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java二叉排序

發布時間: 2021-03-27 12:35:14

Ⅰ java建立二叉排序樹，為什麼最後列印null，沒有建成呢

classBSTreeNode{
intdata;
BSTreeNodeleftchild;
BSTreeNoderightchild;
}
classBSTree{
publicBSTreeNodeinsert(BSTreeNodenode,intkey){
if(node==null){
node=newBSTreeNode();
node.data=key;
node.leftchild=node.rightchild=null;
}elseif(key<node.data)
node.leftchild=insert(node.leftchild,key);
elseif(key>node.data)
node.rightchild=insert(node.rightchild,key);
returnnode;
}
publicBSTreeNodecreateBSTree(){
BSTreeNodenode=null;
int[]data={20,10,30,8,12,25,40};
for(inti=0;i<data.length;i++)
node=insert(node,data[i]);
returnnode;
}
}
publicclassBSTree1{
publicstaticvoidmain(String[]args){
BSTreetree=newBSTree();
BSTreeNoderoot=tree.createBSTree();
System.out.println(root);
}
}

Ⅱ 二叉排序樹的插入與查找求Java大神幫忙


//二叉樹結構類Btree.java
publicclassBtree{
privateintdata;
privateBtreeleft;
privateBtreeright;
publicvoidsetData(intdata){
this.data=data;
}
publicintgetData(){
returndata;
}
publicvoidsetLeft(Btreebtree){
this.left=btree;
}
publicvoidsetRight(Btreebtree){
this.right=btree;
}
publicBtreegetLeft(){
returnleft;
}
publicBtreegetRight(){
returnright;
}
publicBtree(){
super();
}
}
//工具類，二叉樹創建，查找，遍歷，排序。都在這了Tools.java
publicclassTools{
publicBtreecreate(intdata){
Btreebtree=newBtree();
btree.setData(data);
returnbtree;
}
publicvoidadd(Btreebtree,intdata){
if(data<btree.getData()){
if(btree.getLeft()!=null){
btree=btree.getLeft();
add(btree,data);
}else{
btree.setLeft(create(data));
}
}else{
if(btree.getRight()!=null){
btree=btree.getRight();
add(btree,data);
}else{
btree.setRight(create(data));
}
}
}
//中序遍歷
publicvoidmidSerch(Btreebtree){
if(btree.getLeft()!=null)
midSerch(btree.getLeft());
System.out.print(btree.getData()+"");
if(btree.getRight()!=null)
midSerch(btree.getRight());
}
//二叉樹查找
publicvoidfind(Btreebtree,intdata){
if(btree.getData()>data)
{
if(btree.getLeft()==null)
{
System.out.println("沒有這種結果，搜索完畢");
return;
}else
find(btree.getLeft(),data);
}elseif(btree.getData()==data)
{
System.out.println("查找成功，查找到的數據是"+data);
return;
}else
{
if(btree.getRight()==null){
System.out.println("沒有這種結果，搜索完畢");
return;
}
else
find(btree.getRight(),data);
}
}
}
//主類，與注釋的自己看MidSerchBtree.java
publicclassMidSerchBtree{
publicstaticvoidmain(Stringargs[]){
Toolstools=newTools();
intdatas[]={6,4,3,7,8,9,2,1,5,8,9,12,23,45,3,7,5};
Btreebtree=tools.create(datas[0]);
for(inti=1;i<datas.length;i++){//第一個初始化插入作為根節點了
tools.add(btree,datas[i]);
}
tools.midSerch(btree);//中根遍歷小的插入左邊，大的在右邊，所以，中序遍歷一遍就是排序
tools.find(btree,56);
}
}

Ⅲ 數據結構（Java）:給出在二叉排序樹上查找鍵值為K的演算法函數

p* find(p* r, int v){if(v==p->data)return p;if(v < p->data) return find(p->lchild, v);elsereturn find(p->rchild, v);}// 大概思路就是這樣當前節點是要查找的值直接返回小於查找左子樹大於查找右子數
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
java版：
node search(node p, data key) // data是你的鍵值類型
{
if(p == null || p.key == key)
return p;
if(p.key < key)
return search(p.rchild, key);
else
return search(p.lchild, key);
}

Ⅳ 計算機C語言數據結構javaIT二叉排序樹的構造

字元串的大小是逐字元比較，比較字元的ascii碼。
排序樹，可以是左樹比根節點大，右樹比它小，或者反過來也行。這樣就是有序的，可以從根開始查找串。

Ⅳ JAVA如何將英文字母進行二叉樹排序

如果僅限於java，而且是實際應用，java里有一個叫做TreeSet的東西，是個有序的樹結構。Sring類型的英文字元可在裡面自排序。

如果是考試，應該是靠你如何實現一個類似於TreeSet的東西

Ⅵ java寫的關於二叉排序樹的刪除操作的問題

package ggg;

import java.math.BigDecimal;
import java.util.Scanner;
import java.util.Random;
import java.util.Stack;

