二叉树完成删除结点的操作.
规定:
1)如果删除的节点是叶子节点,则删除该节点
2)如果删除的节点是非叶子节点,则删除该子树
思路
首先先处理:
考虑如果树是空树root,如果只有一个root结点,则等价将二叉树置空
//然后进行下面步骤
1.因为我们的二叉树是单向的,所以我们是判断当前结点的子结点是否需要删除结点,而不能去判断当前这个结点是不是需要删除结点
2.如果当前结点的左子结点不为空,并且左子结点就是要删除结点,就将this.left=null;
并且就返回(结束递归删除)
3.如果当前结点的右子结点不为空,并且右子结点就是要删除结点,就将this.right=null ;
并且就返回(结束递归删除)
4.如果第2和第3步没有删除结点,那么我们就需要向左子树进行递归删除
5.如果第4步也没有删除结点,则应当向右子树进行递归删除.
public class BinaryTreeDemo {
public static void main(String[] args) {
BinaryTree binaryTree = new BinaryTree();
HeroNode root = new HeroNode("宋江",1);
HeroNode hero2 = new HeroNode("无用",2);
HeroNode hero3 = new HeroNode("卢俊义",3);
HeroNode hero4 = new HeroNode("林冲",4);
HeroNode hero5 = new HeroNode("关胜",5);
//说明,我们先手动创建二叉树,后面我们学习递归的方式创建二叉树
root.setLeft(hero2);
root.setRight(hero3);
hero3.setLeft(hero5);
hero3.setRight(hero4);
binaryTree.setRoot(root);
/*
//前序遍历
System.out.println("前序遍历");
binaryTree.preOrder();
//中序遍历
System.out.println("中序遍历");
binaryTree.infixOrder();
//后序遍历
System.out.println("后序遍历");
binaryTree.postOrder();
//前序查找
HeroNode resHeroNode = binaryTree.preOrderSearch(5);
if (resHeroNode != null){
System.out.printf("找到了编号为%d的%s英雄",resHeroNode.getNo(),resHeroNode.getName());
}else {
System.out.println("没有找到");
}
//中序查找
resHeroNode = binaryTree.infixOrderSearch(5);
if (resHeroNode != null){
System.out.printf("找到了编号为%d的%s英雄",resHeroNode.getNo(),resHeroNode.getName());
}else {
System.out.println("没有找到");
}
//后序查找
resHeroNode = binaryTree.postOrderSearch(5);
if (resHeroNode != null){
System.out.printf("找到了编号为%d的%s英雄",resHeroNode.getNo(),resHeroNode.getName());
}else {
System.out.println("没有找到");
}
*/
//删除前
System.out.println("删除前,遍历");
binaryTree.preOrder();
binaryTree.delNode(5);
System.out.println("删除后");
binaryTree.preOrder();
}
}
//定义BinaryTree 二叉树
class BinaryTree{
private HeroNode root;
public void setRoot(HeroNode root) {
this.root = root;
}
//删除节点
public void delNode(int no){
if (root!=null){
if (root.getNo()==no){
root = null;
}else{
root.delNode(no);
}
}else {
System.out.println("空树不能删除");
}
}
//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");
}
}
//前序查找
public HeroNode preOrderSearch(int no){
if (root != null){
return root.preOrderSearch(no);
}else{
return null;
}
}
//中序查找
public HeroNode infixOrderSearch(int no){
if (root != null){
return root.infixOrderSearch(no);
}else{
return null;
}
}
//后序查找
public HeroNode postOrderSearch(int no){
if (root != null){
return root.postOrderSearch(no);
}else{
return null;
}
}
}
class HeroNode{
private String name;
private int no;
private HeroNode left; //默认null
private HeroNode right; //默认null
public HeroNode(String name, int no) {
this.name = name;
this.no = no;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getNo() {
return no;
}
public void setNo(int no) {
this.no = no;
}
public void setLeft(HeroNode left) {
this.left = left;
}
public void setRight(HeroNode right) {
this.right = right;
}
public HeroNode getLeft() {
return left;
}
public HeroNode getRight() {
return right;
}
@Override
public String toString() {
return "HeroNode{" +
"name='" + name + '\'' +
", no=" + no +
'}';
}
//编写前序遍历
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();
}
//递归向右子树前去遍历
if (this.right != null){
this.right.preOrder();
}
}
//编写中序遍历
public void infixOrder(){
//递归向左子树中序遍历
if (this.left != null) {
this.left.infixOrder();
}
//输出父节点
System.out.println(this);
//递归向右子树中序遍历
if (this.right != null){
this.right.infixOrder();
}
}
//编写后序遍历
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 resHeroNode = null;
//判断当前节点的左子节点是否为空,如果不为空,则递归前序查找
if (this.left != null){
resHeroNode = this.left.preOrderSearch(no);
}
if (resHeroNode != null){//说明找到了
return resHeroNode;
}
//左递归前序查找,找到节点,则返回,继续判断
if (this.right != null){
resHeroNode = this.right.preOrderSearch(no);
}
return resHeroNode;
}
//中序遍历查找
public HeroNode infixOrderSearch(int no){
HeroNode resHerNode = null;
//判断当前节点的左子节点是否为空,如果不为空,则递归中序查找
if (this.left != null){
resHerNode = this.left.infixOrderSearch(no);
}
if (resHerNode != null){
return resHerNode;
}
//如果找到,则返回,如果没有找到,就和当前结点比较,如果是则返回当前节点
if (this.no == no){
return this;
}
//否则继续进行右递归的中序查找
if (this.right != null){
resHerNode = this.right.infixOrderSearch(no);
}
return resHerNode;
}
//后序查找
public HeroNode postOrderSearch(int no){
HeroNode resHeroNode = null;
//判断当前节点的左子节点是否为空,如果不为空,则递归后序查找
if (this.left != null){
resHeroNode = this.left.postOrderSearch(no);
}
if (resHeroNode != null){//说明在左子树找到
return resHeroNode;
}
//如果左子树没有找到,则向右子树递归进行后序遍历查找
if (this.right != null){
resHeroNode = this.right.postOrderSearch(no);
}
if (resHeroNode != null){
return resHeroNode;
}
//左右子树都没有找到, 就比较当前结点是不是
if (this.no == no){
return this;
}
return resHeroNode;
}
//删除节点
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);
}
}
}