Ⅰ java中的二叉樹是什麼意思
二叉樹的相關操作，包括創建，中序、先序、後序（遞歸和非遞歸），其中重點的是java在先序創建二叉樹和後序非遞歸遍歷的的實現。
Ⅱ 如何用java實現二叉樹
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 List<Node> nodeList = null;
private static class Node {
Node leftChild;
Node rightChild;
int data;
Node(int newData) {
leftChild = null;
rightChild = null;
data = newData;
}
}
// 創建二叉樹
public void createBintree() {
nodeList = new LinkedList<Node>();
// 將數組的值轉換為node
for (int nodeIndex = 0; nodeIndex < array.length; nodeIndex++) {
nodeList.add(new Node(array[nodeIndex]));
}
// 對除最後一個父節點按照父節點和孩子節點的數字關系建立二叉樹
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);
}
// 最後一個父節點
int lastParentIndex = array.length / 2 - 1;
// 左孩子
nodeList.get(lastParentIndex).leftChild = nodeList.get(lastParentIndex * 2 + 1);
// 如果為奇數，建立右孩子
if (array.length % 2 == 1) {
nodeList.get(lastParentIndex).rightChild = nodeList.get(lastParentIndex * 2 + 2);
}
}
// 前序遍歷
public static void preOrderTraverse(Node node) {
if (node == null) {
return;
}
System.out.print(node.data + " ");
preOrderTraverse(node.leftChild);
preOrderTraverse(node.rightChild);
}
// 中序遍歷
public static void inOrderTraverse(Node node) {
if (node == null) {
return;
}
inOrderTraverse(node.leftChild);
System.out.print(node.data + " ");
inOrderTraverse(node.rightChild);
}
// 後序遍歷
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("前序遍歷：");
preOrderTraverse(root);
System.out.println();
System.out.println("中序遍歷：");
inOrderTraverse(root);
System.out.println();
System.out.println("後序遍歷：");
postOrderTraverse(root);
}
}
輸出結果：
前序遍歷：
1 2 4 8 9 5 3 6 7
中序遍歷：
8 4 9 2 5 1 6 3 7
後序遍歷：
8 9 4 5 2 6 7 3 1
Ⅲ 關於java中的二叉樹
感覺你的問題好奇怪
你是怎麼知道node是指向誰的？
既然你知道node指向誰了，那它就可以訪問它裡面的成員變數，leftChild不是它的成員變數嗎？它只要知道它的leftChild存在就行了，leftChild指向誰它不管
////////////////////////////
你的意思是再往下一層？
Ⅳ 用java怎麼構造一個二叉樹呢
二叉樹的相關操作，包括創建，中序、先序、後序（遞歸和非遞歸），其中重點的是java在先序創建二叉樹和後序非遞歸遍歷的的實現。
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<T> {
private Node<T> root;
public Tree() {
}
public Tree(Node<T> root) {
this.root = root;
}
//創建二叉樹
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);
}
//先序遍歷創建二叉樹
private Node<T> createTree(Node<T> node,Scanner scn) {
String temp = scn.next();
if (temp.trim().equals("#")) {
return null;
} else {
node = new Node<T>((T)temp);
node.setLeft(createTree(node.getLeft(), scn));
node.setRight(createTree(node.getRight(), scn));
return node;
}
}
//中序遍歷(遞歸)
public void inOrderTraverse() {
inOrderTraverse(root);
}
public void inOrderTraverse(Node<T> node) {
if (node != null) {
inOrderTraverse(node.getLeft());
System.out.println(node.getValue());
inOrderTraverse(node.getRight());
}
}
//中序遍歷(非遞歸)
public void nrInOrderTraverse() {
Stack<Node<T>> stack = new Stack<Node<T>>();
Node<T> 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();
}
}
//先序遍歷(遞歸)
public void preOrderTraverse() {
preOrderTraverse(root);
}
public void preOrderTraverse(Node<T> node) {
if (node != null) {
System.out.println(node.getValue());
preOrderTraverse(node.getLeft());
preOrderTraverse(node.getRight());
}
}
//先序遍歷（非遞歸）
public void nrPreOrderTraverse() {
Stack<Node<T>> stack = new Stack<Node<T>>();
Node<T> 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();
}
}
//後序遍歷(遞歸)
public void postOrderTraverse() {
postOrderTraverse(root);
}
public void postOrderTraverse(Node<T> node) {
if (node != null) {
postOrderTraverse(node.getLeft());
postOrderTraverse(node.getRight());
System.out.println(node.getValue());
}
}
//後續遍歷(非遞歸)
public void nrPostOrderTraverse() {
