二叉树
1.为什么需要树这种数据结构
- 数组存储方式的分析�优点：通过下标方式访问元素，速度快。对于有序数组，还可使用二分查找提高检索速度。�缺点：如果要检索具体某个值，或者插入值(按一定顺序)会整体移动，效率较低
- 链式存储方式的分析�优点：在一定程度上对数组存储方式有优化(比如：插入一个数值节点，只需要将插入节点，链接到链表中即可， 删除效率也很好)。�缺点：在进行检索时，效率仍然较低，比如(检索某个值，需要从头节点开始遍历
- 树存储方式的分析�能提高数据存储，读取的效率, 比如利用 二叉排序树(Binary Sort Tree)，既可以保证数据的检索速度，同时也可以保证数据的插入，删除，修改的速度。
2.树示意图
3.二叉树的概念
- 树有很多种，每个节点最多只能有两个子节点的一种形式称为二叉树。
- 二叉树的子节点分为左节点和右节点。
- 如果该二叉树的所有叶子节点都在最后一层，并且结点总数= 2^n -1 , n 为层数，则我们称为满二叉树。
- 如果该二叉树的所有叶子节点都在最后一层或者倒数第二层，而且最后一层的叶子节点在左边连续，倒数第二层的叶子节点在右边连续，我们称为完全二叉树。
3.1二叉树遍历
前序遍历: 先输出父节点，再遍历左子树和右子树
中序遍历: 先遍历左子树，再输出父节点，再遍历右子树
后序遍历: 先遍历左子树，再遍历右子树，最后输出父节点
小结: 看输出父节点的顺序，就确定是前序，中序还是后序
3.2代码实现
package cn.smallmartial.tree;
/**
* @Author smallmartial
* @Date 2019/6/15
* @Email smallmarital@qq.com
*/
public class BinaryTreeDemo {
public static void main(String[] args) {
//创建一个二叉树
BinaryTree binaryTree = new BinaryTree();
HeroNode root = new HeroNode(1, "doodou");
HeroNode heroNode2 = new HeroNode(2, "smallmartial");
HeroNode heroNode3 = new HeroNode(3, "张三");
HeroNode heroNode4 = new HeroNode(4, "李四");
//先手动创建二叉树
root.setLeft(heroNode2);
root.setRiht(heroNode3);
heroNode3.setRiht(heroNode4);
binaryTree.setRoot(root);
System.out.println("前序遍历");
binaryTree.preOrder();
System.out.println("中序遍历");
binaryTree.infixOrder();
System.out.println("后续序遍历");
binaryTree.postOrder();
}
}
//创建二叉树
class BinaryTree{
private HeroNode root;
public void setRoot(HeroNode root){
this.root = root;
}
//前序遍历
public void preOrder(){
if (this.root != null){
this.root.proOrder();
}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");
}
}
//后续遍历
public void postOrder(){
if (this.root != null){
this.root.postOrder();
}else {
System.out.println("二叉树为空，无法遍历");
}
}
}
class HeroNode{
private int no;
private String name;
private HeroNode left;
private HeroNode right;
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 getRiht() {
return right;
}
public void setRiht(HeroNode riht) {
this.right = riht;
}
@Override
public String toString() {
return "HeroNode{" +
"no=" + no +
", name='" + name + '\'' +
", left=" + left +
", riht=" + right +
'}';
}
//前序遍历
public void proOrder(){
System.out.println(this);
//递归向左子树前序遍历
if (this.left != null){
this.left.proOrder();
}
//递归向右子树前序遍历
if (this.right != null){
this.right.proOrder();
}
//中序遍历
}
//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();
}
}
//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);
}
}
运行结果
3.3二叉树-查找指定节点
要求：
请编写前序查找，中序查找和后序查找的方法。
并分别使用三种查找方式，查找 heroNO = 5 的节点
并分析各种查找方式，分别比较了多少次-
代码
package cn.smallmartial.tree; /** * @Author smallmartial * @Date 2019/6/15 * @Email smallmarital@qq.com */ public class BinaryTreeDemo { public static void main(String[] args) { //创建一个二叉树 BinaryTree binaryTree = new BinaryTree(); HeroNode root = new HeroNode(1, "doodou"); HeroNode heroNode2 = new HeroNode(2, "smallmartial"); HeroNode heroNode3 = new HeroNode(3, "张三"); HeroNode heroNode4 = new HeroNode(4, "李四"); //先手动创建二叉树 root.setLeft(heroNode2); root.setRiht(heroNode3); heroNode3.setRiht(heroNode4); binaryTree.setRoot(root); // // System.out.println("前序遍历"); // binaryTree.preOrder(); // // System.out.println("中序遍历"); // binaryTree.infixOrder(); // // System.out.println("后续序遍历"); // binaryTree.postOrder(); //前序遍历 System.out.println("前序遍历方式"); HeroNode resNode = binaryTree.preOrderSerch(4); if (resNode != null){ System.out.println("找到了信息为no="+resNode.getNo()+"name="+resNode.getName()); }else { System.out.println("没有找到 no ="+5); } } } //创建二叉树 class BinaryTree{ private HeroNode root; public void setRoot(HeroNode root){ this.root = root; } //前序遍历 public void preOrder(){ if (this.root != null){ this.root.proOrder(); }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"); } } //后续遍历 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 preOrderSerch(int no){ if (root != null){ return root.proOrderserch(no); }else { return null; } } //中序遍历 public HeroNode infixOrderSearch(int no){ if (root != null){ return root.infixOrderSearch(no); }else { return null; } } //后续遍历 public HeroNode postOrderSearch(int no ){ if (root != null){ return root.postOrderSerach(no); }else { return null; } } } class HeroNode{ private int no; private String name; private HeroNode left; private HeroNode right; 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 getRiht() { return right; } public void setRiht(HeroNode riht) { this.right = riht; } @Override public String toString() { return "HeroNode{" + "no=" + no + ", name='" + name + '\'' + ", left=" + left + ", riht=" + right + '}'; } //前序遍历 public void proOrder(){ System.out.println(this); //递归向左子树前序遍历 if (this.left != null){ this.left.proOrder(); } //递归向右子树前序遍历 if (this.right != null){ this.right.proOrder(); } //中序遍历 } //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(); } } //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 proOrderserch(int no){ if (this.no == no){ return this; } HeroNode resNode = null; if (this.left != null){ resNode = this.left.proOrderserch(no); } if (resNode != null){ return resNode; } if (this.right != null){ resNode = this.right.proOrderserch(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 public HeroNode postOrderSerach(int no ){ HeroNode resNode = null; if (this.left != null){ resNode = this.left.infixOrderSearch(no); } if (resNode != null){ return resNode; } if (this.right != null){ resNode = this.right.infixOrderSearch(no); } if (this.no == no){ return this; } return resNode; } }
3.4二叉树的删除
如果删除的节点是叶子节点，则删除该节点
如果删除的节点是非叶子节点，则删除该子树.
测试，删除掉 5号叶子节点 和 3号子树.-
代码
package cn.smallmartial.tree; /** * @Author smallmartial * @Date 2019/6/15 * @Email smallmarital@qq.com */ public class BinaryTreeDemo { public static void main(String[] args) { //创建一个二叉树 BinaryTree binaryTree = new BinaryTree(); HeroNode root = new HeroNode(1, "doodou"); HeroNode heroNode2 = new HeroNode(2, "smallmartial"); HeroNode heroNode3 = new HeroNode(3, "张三"); HeroNode heroNode4 = new HeroNode(4, "李四"); //先手动创建二叉树 root.setLeft(heroNode2); root.setRiht(heroNode3); heroNode3.setRiht(heroNode4); binaryTree.setRoot(root); // // System.out.println("前序遍历"); // binaryTree.preOrder(); // // System.out.println("中序遍历"); // binaryTree.infixOrder(); // // System.out.println("后续序遍历"); // binaryTree.postOrder(); // //前序遍历 // System.out.println("前序遍历方式"); // HeroNode resNode = binaryTree.preOrderSerch(4); // if (resNode != null){ // System.out.println("找到了信息为no="+resNode.getNo()+"name="+resNode.getName()); // }else { // System.out.println("没有找到 no ="+5); // } //测试删除 System.out.println("删除前"); binaryTree.preOrder(); binaryTree.delNode(4); System.out.println("删除后"); binaryTree.preOrder(); } } //创建二叉树 class BinaryTree{ private HeroNode root; public void setRoot(HeroNode root){ this.root = root; } public void delNode(int no){ if (root != null){ if (root.getNo() == no){ root = null; }else { root.delNode(no); } } } //前序遍历 public void preOrder(){ if (this.root != null){ this.root.proOrder(); }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"); } } //后续遍历 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 preOrderSerch(int no){ if (root != null){ return root.proOrderserch(no); }else { return null; } } //中序遍历 public HeroNode infixOrderSearch(int no){ if (root != null){ return root.infixOrderSearch(no); }else { return null; } } //后续遍历 public HeroNode postOrderSearch(int no ){ if (root != null){ return root.postOrderSerach(no); }else { return null; } } } class HeroNode{ private int no; private String name; private HeroNode left; private HeroNode right; 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 getRiht() { return right; } public void setRiht(HeroNode riht) { this.right = riht; } @Override public String toString() { return "HeroNode{" + "no=" + no + ", name='" + name + '\'' + ", left=" + left + ", riht=" + right + '}'; } //递归删除节点 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; } //向左子树递归删除 if (this.left != null){ this.left.delNode(no); } //向右递归删除 if (this.right != null){ this.right.delNode(no); } } //前序遍历 public void proOrder(){ System.out.println(this); //递归向左子树前序遍历 if (this.left != null){ this.left.proOrder(); } //递归向右子树前序遍历 if (this.right != null){ this.right.proOrder(); } //中序遍历 } //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(); } } //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 proOrderserch(int no){ if (this.no == no){ return this; } HeroNode resNode = null; if (this.left != null){ resNode = this.left.proOrderserch(no); } if (resNode != null){ return resNode; } if (this.right != null){ resNode = this.right.proOrderserch(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 public HeroNode postOrderSerach(int no ){ HeroNode resNode = null; if (this.left != null){ resNode = this.left.infixOrderSearch(no); } if (resNode != null){ return resNode; } if (this.right != null){ resNode = this.right.infixOrderSearch(no); } if (this.no == no){ return this; } return resNode; } } 运行结果
1560604802227.png
