Question

我的treeNode设置如下：

public class BTreeNode {
  private int data;
  private BTreeNode left;
  private BTreeNode right;

  public BTreeNode(){}
  public BTreeNode(int dataValue){
    this(dataValue, null, null);
  }
  public BTreeNode(int dataValue, BTreeNode leftValue, BTreeNode rightValue){
    data = dataValue;
    left = leftValue;
    right = rightValue;
  }
  public void setData(int dataValue){
    data = dataValue;
  }
  public int getData(){
    return data;
  }
  public void setLeft(BTreeNode leftValue){
    left = leftValue;
  }
  public BTreeNode getLeft(){
    return left;
  }
  public void setRight(BTreeNode rightValue){
    right = rightValue;
  }
  public BTreeNode getRight(){
    return right;
  }
  public String toString(){
    return Integer.toString(getData());
  }
}

和我的二进制搜索树代码是：

public class BinarySearchTree {

private BTreeNode root;

public BinarySearchTree(){}

public void add(int n){
    if (root == null) {
        root = new BTreeNode(n);
        return;
    }else{
        BTreeNode current = root;
        while(current != null){
            if(n <= current.getData()){
                if (current.getLeft() == null) {
                    current.setLeft( new BTreeNode(n));
                    return;
                }
                current = current.getLeft();
            }else{
                if (current.getRight() == null) {
                    current.setRight(new BTreeNode(n));
                    return;
                }
                current = current.getRight();
            }
        }
    }
}

public void remove(int n){
    remove(root, n);
}

private BTreeNode remove(BTreeNode root, int n){
    if(root == null) return root;
    int value = root.getData();
    BTreeNode left = root.getLeft();
    BTreeNode right = root.getRight();
    if( n > value) root.setRight(remove(right, n));
    else if( n < value) root.setLeft(remove(left, n));
    else {
        //case 1: no children at all
        if (left == null && right == null) {
            return null;
        }
        //case 2: one child
        else if (left == null && right != null) {
            return right;


        } else if (right == null && left != null) {
            return left;
        }
        //case 3: two child
        else {
            int min = findRightMin(right);
            root.setData(min);
            remove(right, min);
            return root;
        }
    }
    return root;
}



private int findRightMin(BTreeNode root){
    if (root.getLeft() == null) return root.getData();
    else return findRightMin(root.getLeft());
}


public int getHeight(){
    return getHeight(root);
}

private int getHeight(BTreeNode root){
    if(root == null) return 0;
    return 1 + Math.max(getHeight(root.getLeft()), getHeight(root.getRight()));
}

public String toString(){
    return "{" + toString(root) + "}";
}

private String toString(BTreeNode root){

    if(root == null){
        return "";
    }else{

        String left =  toString(root.getLeft());
        String rootNode = Integer.toString(root.getData()) + ", ";
        String right = toString(root.getRight());

        return left + rootNode + right;
    }

}

public static void main(String[] args){

    BinarySearchTree tree = new BinarySearchTree();
    tree.add(10);
    tree.add(8);
    tree.add(9);
    tree.add(12);
    tree.add(15);
    tree.add(1);
    tree.add(5);
    tree.add(7);
    tree.add(6);

    System.out.println(tree.toString());
    System.out.println(tree.getHeight());
    tree.remove(8);
    System.out.println(tree.toString());
}

}

当我运行此代码时，控制台输出

{1, 5, 6, 7, 8, 9, 10, 12, 15, }
6
{1, 5, 6, 7, 9, 9, 10, 12, 15, }

代码找到了我想要删除并替换它的节点的右子节点，但没有删除节点。我在这里找不到哪个部分是错的。请帮忙！

Answer 1

您忘了为案例3设置right的{{1}}：

root

输出：

        // case 3: two child
            else {
                int min = findRightMin(right);
                root.setData(min);
                root.setRight(remove(right, min)); //this one
                return root;
            }
        }

请再试一次。我没有测试其他案例。

Answer 2

你错过了root.setRight

e0 5d

二叉搜索树的java实现删除树节点不起作用

2 个答案: