删除方法二叉搜索树 [英] Remove method binary search tree

问题描述

我正在尝试为我一直在研究的BST结构实现一个remove方法。以下是包含find，insert和remove方法的代码：

I am trying to implement a remove method for the BST structure that I have been working on. Here is the code with find, insert, and remove methods:

public class BST {
    BSTNode root = new BSTNode("root");

    public void insert(BSTNode root, String title){
        if(root.title!=null){

            if(title==root.title){
                //return already in the catalog
            }
            else if(title.compareTo(root.title)<0){

                if(root.leftChild==null){
                    root.leftChild = new BSTNode(title);
                }
                else{
                    insert(root.leftChild,title);
                }
            }
            else if(title.compareTo(root.title)>0){
                if(root.rightChild==null){
                    root.rightChild = new BSTNode(title);
                }
                else{
                    insert(root.rightChild,title);
                }
            }
        }
    }

    public void find(BSTNode root, String title){
        if(root!= null){
            if(title==root.title){
                //return(true);
            }
            else if(title.compareTo(root.title)<0){
                find(root.leftChild, title);
            }
            else{
                find(root.rightChild, title);
            }
        }
        else{
            //return false;
        }
    }

    public void remove(BSTNode root, String title){
        if(root==null){
            return false;
        }
        if(title==root.title){
            if(root.leftChild==null){
                root = root.rightChild;
            }
            else if(root.rightChild==null){
                root = root.leftChild;
            }
            else{
                //code if 2 chlidren remove
            }
        }
        else if(title.compareTo(root.title)<0){
            remove(root.leftChild, title);
        }
        else{
            remove(root.rightChild, title);
        }
    }
}

我被告知我可以使用insert方法来帮助我使用remove方法，但我只是没有看到我如何抓取最小/最大元素，然后用该值替换我正在删除的元素，然后递归删除我更换的节点值，同时仍保持O（logn）复杂性。任何人都有我错过的任何想法或公然漏洞，或者其他任何有用的因为我对这个问题感到头疼？

I was told that I could use the insert method to help me with the remove method, but I am just not seeing how I can grab the smallest/largest element, and then replace the one I am deleting with that value, then recursively delete the node that I took the replacement value, while still maintaining O(logn) complexity. Anyone have any ideas or blatant holes I missed, or anything else helpful as I bang my head about this issue?

编辑：
我用答案的想法来拿出这个，我相信它会起作用，但我得到一个错误，我的方法（不仅仅是删除）必须返回字符串，这里是代码的样子，我认为这是返回语句？

I used the answers ideas to come up with this, which I believe will work but I'm getting an error that my methods (not just the remove) must return Strings, here is what the code looks like, I thought that's the return statements??

public String remove(BSTNode root, String title){
            if(root==null){
                return("empty root");
            }
            if(title==root.title){
                    if(root.leftChild==null){
                            if(root.rightChild==null){
                                    root.title = null;
                                    return(title+ "was removed");
                            }
                            else{
                            root = root.rightChild;
                            return(title+ "was removed");
                            }
                    }
                    else if(root.rightChild==null){
                            root = root.leftChild;
                            return(title+ "was removed");
                    }
                    else{
                            String minTitle = minTitle(root);
                            root.title = minTitle;
                            remove(root.leftChild,minTitle);
                            return(title+ "was removed");
                    }
            }
            else if(title.compareTo(root.title)<0){
                    remove(root.leftChild, title);
            }
            else{
                    remove(root.rightChild, title);
            }
    }

推荐答案

public void remove (String key, BSTNode pos)
    {
        if (pos == null) return;
        if (key.compareTo(pos.key)<0)
            remove (key, pos.leftChild);
        else if (key.compareTo(pos.key)>0)
            remove (key, pos.rightChild);
        else {
            if (pos.leftChild != null && pos.rightChild != null)
            {
                /* pos has two children */
                BSTNode maxFromLeft = findMax (pos.leftChild); //need to make a findMax helper 
                //"Replacing "  pos.key " with " maxFromLeft.key
                pos.key = maxFromLeft.key;
                remove (maxFromLeft.key, pos.leftChild);
            }
            else if(pos.leftChild != null) {
                /* node pointed by pos has at most one child */
                BSTNode trash = pos;
                //"Promoting " pos.leftChild.key " to replace " pos.key
                pos = pos.leftChild;
                trash = null;
            }
            else if(pos.rightChild != null) {
                /* node pointed by pos has at most one child */
                BSTNode trash = pos;
                /* "Promoting " pos.rightChild.key" to replace " pos.key */
                pos = pos.rightChild;
                trash = null;
            }
            else {
                pos = null;
            }
        }
    }

这是对不平衡树的删除。我有C ++代码，所以我很快翻译了。可能会有一些小错误。您编码的树是否必须平衡？如果需要，我也有平衡删除。根据你问题的措辞，我不太确定。还要确保为findMax（）添加私有帮助函数

This is the remove for an unbalanced tree. I had the code in C++ so I have quickly translated. There may be some minor mistakes though. Does the tree you are coding have to be balanced? I also have the balanced remove if need be. I wasn't quite sure based on the wording of your question. Also make sure you add a private helper function for findMax()

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