JAVA二分搜索树

二叉树:

和链表一样,动态数据结构。

二叉树具有唯一根节点

二叉树具有天然的递归结构

二分搜索树是二叉树

二分搜索树的每个节点的值:

1.大于其左子树的所有节点的值

2.小于其右子树的所有节点的值

每一颗子数也是二分搜索树

 

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
public class BST<E extends Comparable<E>> {
    private class Node{
        public E e;
        public Node left,right;
        public Node(E e){
            this.e=e;
            left=null;
            right=null;
        }
    }
    private Node root;
    private int size;
    public BST(){
        root=null;
        size=0;
    }
    public int size(){
        return size;
    }
    public boolean isEmpty(){
        return size==0;
    }
 
    public void add(E e){
        if(root==null){
            root=new Node(e);
            size++;
        }else{
            add(root,e);
        }
    }
    向以Node为跟节点的二分搜索树中插入元素E递归算法
    private void add(Node node,E e){
        if(e.equals(node.e)) return ;
        else if(e.compareTo(node.e)<0&&node.left==null){
            node.left=new Node(e);
            size++;
            return ;
        }else if(e.compareTo(node.e)>0&&node.right==null){
            node.right=new Node(e);
            size++;
            return;
    }
    if(e.compareTo(node.e)<0)
        add(node.left,e);
    else
        add(node.right, e);
 
public void add(E e){
        root=add(root, e);
    }
    private Node add(Node node,E e){
        if(node==null){
            size++;
            return new Node(e);
        }
        if(e.compareTo(node.e)<0)
            node.left=add(node.left, e);
        else if(e.compareTo(node.e)>0)
            node.right=add(node.right, e);
        return node;   
    }
    //看二分搜索树中是否包含元素e
    public boolean contains(E e){
        return contains(root,e)
    }
    //以node为根的二分搜索树中是否包含元素e,递归算法
    public boolean contains(Node node,E e){
        if(node==null)
            return false;
        if(e.compareTo(node.e)==0)
            return true;
        else if(e.compareTo(node.e)<0)
            return contains(node.left,e);
        else
            return contains(node.right, e);
    }
 
}

  二分搜索树的前序遍历:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
//二分搜索树的前序遍历
    public void preOrder(){
        preOrder(root);
    }
    //前序遍历以node为根的二分搜索树,递归算法
    private void preOrder(Node node){
        if(node==null)
            return;
        System.out.println(node.e);
        preOrder(node.left);
        preOrder(node.right);
    }
     
    @Override
    public String toString(){
        StringBuilder res=new StringBuilder();
        generateBSTString(root,0,res);
        return res.toString();
    }
    //生成node为根节点,深度为depth的描述二叉树的字符串
    private void generateBSTString(Node node,int dept,StringBuilder res){
        if(node==null){
            res.append(generateDepthString(dept)+"null\n");
            return;
        }
        res.append(generateDepthString(dept)+node.e+"\n");
        generateBSTString(node.left,dept+1,res);
        generateBSTString(node.right,dept+1,res);
    }
    private String generateDepthString(int dept) {
        StringBuilder res=new StringBuilder();
        for(int i=0;i<dept;i++)
            res.append("--");
        return res.toString();     
    }

  测试:

1
2
3
4
5
6
7
8
9
10
11
public class Main {
     public static void main(String[] args){
         BST<Integer> bst=new BST<>();
         int[] nums={5,3,6,8,4,2};
         for(int num:nums)
             bst.add(num);
         bst.preOrder();
         System.out.println();
         System.out.println(bst);
     }
}

  二分搜索树的中序遍历和后续遍历

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
//二分搜索树的中序遍历
    public void inOrder(){
        inOrder(root);
    }
    //中序遍历以node为根的二分搜索树,递归算法
    private void inOrder(Node node){
        if(node==null)
            return ;
        inOrder(node.left);
        System.out.println(node.e);
        inOrder(node.right);
    }
    //二分搜索树的后续遍历
    public void postOrder(){
        postOrder(root);
    }
    //后续遍历以node为根的二分搜索树,递归算法
    private void postOrder(Node node){
        if(node==null)
            return ;
        postOrder(node.left);
        postOrder(node.right);
         
        System.out.println(node.e);
    }

  测试:

1
2
3
4
5
6
bst.preOrder();
         System.out.println();
         bst.inOrder();
         System.out.println();
         bst.postOrder();
         System.out.println();

  

 

1
2
3
4
5
6
7
8
9
10
11
12
13
//二分搜索树的非递归前序遍历
    public void preOrderNR(){
        Stack<Node> stack=new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
              Node cur=stack.pop();
              System.out.println(cur.e);
              if(cur.right!=null)
                  stack.push(cur.right);
              if(cur.left!=null)
                  stack.push(cur.left);
        }
    }
1
2
3
4
5
6
7
8
9
10
11
12
13
//二分搜索树的层序遍历
public void levelOrder(){
    Queue<Node> queue=new LinkedList<>();
    queue.add(root);
    while (!queue.isEmpty()) {
         Node cur=queue.remove();
         System.out.println(cur.e);
         if(cur.left!=null)
             queue.add(cur.left);
         if(cur.right!=null)
             queue.add(cur.right);
    }
}

