BST：在 Java 中删除具有 2 个子节点的节点

Posted 2023-02-22

【中文标题】BST：在 Java 中删除具有 2 个子节点的节点

【英文标题】：BST: deleting node with 2 children in Java

【发布时间】：2016-05-04 14:25:32

【问题描述】：

我在 BST 中实现了用于删除节点的递归算法，但如果要删除的节点有两个子节点，它似乎无法正常工作。这是用于删除节点的方法的代码：

public boolean delete(int val)



    Node nodeToBeDeleted = find(val);
    if(nodeToBeDeleted != null)
    
        //case 1: node has no children
        if(nodeToBeDeleted.leftChild == null && nodeToBeDeleted.rightChild == null)
            deleteCase1(nodeToBeDeleted);

        //case 3: node has two children
        else if(nodeToBeDeleted.leftChild != null && nodeToBeDeleted.rightChild != null)
        
            deleteCase3(nodeToBeDeleted);
        

        //case 2: node has one child
        else if(nodeToBeDeleted.leftChild != null)
        
            //case 2 where left child should be deleted
            deleteCase2(nodeToBeDeleted);
        

        else if(nodeToBeDeleted.rightChild != null)
        
            //case 2 where right child should be deleted
            deleteCase2(nodeToBeDeleted);
        

        return true;
    
    else
        return false;

这里还有deleteCase1、deleteCase2和deleteCase3方法：

private void deleteCase1(Node nodeToBeDeleted)

        //check if node to be deleted is a left or a right child of the parent of the node to be deleted
        if(nodeToBeDeleted.parent.leftChild == nodeToBeDeleted)
        
            nodeToBeDeleted.parent.leftChild = null;
        
        else if(nodeToBeDeleted.parent.rightChild == nodeToBeDeleted)
        
            nodeToBeDeleted.parent.rightChild = null;
        

这里找方法：

public Node find(int val)

    if(root != null)
    
        return findNode(root, new Node(val));
    

    return null;


private Node findNode(Node search, Node node)

    if(search == null)
        return null;

    if(search.data == node.data)
    
        return search;
    
    else
    
        Node returnNode = findNode(search.leftChild, node);

        if(returnNode == null)
        
            returnNode = findNode(search.rightChild, node);
        

        return returnNode;
    

minLeftTreversal 方法：

private Node minLeftTreversal(Node node)

    if(node.leftChild == null)
        return node;

    return minLeftTreversal(node.leftChild);

树的结构如下所示： enter image description here

如果我删除 75，算法会起作用，但如果我尝试删除 25，它就会搞砸。

提前谢谢你！

【参考方案1】：

public boolean delete(int val) 中的第一个 if 语句缺少 和

        //case 1: node has no children
        if(nodeToBeDeleted.leftChild == null && nodeToBeDeleted.rightChild == null)
         // <---- ADD THIS
            deleteCase1(nodeToBeDeleted);
         // <---- AND ADD THIS
        //case 3: node has two children
        else if(nodeToBeDeleted.leftChild != null && nodeToBeDeleted.rightChild != null)
        
            deleteCase3(nodeToBeDeleted);