java - [HELP]BST 递归插入时 Java StackOverflow 错误

Question

我正在尝试用递归Insert方法编写BST，但似乎我被困在程序无法跳出的行上。

如果在从 Main 调用 insert 时对元素键进行了排序，则它可以工作；如果它们未排序，则它不起作用，我不明白为什么。

public static void main(String[] args) {

    BST bst = new BST();

    bst.insert(2,"Val_0");
    bst.insert(1,"Val_1");
}

public class BSTNode {

public int key;
public String val;
public BSTNode left, right, parent;

public BSTNode(int key, String val) {
    this.key = key;
    this.val = val;
    this.left = new BSTNode();
    this.right = new BSTNode();
    this.parent = new BSTNode();
}

public BSTNode() {
    // TODO Auto-generated constructor stub
}

}

public class BST {

private BSTNode root;
public BST() {
    this.root = null;   
} 

    public void insert(int key, String val) {

    root = insertRec(new BSTNode(key,val));
    }

    private BSTNode insertRec(BSTNode node) {

    if (root == null) {
        root = node;
        return root;
    }

    if (node.key < root.key) {
        root.left = insertRec(root.left);
        root.left.parent = root;



    }if( node.key > root.key) {
        root.right = insertRec(root.right);
        root.left.parent = root;
    }


    return node;
     }
}

错误:线程“main”java.lang.StackOverflowError中出现异常，调试器在node.key < root.key处显示循环

Answer 1

已修复:我的构造函数每次都与插入方法一起为左右和父级创建元素。

public class BSTNode {

public int key;
public String val;
public BSTNode left, right, parent;

public BSTNode(int key, String val) {
    this.key = key;
    this.val = val;
    this.left = null;
    this.right = null;
    this.parent = null;
}

------//----------还稍微改变了方法:

public void insert(int key, String val) {
    root = insertRec(root, new BSTNode(key,val));

}

private BSTNode insertRec(BSTNode currentParent,BSTNode node) {
    if (currentParent == null) {
        return node;
    }

    if (node.key < currentParent.key) {
        currentParent.left = insertRec(currentParent.left,node);
        currentParent.left.parent = currentParent;

    }if(node.key > currentParent.key) {
        currentParent.right = insertRec(currentParent.right,node);
        currentParent.right.parent = currentParent;
    }


    return currentParent;
}