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!!(这里我debug很久才理解过来) 这里8的前驱为null，所以8的leftType=1，但是6是没有后继的或者说后继为null但是rightType为0(因为后继是在下一个节点来进行连接的，6没有下一个节点，所以不能实现后继的线索化，所以rightType=0)

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   public void threadedNodes(HeroNode node){//如果node == null，就不能线索化if (node == null){return;}//（1）先线索化左子树threadedNodes(node.getLeft());//（2）线索化当前节点//先处理当前节点的前驱节点//以8节点来理解//8节点的left = null，8节点的leftType = 1if (node.getLeft() == null){//让当前节点的左指针指向前驱节点node.setLeft(pre);//修改当前节点的做指针的类型node.setLeftType(1);}//处理后继节点,处理8的后继节点，让pre的right指向nodeif (pre != null && pre.getRight() == null){//让前驱节点的右指针指向当前节点pre.setRight(node);//修改前驱节点的右指针类型pre.setLeftType(1);}//！！！！每处理一个节点，让当前节点是下一个节点的前驱结点pre = node;//（3）线索化右子树threadedNodes(node.getRight());}

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   //遍历线索化二叉树的方法public void threadedList(){//定义一个遍历存储当前遍历的节点，从root开始HeroNode node = root;while (node != null){//循环找到lefeYype == 1的节点，第一个找到的就是8//后面随着遍历而变化，因为放leftType==1时说明该节点是按照线索化处理后的有效节点while (node.getLeftType() == 0){node = node.getLeft();}//打印当前节点System.out.println(node);//如果当前节点的右指针指向的是后继节点，就一直输出while (node.getRightType() == 1){//获取到当前节点的后继节点node = node.getRight();System.out.println(node);}//替换这个遍历的节点node = node.getRight();}}

