数据结构与算法---线索化二叉树(Threaded BinaryTree)

 1 class BinaryTree {
 2 private HeroNode root;
 3 privateHeroNode pre=null;
 4 public void threadNodes() {
 5 threadNodes(root);
 6 }
 7 public void threadNodes(HeroNode node) {
 8 if(node==null) {
 9 return;
10 }
11 threadNodes(node.getLeft());
12 if(node.getLeft()==null){
13 node.setLeft(pre);
14 node.setLeftType(1);
15 }
16 if(pre!=null&&pre.getRight()==null) {
17 pre.setRight(node);
18 pre.setRightType(1); }
19 pre=node;
20 threadNodes(node.getRight());}}

 1 public class BinaryTreeDemo {
 2 public static void main(String[] args) {
 3 BinaryTree binaryTree = new BinaryTree();
 4 HeroNode root = new HeroNode(1, "jack");
 5 HeroNode node1 = new HeroNode(3, "tom");
 6 HeroNode node2 = new HeroNode(6, "mike");
 7 
 8 root.setLeftNode(node1);
 9 root.setRightNode(node2);
10 binaryTree.setRoot(root);
11 
12 HeroNode node3 = new HeroNode(8, "林冲");
13 HeroNode node4 = new HeroNode(10, "关胜");
14 node1.setLeftNode(node3);
15 node1.setRightNode(node4);
16 HeroNode node5 = new HeroNode(14, "jerry");
17 node2.setLeftNode(node5);
18 
19 System.out.println("---中序---");
20 binaryTree.infixOrder();
21 //中序线索化二叉树
22 binaryTree.threadNodes();
23 
24 HeroNode afterHeroNode10 = node4.getRight();
25 System.out.println(afterHeroNode10);// no=1 name=jack ok了
26 }}

 1 //遍历线索化二叉树的方法
 2     public void threadedList() {
 3         //定义一个变量，存储当前遍历的结点，从root开始
 4         HeroNode node = root;
 5         while(node != null) {
 6             //循环的找到leftType == 1的结点，第一个找到就是8结点
 7             //后面随着遍历而变化,因为当leftType==1时，说明该结点是按照线索化
 8             //处理后的有效结点
 9             while(node.getLeftType() == 0) {
10                 node = node.getLeft();
11             }
12             
13             //打印当前这个结点
14             System.out.println(node);
15             //如果当前结点的右指针指向的是后继结点,就一直输出
16             while(node.getRightType() == 1) {
17                 //获取到当前结点的后继结点
18                 node = node.getRight();
19                 System.out.println(node);
20             }
21             //替换这个遍历的结点
22             node = node.getRight();
23             
24         }
25     }

 1 public static void main(String[] args) {
 2         //测试一把中序线索二叉树的功能
 3         HeroNode root = new HeroNode(1, "tom");
 4         HeroNode node2 = new HeroNode(3, "jack");
 5         HeroNode node3 = new HeroNode(6, "smith");
 6         HeroNode node4 = new HeroNode(8, "mary");
 7         HeroNode node5 = new HeroNode(10, "king");
 8         HeroNode node6 = new HeroNode(14, "dim");
 9         
10         //二叉树，后面我们要递归创建, 现在简单处理使用手动创建
11         root.setLeft(node2);
12         root.setRight(node3);
13         node2.setLeft(node4);
14         node2.setRight(node5);
15         node3.setLeft(node6);
16         
17         //测试中序线索化
18         ThreadedBinaryTree threadedBinaryTree = new ThreadedBinaryTree();
19         threadedBinaryTree.setRoot(root);
20         threadedBinaryTree.threadedNodes();
21         
22         //测试: 以10号节点测试
23         HeroNode leftNode = node5.getLeft();
24         HeroNode rightNode = node5.getRight();
25         System.out.println("10号结点的前驱结点是 ="  + leftNode); //3
26         System.out.println("10号结点的后继结点是="  + rightNode); //1
27         
28         //当线索化二叉树后，能在使用原来的遍历方法
29         //threadedBinaryTree.infixOrder();
30         System.out.println("使用线索化的方式遍历 线索化二叉树");
31         threadedBinaryTree.threadedList(); // 8, 3, 10, 1, 14, 6
32         
33     }