在JavaFX TableView中，如何根据某些按钮单击事件更改复选框行的颜色

TableView<Item> tableView = new TableView<>();
TableColumn<Item, Boolean> checkBoxColumn = new TableColumn<>("选择");
checkBoxColumn.setCellValueFactory(new PropertyValueFactory<>("selected"));
TableColumn<Item, String> nameColumn = new TableColumn<>("名称");
nameColumn.setCellValueFactory(new PropertyValueFactory<>("name"));
tableView.getColumns().addAll(checkBoxColumn, nameColumn);
tableView.setItems(data); // 设置数据模型

checkBoxColumn.setCellFactory(column -> new TableCell<Item, Boolean>() {
    @Override
    protected void updateItem(Boolean item, boolean empty) {
        super.updateItem(item, empty);
        if (empty || item == null) {
            setText(null);
            setStyle("");
        } else {
            setText(null);
            TableRow<Item> currentRow = getTableRow();
            if (item) {
                currentRow.setStyle("-fx-background-color: green;");
            } else {
                currentRow.setStyle("");
            }
        }
    }
});

button.setOnAction(event -> {
    Item selectedItem = tableView.getSelectionModel().getSelectedItem();
    if (selectedItem != null) {
        selectedItem.setSelected(!selectedItem.isSelected());
    }
});