完整代码（后序遍历线索二叉树没写出来，网上找了一些资料看完感觉还是有一点难度，暂时先不深入了）

package tree.threadedbinarytree;public class ThreadedBrinaryTreeDemo {public static void main(String[] args) {//测试中序线索二叉树的功能HeroNode root = new HeroNode(1, "tom");HeroNode node2 = new HeroNode(3, "jack");HeroNode node3 = new HeroNode(6, "smith");HeroNode node4 = new HeroNode(8, "mary");HeroNode node5 = new HeroNode(10, "king");HeroNode node6 = new HeroNode(14, "dmi");//二叉树，我们再递归创建，目前简单手动创建root.setLeft(node2);root.setRight(node3);node2.setLeft(node4);node2.setRight(node5);node3.setLeft(node6);//测试线索化ThreadedBrinaryTree tBrinaryTree = new ThreadedBrinaryTree();tBrinaryTree.setRoot(root);tBrinaryTree.threadedNodes();//测试以10号节点做测试
//        HeroNode left = node3.getLeft();
//        System.out.println("10号节点的前驱节点是："+ left);
//        HeroNode right = node3.getRight();
//        System.out.println("10号节点的后继节点是："+ right);//        System.out.println("使用线索化的方法中序遍历线索化 二叉树");
//        tBrinaryTree.threadedInfixOrder();//        System.out.println("使用线索化的方法前序遍历线索化 二叉树");
//        tBrinaryTree.threadedPreOrder();}
}//定义ThreadedBinaryTree
class ThreadedBrinaryTree{private HeroNode root;//为了实现线索化，需要创建一个指向当前节点的前驱节点的引用//在递归线索化时，pre总是保留前一个节点private HeroNode pre = null;public void setRoot(HeroNode root){this.root = root;}//重载threadedNodespublic void threadedNodes(){this.threadedNodes(root);}//遍历线索化二叉树的前序遍历方法public void threadedPreOrder(){HeroNode node = root;while (node != null){//打印当前节点System.out.println(node);//向左循环有前驱while(node.getLeftType() == 0){node = node.getLeft();System.out.println(node);}while (node.getRightType() == 1){node = node.getRight();}node = node.getRight();}}//遍历线索化二叉树的中序遍历方法public void threadedInfixOrder(){//定义一个遍历存储当前遍历的节点，从root开始HeroNode node = root;while (node != null){//循环找到lefeYype == 1的节点，第一个找到的就是8//后面随着遍历而变化，因为放leftType==1时说明该节点是按照线索化处理后的有效节点while (node.getLeftType() == 0){node = node.getLeft();}//打印当前节点System.out.println(node);//如果当前节点的右指针指向的是后继节点，就一直输出while (node.getRightType() == 1){//获取到当前节点的后继节点node = node.getRight();System.out.println(node);}//替换这个遍历的节点node = node.getRight();}}//编写对二叉树进行中序线索化的方法public void threadedNodes(HeroNode node){//如果node == null，就不能线索化if (node == null){return;}//（1）先线索化左子树threadedNodes(node.getLeft());//（2）线索化当前节点//先处理当前节点的前驱节点//以8节点来理解//8节点的left = null，8节点的leftType = 1if (node.getLeft() == null){//让当前节点的左指针指向前驱节点node.setLeft(pre);//修改当前节点的做指针的类型node.setLeftType(1);}//处理后继节点,处理8的后继节点，让pre的right指向nodeif (pre != null && pre.getRight() == null){//让前驱节点的右指针指向当前节点pre.setRight(node);//修改前驱节点的右指针类型pre.setRightType(1);}//！！！！每处理一个节点，让当前节点是下一个节点的前驱结点pre = node;//（3）线索化右子树threadedNodes(node.getRight());}//前序遍历public void preOrder(){if (this.root != null){this.root.preOrder();}else {System.out.println("二叉树为空无法遍历");}}//中序遍历public void infixOrder(){if (this.root != null){this.root.infixOrder();}else {System.out.println("二叉树为空无法遍历");}}//后序遍历public void postOrder(){if (this.root != null){this.root.postOrder();}else {System.out.println("二叉树为空无法遍历");}}//前序查找public HeroNode preOrederSearch(int no){if (root != null){return root.preOrderSearch(no);}else {return null;}}//中序查找public HeroNode infixOrderSeach(int no){if (root != null){return root.infixOrderSearch(no);}else {return null;}}//后序查找public HeroNode postOrderSeach(int no){if (root != null){return root.postOrderSearch(no);}else {return null;}}//删除节点public void delNode(int no){if (root != null){//判断root是不是要删除的节点if (root.getNo() == no){root = null;}else {root.delNode(no);}}}}
//创建节点
class HeroNode{private int no;private String name;private HeroNode left;//默认nullprivate HeroNode right;//默认null;//说明//1.如果leftType == 0 表示指向的是左子树，如果是1表示指向前驱节点//1.如果rightType == 0 表示指向的是右子树，如果是1表示指向后继节点private int leftType;private int rightType;public int getLeftType() {return leftType;}public void setLeftType(int leftType) {this.leftType = leftType;}public int getRightType() {return rightType;}public void setRightType(int rightType) {this.rightType = rightType;}public HeroNode(int no, String name) {this.no = no;this.name = name;}public int getNo() {return no;}public void setNo(int no) {this.no = no;}public String getName() {return name;}public void setName(String name) {this.name = name;}public HeroNode getLeft() {return left;}public void setLeft(HeroNode left) {this.left = left;}public HeroNode getRight() {return right;}public void setRight(HeroNode right) {this.right = right;}@Overridepublic String toString() {return "HeroNode{" +"no=" + no +", name='" + name + '\'' +'}';}//编写前序遍历方法public void preOrder(){System.out.println(this);//先输出父节点//递归向左子树前序遍历if (this.left != null){this.left.preOrder();}//递归向右子树前序遍历if (this.right != null){this.right.preOrder();}}//编写中序遍历方法public void infixOrder(){//递归向左子树前序遍历if (this.left != null){this.left.infixOrder();}System.out.println(this);//输出父节点//递归向右子树前序遍历if (this.right != null){this.right.infixOrder();}}//编写后序遍历方法public void postOrder(){if (this.left != null){this.left.postOrder();}if (this.right != null){this.right.postOrder();}System.out.println(this);}public static int i = 1, j = 1, k =1;//编写前序查找方法public HeroNode preOrderSearch(int no){System.out.println("前序遍历"+(i++)+"次");if (this.no == no){return this;}HeroNode heroNode = null;if (this.left != null){heroNode = this.left.preOrderSearch(no);}//不等于空说明在左边找到了if (heroNode != null){return heroNode;}if (this.right != null){heroNode = this.right.preOrderSearch(no);}return heroNode;}//中序遍历查找public HeroNode infixOrderSearch(int no){HeroNode heroNode = null;//先判断当前节点的左子节点是否为空，不为空继续进行中序查找if (this.left != null){heroNode = this.left.infixOrderSearch(no);}if (heroNode != null){return heroNode;}System.out.println("中序遍历"+(j++)+"次");if (this.no == no){return this;}if (this.right != null){heroNode = this.right.infixOrderSearch(no);}return heroNode;}//后序遍历查找public HeroNode postOrderSearch(int no){HeroNode heroNode = null;//判断当前节点的左子节点是否为空，不为空，则递归后序遍历查找if (this.left != null){heroNode = this.left.postOrderSearch(no);}if (heroNode != null){return heroNode;}//判断当前节点的右子节点是否为空，不为空，则递归后序遍历查找if (this.right != null){heroNode = this.right.postOrderSearch(no);}if (heroNode != null){return heroNode;}System.out.println("后序遍历"+(k++)+"次");//左右子树都没有找到，比较当前节点是不是if (this.no == no){return this;}return heroNode;}//递归删除节点//规定：如果是叶子节点就删除节点，如果非叶子节点就删除子树public void delNode(int no){if (this.left !=null && this.left.no == no){this.left = null;return;}if (this.right != null && this.right.no == no){this.right = null;return;}if (this.left != null){this.left.delNode(no);}if (this.right != null){this.right.delNode(no);}}
}