在二叉树中找到一个节点的后继节点

【题目】 现在有一种新的二叉树节点类型如下：
public class Node {
public int value;
public Node left;
public Node right;
public Node parent;
public Node(int data) { this.value = data; }
}
该结构比普通二叉树节点结构多了一个指向父节点的parent指针。假
设有一 棵Node类型的节点组成的二叉树，树中每个节点的parent指针
都正确地指向 自己的父节点，头节点的parent指向null。只给一个在
二叉树中的某个节点 node，请实现返回node的后继节点的函数。在二
叉树的中序遍历的序列中， node的下一个节点叫作node的后继节点。

public static Node getSuccessorNode(Node node) {
        if (node == null) {
            return null;
        }
        if (node.right != null) { //当前节点存在右子树
            return getLeftMost(node.right); //找右子树最左的节点
        } else { // 没有右子树
            Node parent = node.parent; //找父节点
            while (parent != null && parent.left != node) {
                //当前节点没有父节点 or 当前节点是其父节点的左孩子时，循环结束
                node = parent;
                parent = node.parent;
            }
            return parent;
        }
    }

完整代码：

public class SuccessorNode {

    public static class Node {
        public int value;
        public Node left;
        public Node right;
        public Node parent;

        public Node(int data) {
            this.value = data;
        }
    }

    public static Node getSuccessorNode(Node node) {
        if (node == null) {
            return null;
        }
        if (node.right != null) { //当前节点存在右子树
            return getLeftMost(node.right); //找右子树最左的节点
        } else { // 没有右子树
            Node parent = node.parent; //找父节点
            while (parent != null && parent.left != node) {
                node = parent;
                parent = node.parent;
            }
            return parent;
        }
    }

    public static Node getLeftMost(Node node) {
        if (node == null) {
            return null;
        }
        while (node.left != null) {
            node = node.left;
        }
        return node;
    }

    public static void main(String[] args) {
        Node head = new Node(6);
        head.parent = null;
        head.left = new Node(3);
        head.left.parent = head;
        head.left.left = new Node(1);
        head.left.left.parent = head.left;
        head.left.left.right = new Node(2);
        head.left.left.right.parent = head.left.left;
        head.left.right = new Node(4);
        head.left.right.parent = head.left;
        head.left.right.right = new Node(5);
        head.left.right.right.parent = head.left.right;
        head.right = new Node(9);
        head.right.parent = head;
        head.right.left = new Node(8);
        head.right.left.parent = head.right;
        head.right.left.left = new Node(7);
        head.right.left.left.parent = head.right.left;
        head.right.right = new Node(10);
        head.right.right.parent = head.right;

        Node test = head.left.left;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.left.left.right;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.left;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.left.right;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.left.right.right;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.right.left.left;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.right.left;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.right;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.right.right; // 10's next is null
        System.out.println(test.value + " next: " + getSuccessorNode(test));
    }

}