将节点插入 BST 时出现无限循环

Question

我有一种根据字母顺序在 BST 中插入节点的方法，但是当我比较 2 个字符串时我有一个无限循环我认为当它通过比较时值永远不会改变，所以它再次与相同值导致无限循环。我认为aux 和Tnodes 没有使用递归方法更新值，因此它会一遍又一遍地比较相同的值。

class BST {

    BSTNode root;

    public BST() {
        root = null;
    }

    BSTNode aux = new BSTNode();

    BSTNode insertNames(BSTNode T , int data, String name, double salary) {
        if (root == null) {
            T = new BSTNode();  
            T.setName(name);
            root = T;
        } else {
            aux = root;

            if (name.compareTo(aux.getName()) < 0)
                aux.setLeft(insertNames(aux.getLeft(),data, name, salary));
            else if (name.compareTo(aux.getName()) >= 0)
                aux.setRight(insertNames(aux.getRight(),data, name, salary));
        }

        return T;
    }    
}

class Main{
public static void main(String[] args){

        BST alpha=new BST();
        BSTNode root = new BSTNode();
        alpha.insertNames(root, 0, "Roy", 0);
        alpha.insertNames(root, 0, "Joseph", 0);
}
}

Answer 1

package com.gati.dsalgo.string;

class BST {

    BSTNode root;

    public BST() {
        root = null;
    }

    void insertNames(int data, String name, double salary) {
        root = insertNames(root, data, name, salary);
    }

    BSTNode insertNames(BSTNode root, int data, String name, double salary) {
        if (root == null) {
            root = new BSTNode();
            root.setName(name);
            return root;
        }

        if (name.compareTo(root.getName()) < 0)
            root.setLeft(insertNames(root.getLeft(), data, name, salary));
        else if (name.compareTo(root.getName()) >= 0)
            root.setRight(insertNames(root.getRight(), data, name, salary));

        return root;
    }
}

public class Main1 {
    public static void main(String[] args) {

        BST alpha = new BST();
        alpha.insertNames(0, "Roy", 0);
        alpha.insertNames(0, "Joseph", 0);
        System.out.println("hello");
    }
}

class BSTNode {
    private String name;
    BSTNode left;
    BSTNode right;

    public void setName(String name) {
        this.name = name;

    }

    public void setRight(BSTNode right) {
        this.right = right;
    }

    public void setLeft(BSTNode left) {
        this.left = left;
    }

    public BSTNode getRight() {

        return right;
    }

    public BSTNode getLeft() {
        return left;
    }

    public String getName() {

        return name;
    }

}

Answer 2

请在递归结束逻辑中返回节点。

       /* If the tree is empty, return a new node */
if (root == null) {
    root = new BSTNode();
    root.setName(name);
    return root;
}

/* Otherwise, recur down the tree */
if (name.compareTo(root.getName()) < 0)
    root.setLeft(insertNames(root.getLeft(), data, name, salary));
else if (name.compareTo(aux.getName()) >= 0)
    root.setRight(insertNames(root.getRight(), data, name, salary));

/* return the (unchanged) node pointer */
return root;

也可以迭代求解

 if (localRoot == null) {
    newNode = new Node < V > (value, null);
    root = newNode;
    size++;
    return true;
}
if (comparator != null) {
//Some code
} else {
    Comparable << ? super V > v = (Comparable << ? super V > ) value;
    while (localRoot != null) {
        parent = localRoot;
        cmp = v.compareTo(localRoot.getValue());
        if (cmp < 0) {
            localRoot = localRoot.getLeftChield();
        } else if (cmp > 0) {
            localRoot = localRoot.getRightChield();
        } else {
            localRoot.incrementBy(nCopies);
            return true;
        }
    }
    newNode = new Node < V > (value, parent);
    if (cmp < 0) {
        parent.setLeftChield(newNode);
    } else if (cmp > 0) {
        parent.setRightChield(newNode);
    }
    size++;
    return true;
}
return false;