我有用于BinaryTree创建和遍历的代码
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();
}
}它编译并运行。但我认为顺序遍历存在问题。
我创建了以下树，
2 1 3
但它只打印2个。
答案 0 :(得分：11)
我试过按照你的方式解决问题，我已经粘贴了下面的解决方案。虽然我没有彻底测试过，所以它可能会在一些边缘情况下失败。但我已经测试了一个案例。如果在某些情况下失败，请告诉我。我希望其他人能帮助我更好地回答这个问题。我同意这个解决方案不是编写二叉树最理想的方法，但是如果有人正在练习它就不会受到这样的影响..
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] ：更新了代码以使用数组生成二叉树。这将减少用户互动。
答案 1 :(得分：1)
使用所有遍历类型和测试用例实现Java中二进制树的最佳方法，如下所示
package com.nitin.tree;
public class Tree
{
private Node parent;
private int data;
private int size = 0;
public Tree() {
parent = new Node(data);
}
public void add(int data) {
if (size == 0) {
parent.data = data;
size++;
} else {
add(parent, new Node(data));
}
}
private void add(Node root, Node newNode) {
if (root == null) {
return;
}
if (newNode.data < root.data) {
if (root.left == null) {
root.left = newNode;
size++;
} else {
add(root.left, newNode);
}
} else {
if (root.right == null) {
root.right = newNode;
size++;
} else {
add(root.right, newNode);
}
}
}
public int getLow() {
Node current = parent;
while (current.left != null) {
current = current.left;
}
return current.data;
}
public int getHigh() {
Node current = parent;
while (current.right != null) {
current = current.right;
}
return current.data;
}
private void in(Node node) {
if (node != null) {
in(node.left);
System.out.print(node.data + " ");
in(node.right);
}
}
private void pre(Node node) {
if (node != null) {
System.out.print(node.data + " ");
pre(node.left);
pre(node.right);
}
}
private void post(Node node) {
if (node != null) {
post(node.left);
post(node.right);
System.out.print(node.data + " ");
}
}
public void preorder() {
System.out.print("Preorder Traversal->");
pre(parent);
System.out.println();
}
public void postorder() {
System.out.print("Postorder Traversal->");
post(parent);
System.out.println();
}
public void inorder() {
System.out.print("Inorder Traversal->");
in(parent);
System.out.println();
}
private class Node {
Node left;
Node right;
int data;
public Node(int data) {
this.data = data;
}
}
public String toString() {
Node current = parent;
System.out.print("Traverse From Left ");
while (current.left != null && current.right != null) {
System.out.print(current.data + "->[" + current.left.data + " " + current.right.data + "] ");
current = current.left;
}
System.out.println();
System.out.print("Traverse From Right ");
current = parent;
while (current.left != null && current.right != null) {
System.out.print(current.data + "->[" + current.left.data + " " + current.right.data + "] ");
current = current.right;
}
return "";
}
public static void main(String af[]) {
Tree t = new Tree();
t.add(40);
t.add(25);
t.add(78);
t.add(10);
t.add(32);
t.add(50);
t.add(93);
t.add(3);
t.add(17);
t.add(30);
t.add(38);
System.out.println(t.getLow());
System.out.println(t.getHigh());
System.out.println("Size-" + t.size);
System.out.println(t);
t.inorder();
t.preorder();
t.postorder();
}
}
答案 2 :(得分：0)
您的问题出在public void createNodes(Node temp, T data)函数中。传入一个与类变量temp同名的参数。首先，我认为你不需要自己的类变量。其次，在此方法中为temp分配仅具有局部效果 - 您放松了temp参数中的信息，但设置temp不会影响其在被调用方法中的值。我建议你重写方法，使它返回指向新创建节点的指针，并将此指针指向本地left的{{1}}和right。这样一来，变化就会传播开来。
答案 3 :(得分：0)
输出树的另一种类型：
public void inorder()
{
inorder(root);
}
protected void visit(BSTNode<T> p)
{
System.out.println("Node: " + p.el + "Left Side:" + (p.left!=null?p.left.el:"null") +
"Right Side:" + (p.right!=null?p.right.el:"null"));
}
答案 4 :(得分：0)
我已经更改了BinaryTree类，如下所示。请特别注意 createNode 方法的更改。
正如之前的帖子中提到的那样，问题是当你的引用作为参数传递给 createNode 方法时，它不会持久存在。这种变化只是局部的。在您创建节点时，您需要在方法本身中返回显式节点引用。
public Node createNode()
{
Integer value = getData();
Node temp = new Node(value);
if(head==null)
{
head = temp;
}
System.out.println("Do you want to add left branch on node("+value+")? Enter y/n");
if(input.next().charAt(0)=='y')
{
temp.left=createNode();
}
System.out.println("Do you want to add right branch on node("+value+")? Enter y/n");
if(input.next().charAt(0)=='y')
{
temp.right=createNode();
}
return temp;
}
以下是结果输出：
Basic operations on a tree:
1. Create tree
2. Insert
3. Search value
4. print list
Else. Exit
Choice:1
Basic operations on a tree:
1. Create tree
2. Insert
3. Search value
4. print list
Else. Exit
Choice:2
Enter the value to insert:
10
Do you want to add left branch on node(10)? Enter y/n
y
Enter the value to insert:
20
Do you want to add left branch on node(20)? Enter y/n
n
Do you want to add right branch on node(20)? Enter y/n
n
Do you want to add right branch on node(10)? Enter y/n
y
Enter the value to insert:
30
Do you want to add left branch on node(30)? Enter y/n
n
Do you want to add right branch on node(30)? Enter y/n
n
Basic operations on a tree:
1. Create tree
2. Insert
3. Search value
4. print list
Else. Exit
Choice:4
20
10
30
我希望这会对以后的某些人有所帮助（即使这已经晚了3年......）。我今天刚开始学习二叉树。我实际上计划将此作为执行更多相关任务的基础！
答案 5 :(得分：0)
我更改了createNode方法以使其正常工作：
public Node 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')
{
newnode.left = createNode(newnode.left, value);
}
System.out.println("If right child exits for ("+value+") enter y else n");
if(input.next().charAt(0) == 'y')
{
newnode.right = createNode(newnode.right, value);
}
return newnode;
}