Java 二叉搜索树

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Java 二叉搜索树

/少司命   2022-05-23 我要评论

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;
            }
        }
    }
 
}

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