无法实例化TreeNode类型

时间:2014-07-19 09:06:19

标签: java swing collections

package com.cnu.ds.tree;

import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;

public class Tree {
    public static void main(String[] args) {
        TreeNode treeNode = new TreeNode();
        treeNode.t = 1;
        treeNode.left = new TreeNode();
        treeNode.left.t = 2;
        treeNode.right = new TreeNode();
        treeNode.right.t = 3;
        treeNode.left.left = new TreeNode();
        treeNode.left.left.t = 4;
        treeNode.left.right = new TreeNode();
        treeNode.left.right.t = 5;
        treeNode.right.left = new TreeNode();
        treeNode.right.left.t = 6;
        treeNode.right.right = new TreeNode();
        treeNode.right.right.t = 7;
        // //////////////////////
        treeNode.left.left.left = new TreeNode();
        treeNode.left.left.left.t = 8;
        treeNode.left.left.right = new TreeNode();
        treeNode.left.left.right.t = 9;

        treeNode.left.right.left = new TreeNode();
        treeNode.left.right.left.t = 10;

        treeNode.left.right.right = new TreeNode();
        treeNode.left.right.right.t = 11;

        treeNode.right.left.left = new TreeNode();
        treeNode.right.left.left.t = 12;
        treeNode.right.left.right = new TreeNode();
        treeNode.right.left.right.t = 13;

        treeNode.right.right.left = new TreeNode();
        treeNode.right.right.left.t = 14;

        treeNode.right.right.right = new TreeNode();
        treeNode.right.right.right.t = 15;

        levelOrder(treeNode);
        levelOrderReverse(treeNode);
    }

    public static void levelOrderReverse(TreeNode root) {
        Queue<TreeNode> treeNodes = new LinkedList<>();
        TreeNode newRoot = root;
        treeNodes.add(root);
        treeNodes.add(null);
        Stack<Integer> stack = new Stack<>();
        Queue<TreeNode> queue = new LinkedList<>();
        boolean flip = false;
        while (!treeNodes.isEmpty()) {
            root = treeNodes.remove();
            if (root == null) {
                if (flip) {
                    while (!queue.isEmpty()) {
                        root = queue.remove();
                        int r = stack.pop();
                        int l = stack.pop();
                        root.left.t = r;
                        root.right.t = l;
                    }
                }
                flip = !flip;
                if (treeNodes.isEmpty()) {
                    break;
                }
                System.out.println();
                treeNodes.add(null);
            } else {
                if (root.left != null) {
                    treeNodes.add(root.left);
                }
                if (null != root.right) {
                    treeNodes.add(root.right);
                }
                if (flip) {
                    stack.push(root.t);
                } else {
                    queue.add(root);
                }
            }
        }
        System.out.println();
        levelOrder(newRoot);
    }

    public static void levelOrder(TreeNode root) {
        Queue<TreeNode> treeNodes = new LinkedList<>();
        treeNodes.add(root);
        treeNodes.add(null);
        Queue<TreeNode> queue = new LinkedList<>();
        while (!treeNodes.isEmpty()) {
            root = treeNodes.remove();
            if (root == null) {
                if (treeNodes.isEmpty()) {
                    break;
                }
                System.out.println();
                treeNodes.add(null);
            } else {
                if (root.left != null) {
                    treeNodes.add(root.left);
                }
                if (null != root.right) {
                    treeNodes.add(root.right);
                }
                queue.add(root);
                System.out.print(root.t + " ");
            }
        }
    }
}

输出: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 3 2 4 5 6 7 15 14 13 12 11 10 9 8

以上是网站中建议的以下问题的代码段的一部分。 反转二叉树的备用级节点。 一个 / \ b c / \ / \ d e f g / \ / \ / \ / \ 我是谁?

修改树: 一个 / \ c b / \ / \ d e f g / \ / \ / \ / \ o n m l k j i h

最初我收到错误,因为“TreeNode无法解析为类型” 然后我导入java.swing后得到错误“无法实例化类型TreeNode”

请建议

4 个答案:

答案 0 :(得分:0)

您无法TreeNode new实例化TreeNode,因为TreeNode node = new SomeClass(); interface。但是你可以说:

SomeClass

其中TreeNode是实现{{1}}接口的类。

答案 1 :(得分:0)

TreeNode是一个接口,接口永远不能使用new运算符或关键字进行实例化。您可以使用TreeNode作为对象引用或句柄,但是您必须使用一些具有new关键字的实现类来创建实例。访问here以了解有关TreeNode及其实现类的更多信息。

答案 2 :(得分:0)

您无法实例化interfaceTreeNode是一个只能实现或引用任何实现TreeNode的类的接口。

答案 3 :(得分:0)

像这样尝试org.primefaces.model.DefaultTreeNode:

TreeNode root = new DefaultTreeNode("SomeID", null);