在Java中二叉树实现 [英] BinaryTree implementation in java
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问题描述
我有这个$ C $下二叉树建立和遍历
I have this code for BinaryTree creation and traversal
class Node
{
Integer data;
Node left;
Node right;
Node()
{
data = null;
left = null;
right = null;
}
}
class BinaryTree
{
Node head;
Scanner input = new Scanner(System.in);
BinaryTree()
{
head = null;
}
public void createNode(Node temp, Integer value)
{
Node newnode= new Node();
value = getData();
newnode.data = value;
temp = newnode;
if(head==null)
{
head = temp;
}
System.out.println("If left child exits for ("+value+") enter y else n");
if(input.next().charAt(0)=='y')
{
createNode(temp.left, value);
}
System.out.println("If right child exits for ("+value+") enter y else n");
if(input.next().charAt(0)=='y')
{
createNode(temp.right, value);
}
}
public Integer getData()
{
out.println("Enter the value to insert:");
return (Integer)input.nextInt();
}
public void print()
{
inorder(head);
}
public void inorder(Node node)
{
if(node!=null)
{
inorder(node.left);
System.out.println(node.data);
inorder(node.right);
}
else
return;
}
}
class BinaryTreeWorker
{
static BinaryTree treeObj = null;
static Scanner input = new Scanner(System.in);
public static void displaymenu()
{
int choice;
do{
out.print("\n Basic operations on a tree:");
out.print("\n 1. Create tree \n 2. Insert \n 3. Search value \n 4. print list\n Else. Exit \n Choice:");
choice = input.nextInt();
switch(choice)
{
case 1:
treeObj = createBTree();
break;
case 2:
treeObj.createNode(null, null);
break;
case 3:
//searchnode();
break;
case 4:
treeObj.print();
break;
default:
return;
}
}while(true);
}
public static BinaryTree createBTree()
{
return new BinaryTree();
}
public static void main(String[] args)
{
displaymenu();
}
}
编译和运行。不过，我觉得有什么不对的序遍历。
It compiles and runs. But I think there is something wrong with the inorder traversal.
我在下面的树，
2
1 3
但只有二打印。
But it prints only 2.
推荐答案
我曾尝试解决问题的方式，我已经贴在下面。该解决方案虽然我没有测试它彻底，因此可能会在一些边缘状态失败..但是我已经测试了它的一个案例。请让我们如果不能在某些情况下我知道了。我想AP preciate其他人在做这个答案更好的帮助。我同意这个解决方案并不是最理想的方式codeA二叉树，但它不会伤害这样，如果有一个人是刚刚练..
I have tried solving the problem your way and I have pasted the solution below.. Though I haven't tested it thoroughly so it might fail in some edge condition.. But I have tested it for one case. Kindly let me know if it fails in some scenario. I would appreciate others help in making this answer better. I agree that this solution is not the most ideal way to code a Binary Tree but it wont hurt this way if some one is just practicing..
import java.util.Scanner;
class Node
{
Integer data;
Node left;
Node right;
Node()
{
data = null;
left = null;
right = null;
}
}
class BinaryTree
{
Node head;
Scanner input = new Scanner(System.in);
BinaryTree()
{
head = null;
}
public void createNode(Node temp,Node newnode)
{
if(head==null)
{
System.out.println("No value exist in tree, the value just entered is set to Root");
head = newnode;
return;
}
if(temp==null)
temp = head;
System.out.println("where you want to insert this value, l for left of ("+temp.data+") ,r for right of ("+temp.data+")");
char inputValue=input.next().charAt(0);
if(inputValue=='l'){
if(temp.left==null)
{
temp.left=newnode;
System.out.println("value got successfully added to left of ("+temp.data+")");
return;
}else {
System.out.println("value left to ("+temp.data+") is occupied 1by ("+temp.left.data+")");
createNode(temp.left,newnode);
}
}
else if(inputValue=='r')
{
if(temp.right==null)
{
temp.right=newnode;
System.out.println("value got successfully added to right of ("+temp.data+")");
return;
}else {
System.out.println("value right to ("+temp.data+") is occupied by ("+temp.right.data+")");
createNode(temp.right,newnode);
}
}else{
System.out.println("incorrect input plz try again , correctly");
return;
}
}
public Node generateTree(){
int [] a = new int[10];
int index = 0;
while(index<a.length){
a[index]=getData();
index++;
}
if(a.length==0 ){
return null;
}
Node newnode= new Node();
/*newnode.left=null;
newnode.right=null;*/
return generateTreeWithArray(newnode,a,0);
}
public Node generateTreeWithArray(Node head,int [] a,int index){
if(index >= a.length)
return null;
System.out.println("at index "+index+" value is "+a[index]);
if(head==null)
head= new Node();
head.data = a[index];
head.left=generateTreeWithArray(head.left,a,index*2+1);
head.right=generateTreeWithArray(head.right,a,index*2+2);
return head;
}
public Integer getData()
{
System.out.println("Enter the value to insert:");
return (Integer)input.nextInt();
}
public void print()
{
inorder(head);
}
public void inorder(Node node)
{
if(node!=null)
{
inorder(node.left);
System.out.println(node.data);
inorder(node.right);
}
else
return;
}
}
public class BinaryTreeWorker
{
static BinaryTree treeObj = null;
static Scanner input = new Scanner(System.in);
public static void displaymenu()
{
int choice;
do{
System.out.print("\n Basic operations on a tree:");
System.out.print("\n 1. Create tree \n 2. Insert \n 3. Search value \n 4. print list\n 5. generate a tree \n Else. Exit \n Choice:");
choice = input.nextInt();
switch(choice)
{
case 1:
treeObj = createBTree();
break;
case 2:
Node newnode= new Node();
newnode.data = getData();
newnode.left=null;
newnode.right=null;
treeObj.createNode(treeObj.head,newnode);
break;
case 3:
//searchnode();
break;
case 4:
System.out.println("inorder traversal of list gives follows");
treeObj.print();
break;
case 5:
Node tempHead = treeObj.generateTree();
System.out.println("inorder traversal of list with head = ("+tempHead.data+")gives follows");
treeObj.inorder(tempHead);
break;
default:
return;
}
}while(true);
}
public static Integer getData()
{
System.out.println("Enter the value to insert:");
return (Integer)input.nextInt();
}
public static BinaryTree createBTree()
{
return new BinaryTree();
}
public static void main(String[] args)
{
displaymenu();
}
}
[更新] ：更新了code生成使用数组二叉树。这将涉及较少的用户交互。
[Update] : Updated the code to generate a binary tree using an array. This will involve less user interaction.
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