode {
int value;        // The data in this node.
TreeNode left;   // Pointer to the left subtree.
TreeNode right;  // Pointer to the right subtree.
TreeNode parent; //Pointer to the parent of the node. 

TreeNode(int value) {
    this.value = value;
    this.right = null;
    this.left = null;
    this.parent = null;
 }

 public void displayNode() { //Displays the value of the node.
    System.out.println(value + " ");
 }

然后我有了构建二叉树的类。它还按预先打印整个树：

public class BTree2 {

TreeNode root;                // the first node in the tree

public boolean isEmpty() // true if no links
{
    return root == null;
}

private TreeNode addRecursive(TreeNode current, int value) {
    if (current == null) {
        return new TreeNode(value);
    }

    if (value < current.value) {
        current.left = addRecursive(current.left, value);
    } else if (value > current.value) {
        current.right = addRecursive(current.right, value);
    } else {
        // value already exists
        return current;
    }

    return current;
}

public void add(int value) {
    root = addRecursive(root, value);
}


void printPreorder(TreeNode node) {
    if (node == null) {
        return;
    }
    System.out.print(node.value + " "); /* first print data of node */
    printPreorder(node.left);   /* then recur on left subtree */
    printPreorder(node.right);  /* now recur on right subtree */
}

void printPreorder() {
    printPreorder(root);
}   

这就是我陷入困境的地方：如何打印特定节点之后的节点，而不仅仅是整个树？我以为会是：

 public TreeNode findPreorder(int key) // find node with given key  
{                            // (assumes non-empty tree)  
    TreeNode current = root;                // start at root  

    while (current.value == key) // while there is a match  
    {
        current = current.left;

        if (key < current.value) // go left?  
        {
            current = current.right;

        } else {
            current = current.right; // or go right?  
        }
        if (current == null) // if no child,  
        {
            return null;                  // didn't find it  
        }
    }
    return current;                     // found it  

}

但那不起作用。这是我主要的测试代码：

  public static void main(String[] args) {

    BTree2 tree = new BTree2(); 


    tree.root = new TreeNode(1); 
    tree.root.left = new TreeNode(2); 
    tree.root.right = new TreeNode(3); 
    tree.root.left.left = new TreeNode(4); 
    tree.root.left.right = new TreeNode(5); 


    System.out.println("Preorder traversal of binary tree is "); 
    tree.printPreorder(); 

    System.out.println("the node after 1 is " + tree.findPreorder(1).value);

} 

我的输出是：

二叉树的前序遍历为1 2 4 5 3

1之后的节点是5

有任何想法吗？谢谢！！