二叉树画法.doc

体系结构作业 学号：00847001 姓名：王川总 班级：软件二班using System;using System.Collections.Generic;using System.ComponentModel;using System.Data;using System.Drawing;using System.Linq;using System.Text;using System.Windows.Forms;/* 界面* /namespace BinaryTree using BinaryTree1; public partial class Form1 : Form Node node; BinaryTrees bt; public Form1() InitializeComponent(); private void Form1_Load(object sender, EventArgs e) button1.Enabled = false; button3.Enabled = false; button2.Enabled = false; private void button5_Click(object sender, EventArgs e) if (textBox1.Text.Length != 0) listBox1.ClearSelected(); char , aa = new char50,50; for (int i = 0; i 50; i+) for (int j = 0; j 50; j+) aai, j = ; node = new Node(); bt = new BinaryTrees(textBox1.Text); node = bt.CreateBT(node); bt.initial(node); bt.Count(node); int x = bt.RootPoint(node); bt.DrawPoint(node,x,0,aa); bt.DrawLine(node, aa); for (int i = 0; i 50; i+) String s=; for (int j = 0; j r_h?l_h:r_h; else return 0; /* * 求出每个结点所在树中高度 */ public void initial(Node t) if (t != null) t.lchildHeigth = heigth(t.lchild); t.rchildHeigth = heigth(t.rchild); initial(t.lchild); initial(t.rchild); /* * 设置每个结点与左右子女结点的距离 ，根据根结点的左？右？标记确定到左，右孩子结点的距离 */ public void Count(Node t) if (t != null) if (t.flag = Left_Right.left) / 当父结点标记是左时 if (t.rchild != null & t.lchild != null) t.rchildDistance = 2; if (t.rchildHeigth = 2) t.lchildDistance = (t.rchildHeigth - 1) * 2+1; else t.lchildDistance = 2; else if (t.lchild = null & t.rchild = null) return; else if (t.lchild != null) t.lchildDistance = 2; t.rchildDistance = 0; else if (t.rchild != null) t.rchildDistance = 2; t.lchildDistance = 0; else if (t.flag = Left_Right.right) / 当父结点标记是右时 if (t.rchild != null & t.lchild != null) t.lchildDistance = 2; if (t.lchildHeigth = 2) t.rchildDistance = (t.lchildHeigth - 1) * 2+1; else t.rchildDistance = 2; else if (t.rchild = null & t.lchild = null) return; else if (t.rchild != null) t.rchildDistance = 2; t.lchildDistance = 0; else if (t.lchild != null) t.lchildDistance = 2; t.rchildDistance = 0; / else if / 递归遍历设置到每一个结点 Count(t.lchild); Count(t.rchild); / if /* * 计算出根结点的x坐标位置 */ public int RootPoint(Node t) int count =0; /* * 由于算法的实现是使得右边的结点总超不过左边的长度， 故只要求出所有的左边的长度即可 */ if (t != null) count = t.lchildDistance; count += RootPoint(t.lchild); return count; else return 0; /* * 画结点 */ public void DrawPoint(Node t, int x, int y,char , point) if (t != null) pointy,x = t.data; t.x = x; t.y= y; DrawPoint(t.lchild, x - t.lchildDistance, y + t.lchildDistance,point); DrawPoint(t.rchild, x + t.rchildDistance, y + t.rchildDistance,point); /* * 画线，树中的线 */ public void DrawLine(Node t, char, point) if (t != null) int x1, y1, x2, y2; x1 = t.x; y1 = t.y; if (t.lchild != null) x2 = t.lchild.x; y2 = t.lchild.y; for (int x = x1-1, y = y1+1; x x2; x-, y+) pointy, x = /; if (t.rchild != nu