  

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
//寻找二分搜索树的最小元素
    public E mininum(){
        if(size==0)
             throw new IllegalArgumentException("BST is empty");
         return mininum(root).e;
    }
    //返回以node为根的二分搜索树的最小值所在的节点
    private Node mininum(Node node){
        if(node.left==null)
            return node;
        return mininum(node.left);
    }
    //寻找二分搜索树的最大元素
    public E maximum(){
        if(size==0)
            throw new IllegalArgumentException("BST is empty");
        return maximum(root).e;
    }
    //返回node为根的二分搜索树的最大值所在的节点
    private Node maximum(Node node){
        if(node.right==null)
            return node;
        return maximum(node.right);
    }
     
    //从二分搜索树中删除最小值所在节点,并返回最小值
    public E removeMin(){
        E ret=mininum();
        root=removeMin(root);
        return ret;
    }
    //删除掉以node为根的二分搜索树中的最小节点
    //返回删除节点后新的二分搜索树的根
    private Node removeMin(Node node){
        if(node.left==null){
            Node rightNode=node.right;
            node.right=null;
            size--;
            return rightNode;
        }
        node.left= removeMin(node.left);
        return node;
    }
     
    //从二分搜索树中删除最大值所在节点
    public E removeMax(){
        E ret=maximum();
        root=removeMax(root);
        return ret;
    }
    //删除掉以node为根的二分搜索树中的最大节点
    //返回删除节点后新的二分搜索树的根
    public Node removeMax(Node node){
        if(node.right==null){
            Node leftNode=node.left;
            node.left=null;
            size--;
            return leftNode;
        }
        node.right=removeMax(node.right);
        return node;
    }

  测试

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
public class Main {
     public static void main(String[] args){
         BST<Integer> bst=new BST<>();
         Random random=new Random();
         int n=1000;
         for(int i=0;i<n;i++)
             bst.add(random.nextInt(10000));
         ArrayList<Integer> nums=new ArrayList<>();
         while(!bst.isEmpty())
             nums.add(bst.removeMin());
         System.out.println(nums);
          
         for(int i=1;i<nums.size();i++)
             if(nums.get(i-1)>nums.get(i))
                 throw new IllegalArgumentException("Error");
         System.out.println("removeMin test completed.");
          
         //test removeMax
         for(int i=0;i<n;i++)
             bst.add(random.nextInt(10000));
         nums=new ArrayList<>();
         while(!bst.isEmpty())
             nums.add(bst.removeMax());
         System.out.println(nums);
          
         for(int i=1;i<nums.size();i++)
             if(nums.get(i-1)<nums.get(i))
                 throw new IllegalArgumentException("Error");
         System.out.println("removeMax test completed.");
 
     }
}

  

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
//从二分搜索树中删除元素为e的节点
    public void remove(E e){
       root=remove(root,e);
    }
    //删除以node为根的二分搜索树中值为e的节点,递归算法
    //返回删除节点后新的二分搜索树的根
    private Node remove(Node node,E e){
        if(node==null)
            return null;
        if(e.compareTo(node.e)<0){
            node.left=remove(node.left, e);
            return node;
        }
        else if(e.compareTo(node.e)>0){
           node.right=  remove(node.right, e);
           return node;
        }else {
            //待删除节点左子树为空的情况
            if(node.left==null){
                Node rightNode=node.right;
                node.right=null;
                size--;
                return rightNode;
            }
            //待删除节点右子数为空的情况
            if(node.right==null){
                Node leftNode=node.left;
                node.left=null;
                size--;
                return leftNode;
            }
            //待删除节点左右子数均不为空的情况
            //找到比待删除节点大的最小节点,即待删除节点右子树的最小节点
            //用这个节点顶替待删除节点的位置
            Node successor=mininum(node.right);
            successor.right=removeMin(node.right);
             
            successor.left=node.left;
             
            node.left=node.right=null;
            return successor;
        }
    }

  

posted @   石shi  阅读(552)  评论(0编辑  收藏  举报
编辑推荐:
· 为什么说在企业级应用开发中,后端往往是效率杀手?
· 用 C# 插值字符串处理器写一个 sscanf
· Java 中堆内存和栈内存上的数据分布和特点
· 开发中对象命名的一点思考
· .NET Core内存结构体系(Windows环境)底层原理浅谈
阅读排行:
· 为什么说在企业级应用开发中,后端往往是效率杀手?
· DeepSeek 解答了困扰我五年的技术问题。时代确实变了!
· 本地部署DeepSeek后,没有好看的交互界面怎么行!
· 趁着过年的时候手搓了一个低代码框架
· 推荐一个DeepSeek 大模型的免费 API 项目!兼容OpenAI接口!
历史上的今天:
2018-03-25 SQL Server索引维护
点击右上角即可分享
微信分享提示