如何打印二叉树图?
答
cat
/\
cat1 cat2
二叉树由
- 根节点
- 左子树
- 一个右子树
要打印这样的树,我们要打印的左边和右边的子树除了另一边(至少有一个空格)之外,然后在它上面打印根节点,集中在两个子节点的中间ees,并用ASCII线连接。为了这个工作,我们需要知道两个子树的宽度。
使用这些想法和递归来创建您的树形图。
这里是一个可能有用的方法说明:
/**
* creates an ASCII-drawing of a binary tree.
* @param node the root node of the tree in question
* @return a String[] of the individual lines of the drawing.
* The first line contains the representation of the root node,
* the last line only leaf nodes, interim lines may contain
* line drawing characters or interior nodes.
*
* All the contained strings have the same length (are padded
* with spaces, where necessary).
*/
String[] drawTree(Node node) {
...
}
要输出的树,你那么只需要做到这一点:
for(String line : drawTree(root)) {
System.out.println(line);
}
那么,我们如何实现我们的drawTree
方法?
- 它会对叶节点(即无子节点)做什么?
- 如果我们有一个非叶节点,那么如何将两个这样的调用(对于左右子树)的结果(即指定的两个字符串数组)结合到另一个字符串数组中,如上所述? (首先看看两个阵列具有相同长度的简单情况,即树木具有相同的深度。)
祝你好运!
答
这是一个完整的,可运行的Demo,它是Scala代码的翻译版本,所以它不是惯用的Java。
有两种实现方式,一种是采用Tree并从中生成JTree,另一种是使用drawString并将树适合JFrame大小。
import java.awt.*;
import javax.swing.*;
import javax.swing.tree.*;
import javax.swing.JTree;
/**
(c) GPLv3 2010-09-24
*/
class MNode {
MNode l; // left
MNode r; // right
int t; // value
public MNode (int t, MNode l, MNode r) {
this.l = l;
this.r = r;
this.t = t;
}
public void add (MNode mn) {
if (l == null && t > mn.t)
l = mn;
else if (t > mn.t)
l.add (mn);
else if (r == null)
r = mn;
else r.add (mn);
}
}
abstract class NodePrinter {
abstract void nodeprint (MNode root);
int max (int a, int b) { return (a > b) ? a : b; }
int depth (MNode n)
{
if (n.l == null && n.r == null) return 1;
if (n.l == null) return 1 + depth (n.r);
if (n.r == null) return 1 + depth (n.l);
return 1 + max (depth (n.l), depth (n.r));
}
}
class SwingPrinter extends NodePrinter {
void nodeprint (MNode root) {
JFrame jf = new JFrame ("Mein Freund, der Baum, ist tot");
jf.setSize (380, 380);
jf.setLocationRelativeTo (null);
JTree jt = new JTree (translate2SwingTree (root));
jf.add (jt);
openSubnodes (0, jt);
jf.setDefaultCloseOperation (WindowConstants.DISPOSE_ON_CLOSE);
jf.setVisible (true);
}
/**
Open current branch.
We need TreePath AND row.
Open the MNode, iterierate with the row one step, and check there,
whether the Branch is a part of the new branch.
@param row the row of the starting MNode.
*/
void openSubnodes (int row, JTree jt) {
TreePath tp = jt.getPathForRow (row);
jt.expandRow (row);
if (tp.isDescendant (jt.getPathForRow (row + 1)))
openSubnodes (row + 1, jt);
}
DefaultMutableTreeNode translate2SwingTree (MNode ast)
{
DefaultMutableTreeNode dmtn = new DefaultMutableTreeNode ("" + ast.t);
if (ast.l != null)
dmtn.add (translate2SwingTree (ast.l));
if (ast.r != null)
dmtn.add (translate2SwingTree (ast.r));
return dmtn;
}
}
class TreeCanvas extends JPanel {
private MNode root;
private NodePrinter np;
public TreeCanvas (MNode root, NodePrinter np) {
this.root = root;
this.np = np;
d = np.depth (root);
rows = (2 * d); // - 1
cols = 2 << d;
}
private int d;
private int rows;
private int cols;
// @override
public void paint (Graphics g) {
Dimension dim = getSize();
int xf = dim.width/cols;
int yf = dim.height/rows;
int fontsize = (xf + yf)/2;
g.setFont (g.getFont().deriveFont (fontsize* 1.5f));
xyPrint (root, dim.width/2, dim.width/2, 1, xf, yf, g);
}
/**
___50 60 70__________________
10 | x0 x0-x1: (50,30) - (60, 10)
20 | /\ x0-x2: (60,10) - (70, 30)
30 | x1 x2
*/
void xyPrint (MNode n, int x, int dx, int y, int xf, int yf, Graphics g) {
Graphics2D g2d = (Graphics2D) g;
g2d.setStroke (new BasicStroke (3.0f));
g.drawString ("" + n.t, x - xf, (y+1) * yf);
g.setColor (Color.BLACK);
if (n.l != null) {
g.drawLine (x - (dx/2) + xf, (y+2) * yf, x, (y+1) * yf); // line:Up
xyPrint (n.l, x - dx/2, dx/2, y + 2, xf, yf, g);
}
if (n.r != null) {
g.drawLine (x + xf, (y+1) * yf, x + (dx/2), (y+2) * yf); // line down
xyPrint (n.r, x + dx/2, dx/2, y + 2, xf, yf, g);
}
}
}
class ColorSwingPrinter extends NodePrinter {
void nodeprint (MNode root) {
JFrame jf = new JFrame ("Rootnode");
jf.setSize (650, 520);
jf.setLocationRelativeTo (null);
jf.add (new TreeCanvas (root, this));
jf.setDefaultCloseOperation (WindowConstants.DISPOSE_ON_CLOSE);
jf.setVisible (true);
}
}
class RootNode extends MNode {
public RootNode (String s)
{
super (Integer.parseInt ("" + s.charAt (0)), null, null);
for (String elem: s.substring (2).split (" "))
{
int i = Integer.parseInt (elem);
MNode mn = new MNode (i, null, null);
super.add (mn);
}
}
}
public class NodePrinterTest {
public static void main (String [] args)
{
String param = "6 7 4 3 8 2 9 5";
/* 6
4 7
3 5 8
2 9
*/
RootNode root = new RootNode (param);
ColorSwingPrinter printer = new ColorSwingPrinter();
printer.nodeprint (root);
SwingPrinter printer2 = new SwingPrinter();
printer2.nodeprint (root);
}
}
打印,当然你的树需要一些适应,因为你的节点很可能将没有公共的属性L,R和T,但你应该可以翻译它们向左()或setLeft()/ getLeft( ) 等等。
我不记得ColorSwingPrinter从哪里得到它的名字 - 它应该被命名为ResizingCanvasPrinter或类似的东西。对不起。
下面是这两种实现的截图:
定心像猫,CAT1较长的文本,CAT2将需要调整。到现在为止,第一个字符被放置在后代的中心