/**
* 二叉排序樹（又稱二叉查找樹）
* （1）可以是一顆空樹
* （2）若左子樹不空，則左子樹上所有的結點的值均小於她的根節點的值
* （3）若右子樹不空，則右子樹上所有的結點的值均大於她的根節點的值
* （4）左、右子樹也分別為二叉排序樹
*
*
* 性能分析：
* 查找性能：
* 含有n個結點的二叉排序樹的平均查找長度和樹的形態有關，
* （最壞情況）當先後插入的關鍵字有序時，構成的二叉排序樹蛻變為單枝樹。查找性能為O(n)
* （最好情況）二叉排序樹的形態和折半查找的判定樹相同，其平均查找長度和log2(n)成正比
*
*
* 插入、刪除性能：
* 插入、刪除操作間復雜度都O(log(n))級的，
* 即經過O(log(n))時間搜索到了需插入刪除節點位置和刪除節點的位置
* 經O(1)級的時間直接插入和刪除
* 與順序表相比，比序順序表插入刪除O(n)(查找時間O(log(n))移動節點時間O(n))要快
* 與無序順序表插入時間O(1)，刪除時間O(n)相比，因為是有序的，所查找速度要快很多
*
*
*
* 作者：小菜鳥
* 創建時間：2014-08-17
*
*/

public class BinarySortTree {

private Node root = null;

/**查找二叉排序樹中是否有key值*/
public boolean searchBST(int key){
Node current = root;
while(current != null){
if(key == current.getValue())
return true;
else if(key < current.getValue())
current = current.getLeft();
else
current = current.getRight();
}
return false;
}

/**向二叉排序樹中插入結點*/
public void insertBST(int key){
Node p = root;
/**記錄查找結點的前一個結點*/
Node prev = null;
/**一直查找下去，直到到達滿足條件的結點位置*/
while(p != null){
prev = p;
if(key < p.getValue())
p = p.getLeft();
else if(key > p.getValue())
p = p.getRight();
else
return;
}
/**prve是要安放結點的父節點，根據結點值得大小，放在相應的位置*/
if(root == null)
root = new Node(key);
else if(key < prev.getValue())
prev.setLeft(new Node(key));
else prev.setRight(new Node(key));
}

/**
* 刪除二叉排序樹中的結點
* 分為三種情況：（刪除結點為*p ，其父結點為*f）
* （1）要刪除的*p結點是葉子結點，只需要修改它的雙親結點的指針為空
* （2）若*p只有左子樹或者只有右子樹，直接讓左子樹/右子樹代替*p
* （3）若*p既有左子樹，又有右子樹
* 用p左子樹中最大的那個值（即最右端S）代替P，刪除s，重接其左子樹
* */
public void deleteBST(int key){
deleteBST(root, key);
}
private boolean deleteBST(Node node, int key) {
if(node == null) return false;
else{
if(key == node.getValue()){
return delete(node);
}
else if(key < node.getValue()){
return deleteBST(node.getLeft(), key);
}
else{
return deleteBST(node.getRight(), key);
}
}
}

private boolean delete(Node node) {
Node temp = null;
/**右子樹空，只需要重接它的左子樹
* 如果是葉子結點，在這里也把葉子結點刪除了
* */
if(node.getRight() == null){
temp = node;
node = node.getLeft();
}
/**左子樹空，重接它的右子樹*/
else if(node.getLeft() == null){
temp = node;
node = node.getRight();
}
/**左右子樹均不為空*/
else{
temp = node;
Node s = node;
/**轉向左子樹，然後向右走到「盡頭」*/
s = s.getLeft();
while(s.getRight() != null){
temp = s;
s = s.getRight();
}
node.setValue(s.getValue());
if(temp != node){
temp.setRight(s.getLeft());
}
else{
temp.setLeft(s.getLeft());
}
}
return true;
}

/**中序非遞歸遍歷二叉樹
* 獲得有序序列
* */
public void nrInOrderTraverse(){
Stack<Node> stack = new Stack<Node>();
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();
}
}

public static void main(String[] args){
BinarySortTree bst = new BinarySortTree();
/**構建的二叉樹沒有相同元素*/
int[] num = {4,7,2,1,10,6,9,3,8,11,2, 0, -2};
for(int i = 0; i < num.length; i++){
bst.insertBST(num[i]);
}
bst.nrInOrderTraverse();
System.out.println(bst.searchBST(10));
bst.deleteBST(2);
bst.nrInOrderTraverse();
}

/**二叉樹的結點定義*/
public class Node{
private int value;
private Node left;
private Node right;

public Node(){
}
public Node(Node left, Node right, int value){
this.left = left;
this.right = right;
this.value = value;
}
public Node(int value){
this(null, null, value);
}

public Node getLeft(){
return this.left;
}
public void setLeft(Node left){
this.left = left;
}
public Node getRight(){
return this.right;
}
public void setRight(Node right){
this.right = right;
}
public int getValue(){
return this.value;
}
public void setValue(int value){
this.value = value;
}
}

}

Ⅶ java二叉排序樹問題

給分!!!!!!:....還行不難寫....其實我很懷念學C時寫樹......
class TreeNode 數據結構
class SortTree 實現樹的建立和中序遍歷

public class TreeNode
{
public TreeNode left;
public TreeNode right;
public Integer value;

TreeNode ( Integer value )
{
this.value = value;
System.out.println(value);
}
}

public class SortTree
{
SortTree(Integer...integers)
{
createSortTree(integers);
}

private TreeNode root;

private void createSortTree(Integer[] integers)
{
for(Integer nodeValue:integers)
{
if(root == null)
{
System.out.println("Create Root");
root = new TreeNode(nodeValue);
}
else if(root.value > nodeValue)
{
insertLeft(root , nodeValue);
}
else
{
insertRight(root , nodeValue);
}
}
}

private void insertLeft(TreeNode root2, Integer nodeValue)
{
if(root2.left == null)
{
System.out.println("Create Left");
root2.left = new TreeNode(nodeValue);
}
else if(root2.left.value > nodeValue)
{
insertLeft(root2.left,nodeValue);
}
else
{
insertRight(root2.left,nodeValue);
}

}

private void insertRight(TreeNode root2, Integer nodeValue)
{
if(root2.right == null)
{
System.out.println("Create Right");
root2.right = new TreeNode(nodeValue);
}
else if(root2.right.value > nodeValue)
{
insertLeft(root2.right,nodeValue);
}
else
{
insertRight(root2.right,nodeValue);
}
}

private void middleTraversing(TreeNode root)
{
if(root.left !=null) /** 第一步訪問左子樹 */
middleTraversing(root.left);
System.out.println(root.value); /** 第二步訪問根結點 */
if(root.right !=null)
middleTraversing(root.right); /** 第三步訪問右子樹 */
}

public TreeNode getRoot()
{
return root;
}

public static void main (String[] args)
{
SortTree tree = new SortTree(new Integer(5),new Integer(6),new Integer(4),new Integer(3),new Integer(7));
System.out.println("MiddleTraversing:");
tree.middleTraversing(tree.getRoot());
}
}

Ⅷ java二叉樹排序問題

importjava.util.TreeSet;
publicclassStudentTest{
	publicstaticvoidmain(String[]args){
		TreeSet<Student>ts=newTreeSet<>();
		for(;ts.size()<10;){
			Stringname="同學"+(char)((Math.round(Math.random()*26+65)));
			intid=(int)(Math.round(Math.random()*80+10));
			floatfl=(int)Math.floor(Math.random()*50+40);
			ts.add(newStudent(name,fl,id));
		}
		for(Studenta:ts){
			System.out.println(a);
		}
	}
}
<Student>{
	privateStringname;
	privateFloathp;
	privateintid;
	publicStudent(Stringname,floathp,intid){
		this.name=name;
		this.hp=hp;
		this.id=id;
	}
	publicStringtoString(){
		return"(name:"+name+"id:"+id+"hp:"+hp+")";
	}
	publicintgetId(){
		returnid;
	}
	publicintcompareTo(Studentstu){
		returnInteger.compare(this.id,stu.getId());
	}
}

Ⅸ java二叉排序樹，已有代碼，如何調通輸出

你好，很高興回答你的問題。

目前已經有了二叉樹以及二叉樹節點的類。

需要一個內main方法，在其中創建節點（容通過節點類的構造方法），構建樹（通過樹的構造方法以及insert方法）。可以執行查詢的方法以及展示的方法。

如果有幫助到你，請點擊點贊。

Ⅹ 二叉排序樹（BST） Java實現

public class Node<E> {
int key; // data used as key value
E data; // other data
Node<E> leftChild; // this node's left child
Node<E> 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 Tree<E> {
private Node<E> root;

public Tree(Node<E> root) {
if (root == null) {
throw new NullPointerException("root can't be null");
}
this.root = root;
}

public Tree(int key, E o) {
this(new Node<E>(key, o));
}

public Node<E> getRoot() {
return root;
}

/**
* find a node by its key
*
* @param key
* @return
*/
public Node<E> find(int key) {
Node<E> 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) {
Node<E> aNode = new Node<E>(key, o);
if (root == null) {
this.root = aNode;
return;
}

Node<E> current = root;
Node<E> 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) {
Node<E> current = root;
Node<E> 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
Node<E> 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 Node<E> getSuccessor(Node<E> delNode) {
Node<E> successorParent = delNode;
Node<E> successor = delNode;
Node<E> 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(Node<E> aNode) {
if (aNode != null) {
inOrder(aNode.leftChild);
aNode.displayNode();
inOrder(aNode.rightChild);
}
}

public void preOrder(Node<E> aNode) {
if (aNode != null) {
aNode.displayNode();
inOrder(aNode.leftChild);
inOrder(aNode.rightChild);
}
}

public void backOrder(Node<E> aNode) {
if (aNode != null) {
inOrder(aNode.leftChild);
inOrder(aNode.rightChild);
aNode.displayNode();
}
}

public Node<E> minimum() {
Node<E> current = this.root;
Node<E> result = null;
while (current != null) {
result = current;
current = current.leftChild;
}
return result;
}

public Node<E> maximum() {
Node<E> current = this.root;
Node<E> result = null;
while (current != null) {
result = current;
current = current.rightChild;
}
return result;
}
}

以前的代碼, 記得沒寫完, 好像就是BST

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