我试图使用awt和swing来绘制二叉树。我可以显示树的节点,但不能绘制从父节点到子节点的线。我使用以下类:Node(代表节点),treeGUI,DrawTree和Main。
这是没有行的输出。
我的意图是显示从父节点到子节点的行。我尝试使用drawLine
类中的Graphics
metohd,这是输出:
方法drawTree定义节点值在屏幕中的位置,并将该节点的位置存储在ArrayLists中。 drawLine方法绘制线条。我认为这些行是这样的,因为这些值按此特定顺序存储在ArrayList中。我尝试了各种方法以正确的方式画线,但都没有成功。我如何画出父母与孩子的界线?
public class TreeGUI extends JFrame {
private JPanel contentPane;
public Node node;
public DrawTree drawer;
/**
* Create the frame.
*/
public TreeGUI(Node node) {
setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
setBounds(100, 100, 500, 500);
contentPane = new JPanel();
contentPane.setBorder(new EmptyBorder(5, 5, 5, 5));
contentPane.setLayout(new BorderLayout(0, 0));
drawer = new DrawTree(node);
contentPane.add(drawer);
setContentPane(contentPane);
this.node = node;
setVisible(true);
}
}
class DrawTree extends JPanel{
public Node node;
public static ArrayList listX = new ArrayList();
public static ArrayList listY = new ArrayList();
public DrawTree(Node node){
this.node = node;
}
@Override
protected void paintComponent(Graphics g) {
// TODO Auto-generated method stub
g.setFont(new Font("Tahoma", Font.BOLD, 20));
DrawTree(g, 0, getWidth(), 0, getHeight() / node.getheight(node), node);
listX.clear();
listY.clear();
}
public void DrawTree(Graphics g, int StartWidth, int EndWidth, int StartHeight, int Level, Node node) {
String data = String.valueOf(node.getValue());
g.setFont(new Font("Tahoma", Font.BOLD, 20));
FontMetrics fm = g.getFontMetrics();
int dataWidth = fm.stringWidth(data);
g.drawString(data, (StartWidth + EndWidth) / 2 - dataWidth / 2, StartHeight + Level / 2);
listX.add((StartWidth + EndWidth) / 2 - dataWidth / 2);
listY.add(StartHeight + Level / 2);
drawLine(g, node);
if (node.getLeft() != null) {
DrawTree(g, StartWidth, (StartWidth + EndWidth) / 2, StartHeight + Level, Level, node.getLeft());
}
if (node.getRight() != null)
DrawTree(g, (StartWidth + EndWidth) / 2, EndWidth, StartHeight + Level, Level, node.getRight());
}
public void drawLine(Graphics g, Node node){
for (int i=1; i < listY.size(); i++)
g.drawLine((int)listX.get(i-1), (int)listY.get(i-1), (int)listX.get(i), (int)listY.get(i));
}
}
梅多多大街
public static void main(String[] args) {
Node raiz = null;
raiz = raiz.insert(raiz, 35);
raiz.insert(raiz, 25);
raiz.insert(raiz, 75);
raiz.insert(raiz, 30);
raiz.insert(raiz, 20);
raiz.insert(raiz, 12);
raiz.insert(raiz, 6);
raiz.insert(raiz, 23);
raiz.insert(raiz, 90);
TreeGUI gui = new TreeGUI(raiz);
}
最佳答案
您可以让DrawTree
函数返回其打印文本的位置。然后让父级从当前位置到子级DrawTree
函数返回的位置画一条线。这将使您摆脱列表。
public Point DrawTree(Graphics g, int StartWidth, int EndWidth, int StartHeight, int Level, Node node)
{
String data = String.valueOf(node.getValue());
g.setFont(new Font("Tahoma", Font.BOLD, 20));
FontMetrics fm = g.getFontMetrics();
int dataWidth = fm.stringWidth(data);
// Calculate position to draw text string
Point textPos = new Point((StartWidth + EndWidth) / 2 - dataWidth / 2, StartHeight + Level / 2);
g.drawString(data, textPos.x, textPos.y);
if (node.getLeft() != null) {
Point child1 = DrawTree(g, StartWidth, (StartWidth + EndWidth) / 2, StartHeight + Level, Level, node.getLeft());
// Draw line from this node to child node
drawLine(g, textPos, child1);
}
if (node.getRight() != null) {
Point child2 = DrawTree(g, (StartWidth + EndWidth) / 2, EndWidth, StartHeight + Level, Level, node.getRight());
// Draw line from this node to child node
drawLine(g, textPos, child2);
}
// Return position for parent to use
return textPos;
}
public void drawLine(Graphics g, Point p1, Point p2)
{
g.drawLine(p1.x, p1.y, p2.x, p2.y);
}