Stack<Node<T>> stack = new Stack<Node<T>>();
Node<T> node = root;
Node<T> preNode = null;//表示最近一次訪問的節點
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();
}
}
}
//按層次遍歷
public void levelTraverse() {
levelTraverse(root);
}
public void levelTraverse(Node<T> node) {
Queue<Node<T>> queue = new LinkedBlockingQueue<Node<T>>();
queue.add(node);
while (!queue.isEmpty()) {
Node<T> temp = queue.poll();
if (temp != null) {
System.out.println(temp.getValue());
queue.add(temp.getLeft());
queue.add(temp.getRight());
}
}
}
}
//樹的節點
class Node<T> {
private Node<T> left;
private Node<T> right;
private T value;
public Node() {
}
public Node(Node<T> left,Node<T> right,T value) {
this.left = left;
this.right = right;
this.value = value;
}
public Node(T value) {
this(null,null,value);
}
public Node<T> getLeft() {
return left;
}
public void setLeft(Node<T> left) {
this.left = left;
}
public Node<T> getRight() {
return right;
}
public void setRight(Node<T> right) {
this.right = right;
}
public T getValue() {
return value;
}
public void setValue(T value) {
this.value = value;
}
}
測試代碼：
package com.algorithm.tree;
public class TreeTest {
/**
* @param args
*/
public static void main(String[] args) {
Tree<Integer> tree = new Tree<Integer>();
tree.buildTree();
System.out.println("中序遍歷");
tree.inOrderTraverse();
tree.nrInOrderTraverse();
System.out.println("後續遍歷");
//tree.nrPostOrderTraverse();
tree.postOrderTraverse();
tree.nrPostOrderTraverse();
System.out.println("先序遍歷");
tree.preOrderTraverse();
tree.nrPreOrderTraverse();
//
}
}
Ⅳ Java二叉樹問題
寫好了，代碼如下，自己運行下，看下是否符合你的要求：
public class Test
{
static Node root;
static class Node
{
Node left;
Node right;
char data;
Node(char data)
{
this.data = data;
}
}
public static void main(String[] args)
{
String content = "welcomeyou";
for(int i=0;i<content.length();i++)
{
root = insert(root, content.charAt(i));
}
System.out.println("先序遍歷結果如下：");
perorder(root);
System.out.println("中序遍歷結果如下：");
inorder(root);
System.out.println("後序遍歷結果如下：");
postorder(root);
}
public static Node insert(Node node, char data)
{
if(node == null)
node = new Node(data);
else
{
if(node.data > data)
node.left = insert(node.left,data);
else
node.right = insert(node.right,data);
}
return node;
}
public static void perorder(Node node)
{
if (node == null)
return;
System.out.println(node.data);
if (node.left != null)
perorder(node.left);
if (node.right != null)
perorder(node.right);
}
public static void inorder(Node node)
{
if (node == null)
return;
if (node.left != null)
inorder(node.left);
System.out.println(node.data);
if (node.right != null)
inorder(node.right);
}
public static void postorder(Node node)
{
if (node == null)
return;
if (node.left != null)
postorder(node.left);
if (node.right != null)
postorder(node.right);
System.out.println(node.data);
}
}
Ⅵ (Java)二叉樹問題！！！！！
這個密鑰的值 是怎麼回事 自己設定的嗎？
Ⅶ java 由字元串構成的二叉樹
java構造二叉樹，可以通過鏈表來構造，如下代碼：
public class BinTree {public final static int MAX=40;BinTree []elements = new BinTree[MAX];//層次遍歷時保存各個節點 int front;//層次遍歷時隊首 int rear;//層次遍歷時隊尾private Object data; //數據元數private BinTree left,right; //指向左,右孩子結點的鏈public BinTree(){}public BinTree(Object data){ //構造有值結點 this.data = data; left = right = null;}public BinTree(Object data,BinTree left,BinTree right){ //構造有值結點 this.data = data; this.left = left; this.right = right;}public String toString(){ return data.toString();}//前序遍歷二叉樹public static void preOrder(BinTree parent){ if(parent == null) return; System.out.print(parent.data+" "); preOrder(parent.left); preOrder(parent.right);}//中序遍歷二叉樹public void inOrder(BinTree parent){ if(parent == null) return; inOrder(parent.left); System.out.print(parent.data+" "); inOrder(parent.right);}//後序遍歷二叉樹public void postOrder(BinTree parent){ if(parent == null) return; postOrder(parent.left); postOrder(parent.right); System.out.print(parent.data+" ");}// 層次遍歷二叉樹 public void LayerOrder(BinTree parent){ elements[0]=parent; front=0;rear=1; while(front<rear) { try { if(elements[front].data!