java一个关于二叉树的简单编程题

定义一个结点类：

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public class Node {

private int value;

private Node leftNode;

private Node rightNode;

public Node getRightNode() {

return rightNode;

}

public void setRightNode(Node rightNode) {

this.rightNode = rightNode;

}

public int getValue() {

return value;

}

public void setValue(int value) {

this.value = value;

}

public Node getLeftNode() {

return leftNode;

}

public void setLeftNode(Node leftNode) {

this.leftNode = leftNode;

}

}

初始化结点树：

public void initNodeTree()

{

int nodeNumber;

HashMapString, Integer map = new HashMapString, Integer();

Node nodeTree = new Node();

Scanner reader = new Scanner(System.in);

nodeNumber = reader.nextInt();

for(int i = 0; i nodeNumber; i++) {

int value = reader.nextInt();

String str = reader.next();

map.put(str, value);

}

if (map.containsKey("#")) {

int value = map.get("#");

nodeTree.setValue(value);

setChildNode(map, value, nodeTree);

}

preTraversal(nodeTree);

}

private void setChildNode(HashMapString, Integer map, int nodeValue, Node parentNode) {

int value = 0;

if (map.containsKey("L" + nodeValue)) {

value = map.get("L" + nodeValue);

Node leftNode = new Node();

leftNode.setValue(value);

parentNode.setLeftNode(leftNode);

setChildNode(map, value, leftNode);

}

if (map.containsKey("R" + nodeValue)) {

value = map.get("R" + nodeValue);

Node rightNode = new Node();

rightNode.setValue(value);

parentNode.setRightNode(rightNode);

setChildNode(map, value, rightNode);

}

}

前序遍历该结点树：

public void preTraversal(Node nodeTree) {

if (nodeTree != null) {

System.out.print(nodeTree.getValue() + "\t");

preTraversal(nodeTree.getLeftNode());

preTraversal(nodeTree.getRightNode());

}

}

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);

}

}

java 构建二叉树

首先我想问为什么要用LinkedList 来建立二叉树呢? LinkedList 是线性表,

树是树形的, 似乎不太合适。

其实也可以用数组完成，而且效率更高.

关键是我觉得你这个输入本身就是一个二叉树啊，

String input = "ABCDE F G";

节点编号从0到8. 层次遍历的话:

对于节点i.

leftChild = input.charAt(2*i+1); //做子树

rightChild = input.charAt(2*i+2);//右子树

如果你要将带有节点信息的树存到LinkedList里面, 先建立一个节点类:

class Node{

public char cValue;

public Node leftChild;

public Node rightChild;

public Node(v){

this.cValue = v;

}

}

然后遍历input,建立各个节点对象.

LinkedList tree = new LinkedList();

for(int i=0;i input.length;i++)

LinkedList.add(new Node(input.charAt(i)));

然后为各个节点设置左右子树:

for(int i=0;iinput.length;i++){

((Node)tree.get(i)).leftChild = (Node)tree.get(2*i+1);

((Node)tree.get(i)).rightChild = (Node)tree.get(2*i+2);

}

这样LinkedList 就存储了整个二叉树. 而第0个元素就是树根，思路大体是这样吧。

用java怎么构造一个二叉树？

二叉树的相关操作，包括创建，中序、先序、后序（递归和非递归），其中重点的是java在先序创建二叉树和后序非递归遍历的的实现。

package com.algorithm.tree;

import java.io.File;

import java.io.FileNotFoundException;

import java.util.Queue;

import java.util.Scanner;

import java.util.Stack;

import java.util.concurrent.LinkedBlockingQueue;

public class Tree {

private Node root;

public Tree() {

}

public Tree(Node root) {

this.root = root;

}

//创建二叉树

public void buildTree() {

Scanner scn = null;

try {

scn = new Scanner(new File("input.txt"));

} catch (FileNotFoundException e) {

// TODO Auto-generated catch block

e.printStackTrace();

}

root = createTree(root,scn);

}

//先序遍历创建二叉树

private Node createTree(Node node,Scanner scn) {

String temp = scn.next();

if (temp.trim().equals("#")) {

return null;

} else {

node = new Node((T)temp);

node.setLeft(createTree(node.getLeft(), scn));

node.setRight(createTree(node.getRight(), scn));

return node;

}

}

//中序遍历(递归)

public void inOrderTraverse() {

inOrderTraverse(root);

}

public void inOrderTraverse(Node node) {

if (node != null) {

inOrderTraverse(node.getLeft());

System.out.println(node.getValue());

inOrderTraverse(node.getRight());

}

}

//中序遍历(非递归)

public void nrInOrderTraverse() {

StackNode stack = new StackNode();

