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import java.util.ArrayList;
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public class TreeNode {
private TreeNode leftNode;
private TreeNode rightNode;
private String nodeName;
public TreeNode getLeftNode() {
return leftNode;
}
public void setLeftNode(TreeNode leftNode) {
this.leftNode = leftNode;
}
public TreeNode getRightNode() {
return rightNode;
}
public void setRightNode(TreeNode rightNode) {
this.rightNode = rightNode;
}
public String getNodeName() {
return nodeName;
}
public void setNodeName(String nodeName) {
this.nodeName = nodeName;
}
public static int level=0;
public static void findNodeByLevel(ArrayListTreeNode nodes){
if(nodes==null||nodes.size()==0){
return ;
}
level++;
ArrayListTreeNode temp = new ArrayList();
for(TreeNode node:nodes){
System.out.println("第"+level+"层:"+node.getNodeName());
if(node.getLeftNode()!=null){
temp.add(node.getLeftNode());
}
if(node.getRightNode()!=null){
temp.add(node.getRightNode());
}
}
nodes.removeAll(nodes);
findNodeByLevel(temp);
}
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
TreeNode root = new TreeNode();
root.setNodeName("root");
TreeNode node1 = new TreeNode();
node1.setNodeName("node1");
TreeNode node3 = new TreeNode();
node3.setNodeName("node3");
TreeNode node7 = new TreeNode();
node7.setNodeName("node7");
TreeNode node8 = new TreeNode();
node8.setNodeName("node8");
TreeNode node4 = new TreeNode();
node4.setNodeName("node4");
TreeNode node2 = new TreeNode();
node2.setNodeName("node2");
TreeNode node5 = new TreeNode();
node5.setNodeName("node5");
TreeNode node6 = new TreeNode();
node6.setNodeName("node6");
root.setLeftNode(node1);
node1.setLeftNode(node3);
node3.setLeftNode(node7);
node3.setRightNode(node8);
node1.setRightNode(node4);
root.setRightNode(node2);
node2.setLeftNode(node5);
node2.setRightNode(node6);
ArrayListTreeNode nodes = new ArrayListTreeNode();
nodes.add(root);
findNodeByLevel(nodes);
}
}
你的程序有诸多问题,你的程序运行时候应该也会报错的吧?
这个写法不是很通用,不过我还是按照你的源码修改成了你想要的结果。
结构上基本一致,可实现基本已经面目全非了。
我用字符串代替了手工输入,你要是喜欢手工输入,你可以把我那个注释掉,用你自己的……
以下是修改后可用的代码:
import java.util.*;
class Node {
Node left;
Node Right;
char data;// 节点数据
void print() {
System.out.println(data + "");
}
public Node() {
this.left = null;
this.Right = null;
}
public Node(char data) {
this.left = null;
this.Right = null;
this.data = data;
}
}
class BTree {
static Node root = new Node();// 为根节点分配空间
static char ch[];// 输入的字符串
static int i = 0;
static Node CreateTree()// 先序建立二叉树
{
Node node = null;
if (ch[i] == '#') {
node = null;
i++;
}else {
node = new Node();
node.data = ch[i];
i++;
node.left = CreateTree();
node.Right = CreateTree();
}
return node;
}
static public void preorder(Node node)// 先序遍历二叉树
{
if (node != null) {
node.print();
preorder(node.left);
preorder(node.Right);
} else {
System.out.println("Tree node is empty");
}
}
}
public class Tree {
public static void main(String args[]) {
Scanner reader = new Scanner(System.in);
BTree.ch = new char[16];
BTree.ch[0] = 'a';
// 读取输入字符数组,以*结尾
BTree.ch = "ABC##DE#G##F###".toCharArray();
//for (int i = 0; (BTree.ch[i] = reader.next().charAt(0)) != '*'; i++){}
BTree.root = BTree.CreateTree();
BTree.preorder(BTree.root);
}
}
你说的是二叉树吧·····
/**
* 二叉树测试二叉树顺序存储在treeLine中,递归前序创建二叉树。另外还有能
* 够前序、中序、后序、按层遍历二叉树的方法以及一个返回遍历结果asString的
* 方法。
*/
public class BitTree {
public static Node2 root;
public static String asString;
//事先存入的数组,符号#表示二叉树结束。
public static final char[] treeLine = {'a','b','c','d','e','f','g',' ',' ','j',' ',' ','i','#'};
//用于标志二叉树节点在数组中的存储位置,以便在创建二叉树时能够找到节点对应的数据。
static int index;
//构造函数
public BitTree() {
System.out.print("测试二叉树的顺序表示为:");
System.out.println(treeLine);
this.index = 0;
root = this.setup(root);
}
//创建二叉树的递归程序
private Node2 setup(Node2 current) {
if (index = treeLine.length) return current;
if (treeLine[index] == '#') return current;
if (treeLine[index] == ' ') return current;
current = new Node2(treeLine[index]);
index = index * 2 + 1;
current.left = setup(current.left);
index ++;
current.right = setup(current.right);
index = index / 2 - 1;
return current;
}
//二叉树是否为空。
public boolean isEmpty() {
if (root == null) return true;
return false;
}
//返回遍历二叉树所得到的字符串。
public String toString(int type) {
if (type == 0) {
asString = "前序遍历:\t";
this.front(root);
}
if (type == 1) {
asString = "中序遍历:\t";
this.middle(root);
}
if (type == 2) {
asString = "后序遍历:\t";
this.rear(root);
}
if (type == 3) {
asString = "按层遍历:\t";
this.level(root);
}
return asString;
}
//前序遍历二叉树的循环算法,每到一个结点先输出,再压栈,然后访问它的左子树,
//出栈,访问其右子树,然后该次循环结束。
private void front(Node2 current) {
StackL stack = new StackL((Object)current);
do {
if (current == null) {
current = (Node2)stack.pop();
current = current.right;
} else {
asString += current.ch;
current = current.left;
}
if (!(current == null)) stack.push((Object)current);
} while (!(stack.isEmpty()));
}
//中序遍历二叉树
private void middle(Node2 current) {
if (current == null) return;
middle(current.left);
asString += current.ch;
middle(current.right);
}
//后序遍历二叉树的递归算法
private void rear(Node2 current) {
if (current == null) return;
rear(current.left);
rear(current.right);
asString += current.ch;
}
}
/**
* 二叉树所使用的节点类。包括一个值域两个链域
*/
public class Node2 {
char ch;
Node2 left;
Node2 right;
//构造函数
public Node2(char c) {
this.ch = c;
this.left = null;
this.right = null;
}
//设置节点的值
public void setChar(char c) {
this.ch = c;
}
//返回节点的值
public char getChar() {
return ch;
}
//设置节点的左孩子
public void setLeft(Node2 left) {
this.left = left;
}
//设置节点的右孩子
public void setRight (Node2 right) {
this.right = right;
}
//如果是叶节点返回true
public boolean isLeaf() {
if ((this.left == null) (this.right == null)) return true;
return false;
}
}
一个作业题,里面有你要的东西。
主函数自己写吧。当然其它地方也有要改的。