标签:setname root 子节点 tree 实现 bin nbsp binary rgba
package com.dai.tree.threadedbinarytree; public class ThreadedBinaryTreeDemo { public static void main(String[] args) { //测试中序线索二叉树功能是否正确 HeroNode root = new HeroNode(1, "Tom"); HeroNode node2 = new HeroNode(3, "jack"); HeroNode node3 = new HeroNode(6, "huahua"); HeroNode node4 = new HeroNode(8, "mary"); HeroNode node5 = new HeroNode(10, "king"); HeroNode node6 = new HeroNode(14, "dim"); //手动创建 root.setLeft(node2); root.setRight(node3); node2.setLeft(node4); node2.setRight(node5); node3.setLeft(node6); ThreadedBinaryTree threadedBinaryTree = new ThreadedBinaryTree(); threadedBinaryTree.setRoot(root); threadedBinaryTree.threadedNodes(); //测试10号节点 HeroNode leftNode = node5.getLeft(); HeroNode rightNode = node5.getRight(); System.out.println("10号节点的前驱节点是:" + leftNode); System.out.println("10号节点的后继节点是:" + rightNode); } } //定义threadedBinaryTree 二叉树 class ThreadedBinaryTree{ private HeroNode root; //为实现线索化,需要创建要给指向当前节点的前驱节点的指针 private HeroNode pre = null; //递归线索化时,pre总是保留前一个节点 public void setRoot(HeroNode root) { this.root = root; } //重载线索化二叉树的方法 public void threadedNodes() { this.threadedNodes(root); } //编写对二叉树中序线索化的代码 /** * * @param node:当前需要线索化的节点 */ public void threadedNodes(HeroNode node) { if(node == null) { return ; //节点为空直接返回 } //线索化左子树 threadedNodes(node.getLeft()); if(node.getLeft() == null) { node.setLeft(pre); node.setLeftType(1); } if(pre != null && pre.getRight() == null) { pre.setRight(node); pre.setRightType(1); } pre = node; //线索化右子树 threadedNodes(node.getRight()); } //删除节点 public void delNode(int no) { if(root != null) { //如果只有一个root节点,这里需要判断root是不是要删除的节点。 if(root.getNo() == no) { root = null; }else { //递归删除 root.delNode(no); } }else { System.out.println("树空,无法删除"); } } //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"); } } //前序遍历查找 public HeroNode preOrderSearch(int no) { if(root != null) { return root.preOrdersearch(no); }else { return null; } } //中序遍历查找 public HeroNode infixOrderSearch(int no) { if(root != null) { return root.infixOrdersearch(no); }else { return null; } } //后序遍历查找 public HeroNode postOrderSearch(int no) { if(root != null) { return root.postOrdersearch(no); }else { return null; } } } //节点类 class HeroNode{ private int no; private String name; private HeroNode left; private HeroNode right; //leftType为0表示左子树,1表示指向前驱节点 //rightType为0表示右子树,1表示后继 private int leftType; private int rightType; public HeroNode(int no, String name) { this.no = no; this.name = name; } 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 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; } @Override public String toString() { return "HeroNode [no=" + no + ", name=" + name + "]"; } //递归删除节点 //叶子节点则删除,非叶子节点,则要删除子树 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); } } //编写前序遍历的方法 public void preOrder() { System.out.println(this); // //递归向左子数 if(this.left != null) { this.left.preOrder(); } //递归向右前序遍历 if(this.right!=null) { this.right.preOrder(); } } //中序遍历 public void infixOrder() { //递归向左子树中序遍历 if(this.left != null) { this.left.infixOrder(); } System.out.println(this); if(this.right != null) { this.right.infixOrder(); } } //后续遍历 public void postOrder() { if(this.left != null) { this.left.postOrder(); } if(this.right != null) { this.right.postOrder(); } System.out.println(this); } //前序遍历查找 public HeroNode preOrdersearch(int no) { //比较当前节点是不是 if(this.no==no) { return this; } HeroNode resNode =null; if(this.left != null) { resNode = this.left.preOrdersearch(no); } if(resNode != null) { return resNode; } if(this.right != null) { resNode = this.right.preOrdersearch(no); } return resNode; } //中序遍历查找 public HeroNode infixOrdersearch(int no) { HeroNode resNode =null; if(this.left != null) { resNode = this.left.infixOrdersearch(no); } if(resNode != null) { return resNode; } if(this.no==no) { return this; } if(this.right != null) { resNode = this.right.infixOrdersearch(no); } return resNode; } //后序遍历查找 public HeroNode postOrdersearch(int no) { HeroNode resNode =null; if(this.left != null) { resNode = this.left.postOrdersearch(no); } if(resNode != null) { return resNode; } if(this.right != null) { resNode = this.right.postOrdersearch(no); } if(resNode != null) { return resNode; } if(this.no==no) { return this; } return resNode; } }
标签:setname root 子节点 tree 实现 bin nbsp binary rgba
原文地址:https://www.cnblogs.com/shengtudai/p/14454511.html