Node node = root;

while (node != null || !stack.isEmpty()) {

while (node != null) {

stack.push(node);

node = node.getLeft();

}

node = stack.pop();

System.out.println(node.getValue());

node = node.getRight();

}

}

//先序遍历(递归)

public void preOrderTraverse() {

preOrderTraverse(root);

}

public void preOrderTraverse(Node node) {

if (node != null) {

System.out.println(node.getValue());

preOrderTraverse(node.getLeft());

preOrderTraverse(node.getRight());

}

}

//先序遍历（非递归）

public void nrPreOrderTraverse() {

StackNode stack = new StackNode();

Node node = root;

while (node != null || !stack.isEmpty()) {

while (node != null) {

System.out.println(node.getValue());

stack.push(node);

node = node.getLeft();

}

node = stack.pop();

node = node.getRight();

}

}

//后序遍历(递归)

public void postOrderTraverse() {

postOrderTraverse(root);

}

public void postOrderTraverse(Node node) {

if (node != null) {

postOrderTraverse(node.getLeft());

postOrderTraverse(node.getRight());

System.out.println(node.getValue());

}

}

//后续遍历(非递归)

public void nrPostOrderTraverse() {

StackNode stack = new StackNode();

Node node = root;

Node preNode = null;//表示最近一次访问的节点

while (node != null || !stack.isEmpty()) {

while (node != null) {

stack.push(node);

node = node.getLeft();

}

node = stack.peek();

if (node.getRight() == null || node.getRight() == preNode) {

System.out.println(node.getValue());

node = stack.pop();

preNode = node;

node = null;

} else {

node = node.getRight();

}

}

}

//按层次遍历

public void levelTraverse() {

levelTraverse(root);

}

public void levelTraverse(Node node) {

QueueNode queue = new LinkedBlockingQueueNode();

queue.add(node);

while (!queue.isEmpty()) {

Node temp = queue.poll();

if (temp != null) {

System.out.println(temp.getValue());

queue.add(temp.getLeft());

queue.add(temp.getRight());

}

}

}

}

//树的节点

class Node {

private Node left;

private Node right;

private T value;

public Node() {

}

public Node(Node left,Node right,T value) {

this.left = left;

this.right = right;

this.value = value;

}

public Node(T value) {

this(null,null,value);

}

public Node getLeft() {

return left;

}

public void setLeft(Node left) {

this.left = left;

}

public Node getRight() {

return right;

}

public void setRight(Node right) {

this.right = right;

}

public T getValue() {

return value;

}

public void setValue(T value) {

this.value = value;

}

}

测试代码：

package com.algorithm.tree;

public class TreeTest {

/**

* @param args

*/

public static void main(String[] args) {

Tree tree = new Tree();

tree.buildTree();

System.out.println("中序遍历");

tree.inOrderTraverse();

tree.nrInOrderTraverse();

System.out.println("后续遍历");

//tree.nrPostOrderTraverse();

tree.postOrderTraverse();

tree.nrPostOrderTraverse();

System.out.println("先序遍历");

tree.preOrderTraverse();

tree.nrPreOrderTraverse();

//

}

}

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();

}

}

用JAVA写二叉树

/**

* [Tree2.java] Create on 2008-10-20 下午03:03:24

* Copyright (c) 2008 by iTrusChina.

*/

/**

* @author WangXuanmin

* @version 0.10

*/

public class Tree2Bef {

private StringBuffer bef=new StringBuffer();

//传入中序遍历和后序遍历,返回前序遍历字串

public String getBef(String mid, String beh) {

//若节点存在则向bef中添加该节点,继续查询该节点的左子树和右子树

if (root(mid, beh) != -1) {

int rootindex=root(mid, beh);

char root=mid.charAt(rootindex);

bef.append(root);

System.out.println(bef.toString());

String mleft, mright;

mleft = mid.substring(0,rootindex);

mright = mid.substring(rootindex+1);

getBef(mleft,beh);

getBef(mright,beh);

}

//所有节点查询完毕,返回前序遍历值

return bef.toString();

}

//从中序遍历中根据后序遍历查找节点索引值index

private int root(String mid, String beh) {

char[] midc = mid.toCharArray();

char[] behc = beh.toCharArray();

for (int i = behc.length-1; i -1; i--) {

for (int j = 0; j midc.length; j++) {

if (behc[i] == midc[j])

return j;

}

}

return -1;

}

public static void main(String[] args) {

Tree2Bef tree=new Tree2Bef();

String mid="84925163A7B";

String bef="894526AB731";

System.out.println(tree.getBef(mid,bef));

}

}

树结构如图:

1

|-------|

2 3

|---| |---|

4 5 6 7

|-| |-|

8 9 A B

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