Java数据结构二叉搜索树增插，删，创详解

Posted 2022-01-07 /少司命

目录

①概念

②操作-查找

③操作-插入

④操作-删除

⑤性能分析

---

①概念

二叉搜索树又称二叉排序树，它或者是一棵空树**，或者是具有以下性质的二叉树:

若它的左子树不为空，则左子树上所有节点的值都小于根节点的值

若它的右子树不为空，则右子树上所有节点的值都大于根节点的值

它的左右子树也分别为二叉搜索树

②操作-查找

二叉搜索树的查找类似于二分法查找

public Node search(int key) 
        Node cur = root;
        while (cur != null) 
            if(cur.val == key) 
                return cur;
            else if(cur.val < key) 
                cur = cur.right;
            else 
                cur = cur.left;
            
        
        return null;
    

③操作-插入

  public boolean insert(int key) 
        Node node = new Node(key);
        if(root == null) 
            root = node;
            return true;
        

        Node cur = root;
        Node parent = null;

        while(cur != null) 
            if(cur.val == key) 
                return false;
            else if(cur.val < key) 
                parent = cur;
                cur = cur.right;
            else 
                parent = cur;
                cur = cur.left;
            
        
        //parent
        if(parent.val > key) 
            parent.left = node;
        else 
            parent.right = node;
        

        return true;
    

④操作-删除

删除操作较为复杂，但理解了其原理还是比较容易

设待删除结点为 cur, 待删除结点的双亲结点为 parent

1. cur.left == null

1. cur 是 root，则 root = cur.right

2. cur 不是 root，cur 是 parent.left，则 parent.left = cur.right

3. cur 不是 root，cur 是 parent.right，则 parent.right = cur.right

2. cur.right == null

1. cur 是 root，则 root = cur.left

2. cur 不是 root，cur 是 parent.left，则 parent.left = cur.left

3. cur 不是 root，cur 是 parent.right，则 parent.right = cur.left

第二种情况和第一种情况相同，只是方向相反，这里不再画图

3. cur.left != null && cur.right != null

需要使用替换法进行删除，即在它的右子树中寻找中序下的第一个结点(关键码最小)，用它的值填补到被删除节点中，再来处理该结点的删除问题

当我们在左右子树都不为空的情况下进行删除，删除该节点会破坏树的结构，因此用替罪羊的方法来解决，实际删除的过程还是上面的两种情况，这里还是用到了搜索二叉树的性质

public void remove(Node parent,Node cur) 
        if(cur.left == null) 
            if(cur == root) 
                root = cur.right;
            else if(cur == parent.left) 
                parent.left = cur.right;
            else 
                parent.right = cur.right;
            
        else if(cur.right == null) 
            if(cur == root) 
                root = cur.left;
            else if(cur == parent.left) 
                parent.left = cur.left;
            else 
                parent.right = cur.left;
            
        else 
            Node targetParent =  cur;
            Node target = cur.right;
            while (target.left != null) 
                targetParent = target;
                target = target.left;
            
            cur.val = target.val;
            if(target == targetParent.left) 
                targetParent.left = target.right;
            else 
                targetParent.right = target.right;
            
        
    

  public void removeKey(int key) 
        if(root == null) 
            return;
        
        Node cur = root;
        Node parent = null;
        while (cur != null) 
            if(cur.val == key) 
                remove(parent,cur);
                return;
            else if(cur.val < key)
                parent = cur;
                cur = cur.right;
            else 
                parent = cur;
                cur = cur.left;
            
        
    

⑤性能分析

插入和删除操作都必须先查找，查找效率代表了二叉搜索树中各个操作的性能。

对有n个结点的二叉搜索树，若每个元素查找的概率相等，则二叉搜索树平均查找长度是结点在二叉搜索树的深度 的函数，即结点越深，则比较次数越多。

但对于同一个关键码集合，如果各关键码插入的次序不同，可能得到不同结构的二叉搜索树：

最优情况下，二叉搜索树为完全二叉树，其平均比较次数为：

最差情况下，二叉搜索树退化为单支树，其平均比较次数为：

⑥完整代码

public class TextDemo 

    public static class Node 
        public int val;
        public Node left;
        public Node right;

        public Node (int val) 
            this.val = val;
        
    


    public Node root;

    /**
     * 查找
     * @param key
     */
    public Node search(int key) 
        Node cur = root;
        while (cur != null) 
            if(cur.val == key) 
                return cur;
            else if(cur.val < key) 
                cur = cur.right;
            else 
                cur = cur.left;
            
        
        return null;
    

    /**
     *
     * @param key
     * @return
     */
    public boolean insert(int key) 
        Node node = new Node(key);
        if(root == null) 
            root = node;
            return true;
        

        Node cur = root;
        Node parent = null;

        while(cur != null) 
            if(cur.val == key) 
                return false;
            else if(cur.val < key) 
                parent = cur;
                cur = cur.right;
            else 
                parent = cur;
                cur = cur.left;
            
        
        //parent
        if(parent.val > key) 
            parent.left = node;
        else 
            parent.right = node;
        

        return true;
    

    public void remove(Node parent,Node cur) 
        if(cur.left == null) 
            if(cur == root) 
                root = cur.right;
            else if(cur == parent.left) 
                parent.left = cur.right;
            else 
                parent.right = cur.right;
            
        else if(cur.right == null) 
            if(cur == root) 
                root = cur.left;
            else if(cur == parent.left) 
                parent.left = cur.left;
            else 
                parent.right = cur.left;
            
        else 
            Node targetParent =  cur;
            Node target = cur.right;
            while (target.left != null) 
                targetParent = target;
                target = target.left;
            
            cur.val = target.val;
            if(target == targetParent.left) 
                targetParent.left = target.right;
            else 
                targetParent.right = target.right;
            
        
    

    public void removeKey(int key) 
        if(root == null) 
            return;
        
        Node cur = root;
        Node parent = null;
        while (cur != null) 
            if(cur.val == key) 
                remove(parent,cur);
                return;
            else if(cur.val < key)
                parent = cur;
                cur = cur.right;
            else 
                parent = cur;
                cur = cur.left;