Question

我正在尝试为我的BST类实现删除功能。我正在尝试遵循一些标准教程，并且除了我的删除功能之外我很好。我没有收到任何错误，但它没有删除预期的节点。我递归地到达预期的Node，如果它只有一个子节点则返回对方节点，否则我从n.right返回最小值。

任何帮助将不胜感激 - 谢谢！

public class BinaryTree {

    private Node root = null;

    private class Node {

        private Node left;
        private Node right;
        private int data;

        public Node(int data, Node left, Node right) {
            this.data = data;
            this.left = left;
            this.right = right;
        }
    }

    public boolean add(int data) {

        root = add(root, data);
        return true;

    }

    private Node add(Node node, int data) {

        if (node == null) {
            node = new Node(data, null, null);
        }

        else {
            if (data < node.data) {
                node.left = add(node.left, data);
            }
            else {
                node.right = add(node.right, data);
            }
        }
        return node;
    }

    // Helper function

    private Node findMin(Node node) {
        while (node.left != null) {
            node = node.left;
        }
        return node;
    }

    private Node findMax(Node node) {
        while (node.right != null) {
            node = node.right;
        }
        return node;
    }

    private void inOrderTraversal(Node n) {
        if (n == null) return;

        else {
            inOrderTraversal(n.left);
            System.out.println(n.data);
            inOrderTraversal(n.right);
        }
    }

    public void inOrderTraversal() {
        inOrderTraversal(root);
    }

    private Node deleteKey(Node n, int data) {

        if (n == null) return n;

        if (data < n.data) deleteKey(n.left, data);
        else if (data > n.data) deleteKey(n.right, data);

        else {
            if (n.left == null) return n.right;
            else if (n.right == null) return n.left;

            n.data =  findMin(n.right).data;
            n.right = deleteKey(n.right, n.data);
        }
        return n;
    }

    public void deleteKey(int data) {
        root = deleteKey(root, data);
    }

    public static void main(String[] args) {

        BinaryTree bst = new BinaryTree();

        bst.add(2);
        bst.add(8);
        bst.add(1);
        bst.add(7);
        bst.add(3);
        bst.add(11);
        bst.add(1);
        bst.add(21);
        bst.add(10);
        bst.add(12);

        bst.inOrderTraversal();

        System.out.println("----------------");

        bst.deleteKey(3);

        bst.inOrderTraversal();
    }
}

Answer 1

我认为问题出在您的private Node deleteKey(Node n, int data) { if (n == null) return n; if (data < n.data) n.left = deleteKey(n.left, data); // must reassign child here else if (data > n.data) n.right = deleteKey(n.right, data); // must reassign child here else { if (n.left == null) return n.right; else if (n.right == null) return n.left; n.data = findMin(n.right).data; n.right = deleteKey(n.right, n.data); // this was correct } return n; }函数中：

data < n.data

当data > n.data或LOGINID DATE TIME mickey 1/1/2018 0 days 08:54:00.000000000 mickey 1/1/2018 0 days 08:57:00.000000000 mickey 1/1/2018 0 days 09:01:00.000000000 mickey 1/1/2018 0 days 09:19:00.000000000 mickey 1/1/2018 0 days 09:26:00.000000000 mickey 1/1/2018 0 days 09:41:00.000000000 mickey 1/1/2018 0 days 10:03:00.000000000 mickey 1/1/2018 0 days 11:35:00.000000000 mickey 1/1/2018 0 days 11:36:00.000000000 mickey 1/2/2018 0 days 09:54:00.000000000 mickey 1/2/2018 0 days 10:21:00.000000000 minnie 1/1/2018 0 days 09:00:00.000000000 minnie 1/1/2018 0 days 10:20:00.000000000 minnie 1/1/2018 0 days 10:14:00.000000000 minnie 1/1/2018 0 days 10:51:00.000000000 minnie 1/1/2018 0 days 11:05:00.000000000 minnie 1/1/2018 0 days 11:24:00.000000000 minnie 1/2/2018 0 days 02:20:00.000000000 minnie 1/2/2018 0 days 02:25:00.000000000 minnie 1/2/2018 0 days 02:39:00.000000000 This is what I want: LOGINID DATE TIME intense_cnt mickey 1/1/2018 0 days 08:54:00.000000000 4 mickey 1/1/2018 0 days 08:57:00.000000000 4 mickey 1/1/2018 0 days 09:01:00.000000000 3 mickey 1/1/2018 0 days 09:19:00.000000000 3 mickey 1/1/2018 0 days 09:26:00.000000000 2 mickey 1/1/2018 0 days 09:41:00.000000000 2 mickey 1/1/2018 0 days 10:03:00.000000000 1 mickey 1/1/2018 0 days 11:35:00.000000000 2 mickey 1/1/2018 0 days 11:36:00.000000000 2 mickey 1/2/2018 0 days 09:54:00.000000000 2 mickey 1/2/2018 0 days 10:21:00.000000000 1 minnie 1/1/2018 0 days 09:00:00.000000000 1 minnie 1/1/2018 0 days 10:20:00.000000000 2 minnie 1/1/2018 0 days 10:14:00.000000000 2 minnie 1/1/2018 0 days 10:51:00.000000000 2 minnie 1/1/2018 0 days 11:05:00.000000000 2 minnie 1/1/2018 0 days 11:24:00.000000000 1 minnie 1/2/2018 0 days 02:20:00.000000000 3 minnie 1/2/2018 0 days 02:25:00.000000000 3 minnie 1/2/2018 0 days 02:39:00.000000000 3时，您不会更新当前节点的离开。所以这是找到要删除的相应节点，但是它不能在从递归的备份路径上正确地重建节点分支。

BST删除功能不会删除预期节点（Java）

1 个答案: