JAVA--递归分形树

递归分形树 --》 此做法相当于二叉树先序遍历  --  先画出此树干，再递归画出两个枝干。

画枝干时需要求出两个枝干的终点，需要运用一个几何的方法--- 
通过 repaint 调用paintComponent 来补充树干--
atan2 和 atan 的 区别  --  atan（x/y）时，y = 0 就会出错，我开始的代码在n=10左右时就出错了，看了老师的代码然后改为了atan2就可以了。

程序运行如图所示：

代码：

import java.awt.BorderLayout;import java.awt.Color;import java.awt.Graphics;import java.awt.Point;import java.awt.event.ActionEvent;import java.awt.event.ActionListener;import javax.swing.JButton;import javax.swing.JFrame;import javax.swing.JPanel;public class Recursive extends JFrame{private Tree t = new Tree();private JButton jb = new JButton("Increase");public Recursive(){this.add(t);JPanel panel = new JPanel();panel.add(jb);this.add(panel,BorderLayout.SOUTH);jb.addActionListener(new ActionListener() {@Overridepublic void actionPerformed(ActionEvent e) {t.run();}});}public static void main(String[] args) {Recursive f = new Recursive();f.setTitle("递归分形树");f.setSize(600,600);f.setLocationRelativeTo(null);f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);f.setVisible(true);}}class Tree extends JPanel{private int n = 0;private double A,B,C;private double PI = Math.acos(-1.0);protected void paintComponent(Graphics g){super.paintComponent(g);g.setColor(Color.red);Point p1 = new Point(this.getWidth()/2,this.getHeight()-10);Point p2 = new Point(this.getWidth()/2,this.getHeight()/2);display(g,n,p1,p2);}public void run(){n++;repaint();}private void display(Graphics g,int n,Point p1,Point p2){if (n>=0){g.drawLine(p1.x, p1.y, p2.x, p2.y);Point p3 = mid1(p1,p2);Point p4 = mid2(p1,p2);//System.out.println(p1.x+" "+p1.y+" "+p2.x+" "+p2.y);display(g, n-1, p2, p3);display(g, n-1, p2, p4);}}private Point mid1(Point p1,Point p2){Point p = new Point();A = Math.atan2(p2.x-p1.x,p1.y-p2.y);//A = Math.atan((double)(p2.x-p1.x)/(p1.y-p2.y));B = A - PI/6;C = Math.sqrt((p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y))/2;p.x= (int)(p2.x + C*Math.sin(B));p.y= (int)(p2.y - C*Math.cos(B));return p;}private Point mid2(Point p1,Point p2){Point p = new Point();A = Math.atan2(p2.x-p1.x,p1.y-p2.y);B = A + PI/6;C = Math.sqrt((p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y))/2;p.x= (int)(p2.x + C*Math.sin(B));p.y= (int)(p2.y - C*Math.cos(B));if (p.x==0){System.out.println(p1.x+" "+p1.y+" "+p2.x+" "+p2.y+" "+A);}return p;}}