重新整理数据结构与算法(c#系列)—— 树的前中后序遍历[十六]
前言
理论文章:
直接看百度百科。
这个比较简单,直接放c#代码。
正文
建立节点模型:
public class HeroNode
{
private int no;
private string name;
private HeroNode left;
private HeroNode right;
public HeroNode(int no, string name) {
this.no = no;
this.name = name;
}
public int getNo() {
return no;
}
public void setNo(int no)
{
this.no = no;
}
public String getName()
{
return name;
}
public void setName(String name)
{
this.name = name;
}
public HeroNode getLeft()
{
return left;
}
public void setLeft(HeroNode left)
{
this.left = left;
}
public HeroNode getRight()
{
return right;
}
public void setRight(HeroNode right)
{
this.right = right;
}
public override string ToString()
{
return "姓名:" + name + "编号:" + no;
}
//编写前序遍历的方法 是根、左、右
public void preOrder() {
Console.WriteLine(this);
if (this.left != null)
{
this.left.preOrder();
}
if (this.right != null)
{
this.right.preOrder();
}
}
//中序遍历 是左、根、右
public void infixOrder() {
if (this.left != null)
{
this.left.infixOrder();
}
Console.WriteLine(this);
if (this.right != null)
{
this.right.infixOrder();
}
}
// 后续遍历为 左、右、根
public void postOrder()
{
if (this.left != null)
{
this.left.postOrder();
}
if (this.right != null)
{
this.right.postOrder();
}
Console.WriteLine(this);
}
}
建立树模型:
public class BinaryTree
{
private HeroNode root;
public void setRoot(HeroNode root)
{
this.root = root;
}
//前序遍历
public void preOrder()
{
if (this.root != null)
{
this.root.preOrder();
}
else
{
Console.WriteLine("二叉树为空,无法遍历");
}
}
//中序遍历
public void infixOrder()
{
if (this.root != null)
{
this.root.infixOrder();
}
else
{
Console.WriteLine("二叉树为空,无法遍历");
}
}
//后序遍历
public void postOrder()
{
if (this.root != null)
{
this.root.postOrder();
}
else
{
Console.WriteLine("二叉树为空,无法遍历");
}
}
}
测试:
static void Main(string[] args)
{
//先需要创建一颗二叉树
BinaryTree binaryTree = new BinaryTree();
//创建需要的结点
HeroNode root = new HeroNode(1, "宋江");
HeroNode node2 = new HeroNode(2, "吴用");
HeroNode node3 = new HeroNode(3, "卢俊义");
HeroNode node4 = new HeroNode(4, "林冲");
HeroNode node5 = new HeroNode(5, "关胜");
//设置节点
root.setLeft(node2);
root.setRight(node3);
node3.setRight(node4);
node3.setLeft(node5);
binaryTree.setRoot(root);
//前序遍历
Console.WriteLine("前序遍历");
binaryTree.preOrder();
Console.WriteLine("中序遍历");
binaryTree.infixOrder();
Console.WriteLine("后续遍历");
binaryTree.postOrder();
Console.ReadKey();
}
不好意思前面结果有点问题。
结果: