找一个Java程序:关于二叉树的建立和排序

从键盘接受输入(先序)，以二叉链表作为存储结构，建立二叉树(以先序来建立)

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结果不是唯一

二叉排序树（BST） Java实现

public class NodeE {

int key; // data used as key value

E data; // other data

NodeE leftChild; // this node's left child

NodeE rightChild; // this node's right child

public Node(int key, E o) {

this.key = key;

this.data = o;

}

public void displayNode() {

System.out.printf("%d, %s\n", key, data.toString());

}

}

===============================================================

package net.acoder.adt.tree;

public class TreeE {

private NodeE root;

public Tree(NodeE root) {

if (root == null) {

throw new NullPointerException("root can't be null");

}

this.root = root;

}

public Tree(int key, E o) {

this(new NodeE(key, o));

}

public NodeE getRoot() {

return root;

}

/**

* find a node by its key

*

* @param key

* @return

*/

public NodeE find(int key) {

NodeE current = root;

while (current.key != key) {

if (key current.key) {

current = root.leftChild;

} else {

current = root.rightChild;

}

if (current == null) {

return null;

}

}

return current;

}

/**

* insert a node to this tree

*

* @param key

* @param o

*/

public void insert(int key, E o) {

NodeE aNode = new NodeE(key, o);

if (root == null) {

this.root = aNode;

return;

}

NodeE current = root;

NodeE parent = root;

while (true) {

parent = current;

if (key parent.key) {

current = parent.leftChild;

if (current == null) {

parent.leftChild = aNode;

return;

}

} else {

current = parent.rightChild;

if (current == null) {

parent.rightChild = aNode;

return;

}

}

}

}

public boolean delete(int key) {

NodeE current = root;

NodeE parent = root;

boolean isLeftChild = true;

// search for node

while (current.key != key) {

parent = current;

if (key current.key) {

isLeftChild = true;

current = current.leftChild;

} else {

isLeftChild = false;

current = current.rightChild;

}

if (current == null) {

return false;

}

}

// if no children, simply delete it

if (current.leftChild == null current.rightChild == null) {

if (current == parent) {

root = null;

} else if (isLeftChild == true) {

parent.leftChild = null;

} else if (isLeftChild == false) {

parent.rightChild = null;

}

return true;

}

// if no left children, replace with right subtree

if (current.leftChild == null) {

if (current == root) {

root = current.rightChild;

} else if (isLeftChild) {

parent.leftChild = current.rightChild;

} else if (!isLeftChild) {

parent.leftChild = current.rightChild;

}

return true;

}

// if no right children, replace with left subtree

if (current.rightChild == null) {

if (current == root) {

root = current.leftChild;

} else if (isLeftChild) {

parent.leftChild = current.leftChild;

} else if (!isLeftChild) {

parent.leftChild = current.leftChild;

}

return true;

}

// get successor of node to delete

NodeE successor = getSuccessor(current);

if (current == root) {

current = successor;

} else if (isLeftChild) {

parent.leftChild = successor;

} else {

parent.rightChild = successor;

}

successor.leftChild = current.leftChild;

return true;

}

private NodeE getSuccessor(NodeE delNode) {

NodeE successorParent = delNode;

NodeE successor = delNode;

NodeE current = delNode.rightChild;

while (current != null) {

successorParent = successor;

successor = current;

current = current.leftChild;

}

if (successor != delNode.rightChild) {

successorParent.leftChild = successor.rightChild;

successor.rightChild = delNode.rightChild;

}

return successor;

}

public void inOrder(NodeE aNode) {

if (aNode != null) {

inOrder(aNode.leftChild);

aNode.displayNode();

inOrder(aNode.rightChild);

}

}

public void preOrder(NodeE aNode) {

if (aNode != null) {

aNode.displayNode();

inOrder(aNode.leftChild);

inOrder(aNode.rightChild);

}

}

public void backOrder(NodeE aNode) {

if (aNode != null) {

inOrder(aNode.leftChild);

inOrder(aNode.rightChild);

aNode.displayNode();

}

}

public NodeE minimum() {

NodeE current = this.root;

NodeE result = null;

while (current != null) {

result = current;

current = current.leftChild;

}

return result;

}

public NodeE maximum() {

NodeE current = this.root;

NodeE result = null;

while (current != null) {

result = current;

current = current.rightChild;

}

return result;

}

}

以前的代码, 记得没写完, 好像就是BST

java二叉排序树，已有代码，如何调通输出？

你好，很高兴回答你的问题。

目前已经有了二叉树以及二叉树节点的类。

需要一个main方法，在其中创建节点（通过节点类的构造方法），构建树（通过树的构造方法以及insert方法）。可以执行查询的方法以及展示的方法。

如果有帮助到你，请点击采纳。

java二叉树的顺序表实现

做了很多年的程序员，觉得什么树的设计并不是非常实用。二叉树有顺序存储，当一个insert大量同时顺序自增插入的时候，树就会失去平衡。树的一方为了不让塌陷，会增大树的高度。性能会非常不好。以上是题外话。分析需求在写代码。

import java.util.List;

import java.util.LinkedList;

public class Bintrees {

private int[] array = {1, 2, 3, 4, 5, 6, 7, 8, 9};

private static ListNode nodeList = null;

private static class Node {

Node leftChild;

Node rightChild;

int data;

Node(int newData) {

leftChild = null;

rightChild = null;

data = newData;

}

}

// 创建二叉树

public void createBintree() {

nodeList = new LinkedListNode();

// 将数组的值转换为node

for (int nodeIndex = 0; nodeIndex array.length; nodeIndex++) {

nodeList.add(new Node(array[nodeIndex]));

}

// 对除最后一个父节点按照父节点和孩子节点的数字关系建立二叉树

for (int parentIndex = 0; parentIndex array.length / 2 - 1; parentIndex++) {

nodeList.get(parentIndex).leftChild = nodeList.get(parentIndex * 2 + 1);

nodeList.get(parentIndex).rightChild = nodeList.get(parentIndex * 2 + 2);

}

// 最后一个父节点

int lastParentIndex = array.length / 2 - 1;

// 左孩子

nodeList.get(lastParentIndex).leftChild = nodeList.get(lastParentIndex * 2 + 1);

// 如果为奇数，建立右孩子

if (array.length % 2 == 1) {

nodeList.get(lastParentIndex).rightChild = nodeList.get(lastParentIndex * 2 + 2);

}

}

// 前序遍历

public static void preOrderTraverse(Node node) {

if (node == null) {

return;

}

System.out.print(node.data + " ");

preOrderTraverse(node.leftChild);

preOrderTraverse(node.rightChild);

}

// 中序遍历

public static void inOrderTraverse(Node node) {

if (node == null) {

return;

}

inOrderTraverse(node.leftChild);

System.out.print(node.data + " ");

inOrderTraverse(node.rightChild);

}

// 后序遍历

public static void postOrderTraverse(Node node) {

if (node == null) {

return;

}

postOrderTraverse(node.leftChild);

postOrderTraverse(node.rightChild);

System.out.print(node.data + " ");

}

public static void main(String[] args) {

Bintrees binTree = new Bintrees();

binTree.createBintree();

Node root = nodeList.get(0);

System.out.println("前序遍历：");

preOrderTraverse(root);

System.out.println();

System.out.println("中序遍历：");

inOrderTraverse(root);

System.out.println();

System.out.println("后序遍历：");

postOrderTraverse(root);

}

}

写一个java层次遍历二叉树,简单点就可以,我要的是代码,不是纯文字说明

public class BinaryNode {

Object element;

BinaryNode left;

BinaryNode right;

}

import java.util.*;

public class Queue {

protected LinkedList list;

// Postcondition: this Queue object has been initialized.

public Queue() {

list = new LinkedList();

} // default constructor

// Postcondition: the number of elements in this Queue object has been

// returned.

public int size() {

return list.size();

} // method size

// Postcondition: true has been returned if this Queue object has no

// elements. Otherwise, false has been returned.

public boolean isEmpty() {

return list.isEmpty();

} // method isEmpty

// Postconditon: A copy of element has been inserted at the back of this

// Queue object. The averageTime (n) is constant and

// worstTime (n) is O (n).

public void enqueue(Object element) {

list.addLast(element);

} // method enqueue

// Precondition: this Queue object is not empty. Otherwise,

// NoSuchElementException will be thrown.

// Postcondition: The element that was at the front of this Queue object -

// just before this method was called -- has been removed

// from this Queue object and returned.

public Object dequeue() {

return list.removeFirst();

} // method dequeue

// Precondition: this Queue object is not empty. Otherwise,

// NoSuchElementException will be thrown.

// Postcondition: the element at index 0 in this Queue object has been

// returned.

public Object front() {

return list.getFirst();

} // method front

} // Queue class

import java.io.IOException;

public class BinaryTree {

BinaryNode root;

public BinaryTree() {

super();

// TODO 自动生成构造函数存根

root=this.createPre();

}

public BinaryNode createPre()

//按照先序遍历的输入方法，建立二叉树

{

BinaryNode t=null;

char ch;

try {

ch = (char)System.in.read();

if(ch==' ')

t=null;

else

{

t=new BinaryNode();

t.element=(Object)ch;

t.left=createPre();

t.right=createPre();

}

} catch (IOException e) {

// TODO 自动生成 catch 块

e.printStackTrace();

}

return t;

}

public void inOrder()

{

this.inOrder(root);

}

public void inOrder(BinaryNode t)

//中序遍历二叉树

{

if(t!=null)

{

inOrder(t.left);

System.out.print(t.element);

inOrder(t.right);

}

}

public void postOrder()

{

this.postOrder(root);

}

public void postOrder(BinaryNode t)

//后序遍历二叉树

{

if(t!=null)

{

postOrder(t.left);

System.out.print(t.element);

postOrder(t.right);

}

}

public void preOrder()

{

this.preOrder(root);

}

public void preOrder(BinaryNode t)

//前序遍历二叉树

{

if(t!=null)

{

System.out.print(t.element);

preOrder(t.left);

preOrder(t.right);

}

}

public void breadthFirst()

{

Queue treeQueue=new Queue();

BinaryNode p;

if(root!=null)

treeQueue.enqueue(root);

while(!treeQueue.isEmpty())

{

System.out.print(((BinaryNode)(treeQueue.front())).element);

p=(BinaryNode)treeQueue.dequeue();

if(p.left!=null)

treeQueue.enqueue(p.left);

if(p.right!=null)

treeQueue.enqueue(p.right);

}

}

}

public class BinaryTreeTest {

/**

* @param args

*/

public static void main(String[] args) {

// TODO 自动生成方法存根

BinaryTree tree = new BinaryTree();

System.out.println("先序遍历：");

tree.preOrder();

System.out.println();

System.out.println("中序遍历：");

tree.inOrder();

System.out.println();

System.out.println("后序遍历：");

tree.postOrder();

System.out.println();

System.out.println("层次遍历：");

tree.breadthFirst();

System.out.println();

}

}

如何在java中创建二叉排序树

试试这些代码:

private BinaryNodeT remove(T x,BinaryNodeT p,BinaryNodeT parent){

if(p==null){

return null;

}

if(x.compareTo(p.data)0){

return remove(x,p.left,p);

}

if(x.compareTo(p.data)0){

return remove(x,p.right,p);

}

if(p.left!=nullp.right!=null){

BinaryNodeT insucc=p.right;

while(insucc.left!=null){

insucc=insucc.left;

}

p.data=insucc.data;

return remove(p.data,p.right,p);

}

if(parent==null){

if(p.left!=null){

root=p.left;

}

else root=p.right;

return p;

}

if(p==parent.left){

if(p.left!=null){

parent.left=p.left;

}

else{

parent.left=p.right;

}

}

else{

if(p.left!=null){

parent.right=p.left;

}

else{

parent.right=p.right;

}

}

return p;

}

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