在计算机科学中，树是非常重要的数据结构二叉排序树是树的一种特殊情况二叉排序树是，它的每个节点只能有两个子节点，左子树的节点小于其父节点，右子树中的节点大于其父节点. 父节点二进制排序树在搜索中使用非常广泛，并且二进制排序树的一种变体(红色和黑色树)是Java中TreeMap和TreeSet的实现基础. 以下是二进制排序树的定义. 使用了两个类. 一个是Node类，它表示树中的节点，另一个是Name类，它表示节点的数据. Name类实现Comparable接口，以便可以比较节点. 大小.

public class BinarySearchTree{

private Node root;

private int size;

public BinarySearchTree(Node root){

this.root=root;

size++;

}

public int getSize(){

return this.size;

}

public boolean contains(Name name){

return contains(name,this.root);

//return false;

}

private boolean contains(Name n,Node root){

if(root==null){

return false;

}

int compare=n.compareTo(root.element);

if(compare>0){

if(root.right!=null){

return contains(n,root.right);

}else{

return false;

}

}else if(compare<0){

if(root.left!=null){

return contains(n,root.left);

}else{

return false;

}

}else{

return true;

}

}

public boolean insert(Name n){

boolean flag = insert(n,this.root);

if(flag) size++;

return flag;

}

private boolean insert(Name n,Node root){

if(root==null){

this.root=new Node(n);

return true;

}else if(root.element.compareTo(n)>0){

if(root.left!=null){

return insert(n,root.left);

}else{

root.left=new Node(n);

return true;

}

}else if(root.element.compareTo(n)<0){

if(root.right!=null){

return insert(n,root.right);

}else{

root.right=new Node(n);

return true;

}

}else{

root.frequency++;

return true;

}

}

public boolean remove(Name name){

root = remove(name,this.root);

if(root != null){

size--;

return true;

}

return false;

}

private Node remove(Name name,Node root){

int compare = root.element.compareTo(name);

if(compare == 0){

if(root.frequency>1){

root.frequency--;

}else{

/**根据删除节点的类型，分成以下几种情况

**①如果被删除的节点是叶子节点，直接删除

**②如果被删除的节点含有一个子节点，让指向该节点的指针指向他的儿子节点

**③如果被删除的节点含有两个子节点，找到左字数的最大节点，并替换该节点

**/

if(root.left == null && root.right == null){

root = null;

}else if(root.left !=null && root.right == null){

root = root.left;

}else if(root.left == null && root.right != null){

root = root.right;

}else{

//被删除的节点含有两个子节点

Node newRoot = root.left;

while (newRoot.left != null){

newRoot = newRoot.left;//找到左子树的最大节点

}

root.element = newRoot.element;

root.left = remove(root.element,root.left);

}

}

}else if(compare > 0){

if(root.left != null){

root.left = remove(name,root.left);

}else{

return null;

}

}else{

if(root.right != null){

root.right = remove(name,root.right);

}else{

return null;

}

}

return root;

}

public String toString(){

//中序遍历就可以输出树中节点的顺序

return toString(root);

}

private String toString(Node n){

String result = "";

if(n != null){

if(n.left != null){

result += toString(n.left);

}

result += n.element + " ";

if(n.right != null){

result += toString(n.right);

}

}

return result;

}

}

在二进制排序树的操作中，删除节点最难处理，因此必须将其分为许多情况分别处理. 下面是Node类和Name类的定义:

class Node{

public Name element;

public Node left;

public Node right;

public int frequency = 1;

public Node(Name n){

this.element=n;

}

}

class Name implements Comparable{

private String firstName;

private String lastName;

public Name(String firstName,String lastName){

this.firstName=firstName;

this.lastName=lastName;

}

public int compareTo(Name n) {

int result = this.firstName.compareTo(n.firstName);

return result==0?this.lastName.compareTo(n.lastName):result;

}

public String toString(){

return firstName + "-" +lastName;

}

}

最后，对二进制排序树进行测试:

public static void main(String[] args){

//System.out.println("sunzhenxing");

Node root = new Node(new Name("sun","zhenxing5"));

BinarySearchTree bst =new BinarySearchTree(root);

bst.insert(new Name("sun","zhenxing3"));

bst.insert(new Name("sun","zhenxing7"));

bst.insert(new Name("sun","zhenxing2"));

bst.insert(new Name("sun","zhenxing4"));

bst.insert(new Name("sun","zhenxing6"));

bst.insert(new Name("sun","zhenxing8"));

System.out.println(bst);

bst.remove(new Name("sun","zhenxing2"));

System.out.println(bst);

bst.remove(new Name("sun","zhenxing7"));

System.out.println(bst);

}

测试输出为:

sun-zhenxing2 sun-zhenxing3 sun-zhenxing4 sun-zhenxing5 sun-zhenxing6 sun-zhenxing7 sun-zhenxing8

sun-zhenxing3 sun-zhenxing4 sun-zhenxing5 sun-zhenxing6 sun-zhenxing7 sun-zhenxing8

sun-zhenxing3 sun-zhenxing4 sun-zhenxing5 sun-zhenxing6 sun-zhenxing8

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