=null) { System.out.print(elements[front].data + " "); if(elements[front].left!=null) elements[rear++]=elements[front].left; if(elements[front].right!=null) elements[rear++]=elements[front].right; front++; } }catch(Exception e){break;} }}//返回樹的葉節點個數public int leaves(){ if(this == null) return 0; if(left == null&&right == null) return 1; return (left == null ? 0 : left.leaves())+(right == null ? 0 : right.leaves());}//結果返回樹的高度public int height(){ int heightOfTree; if(this == null) return -1; int leftHeight = (left == null ? 0 : left.height()); int rightHeight = (right == null ? 0 : right.height()); heightOfTree = leftHeight<rightHeight?rightHeight:leftHeight; return 1 + heightOfTree;}//如果對象不在樹中,結果返回-1;否則結果返回該對象在樹中所處的層次,規定根節點為第一層public int level(Object object){ int levelInTree; if(this == null) return -1; if(object == data) return 1;//規定根節點為第一層 int leftLevel = (left == null?-1:left.level(object)); int rightLevel = (right == null?-1:right.level(object)); if(leftLevel<0&&rightLevel<0) return -1; levelInTree = leftLevel<rightLevel?rightLevel:leftLevel; return 1+levelInTree; }//將樹中的每個節點的孩子對換位置public void reflect(){ if(this == null) return; if(left != null) left.reflect(); if(right != null) right.reflect(); BinTree temp = left; left = right; right = temp;}// 將樹中的所有節點移走,並輸出移走的節點public void defoliate(){ if(this == null) return; //若本節點是葉節點，則將其移走 if(left==null&&right == null) { System.out.print(this + " "); data = null; return; } //移走左子樹若其存在 if(left!=null){ left.defoliate(); left = null; } //移走本節點，放在中間表示中跟移走... String innerNode += this + " "; data = null; //移走右子樹若其存在 if(right!=null){ right.defoliate(); right = null; }} /*** @param args*/public static void main(String[] args) { // TODO Auto-generated method stub BinTree e = new BinTree("E"); BinTree g = new BinTree("G"); BinTree h = new BinTree("H"); BinTree i = new BinTree("I"); BinTree d = new BinTree("D",null,g); BinTree f = new BinTree("F",h,i); BinTree b = new BinTree("B",d,e); BinTree c = new BinTree("C",f,null); BinTree tree = new BinTree("A",b,c); System.out.println("前序遍歷二叉樹結果: "); tree.preOrder(tree); System.out.println(); System.out.println("中序遍歷二叉樹結果: "); tree.inOrder(tree); System.out.println(); System.out.println("後序遍歷二叉樹結果: "); tree.postOrder(tree); System.out.println(); System.out.println("層次遍歷二叉樹結果: "); tree.LayerOrder(tree); System.out.println(); System.out.println("F所在的層次: "+tree.level("F")); System.out.println("這棵二叉樹的高度: "+tree.height()); System.out.println("--------------------------------------"); tree.reflect(); System.out.println("交換每個節點的孩子節點後......"); System.out.println("前序遍歷二叉樹結果: "); tree.preOrder(tree); System.out.println(); System.out.println("中序遍歷二叉樹結果: "); tree.inOrder(tree); System.out.println(); System.out.println("後序遍歷二叉樹結果: "); tree.postOrder(tree); System.out.println(); System.out.println("層次遍歷二叉樹結果: "); tree.LayerOrder(tree); System.out.println(); System.out.println("F所在的層次: "+tree.level("F")); System.out.println("這棵二叉樹的高度: "+tree.height());
Ⅷ java 構建二叉樹
首先我想問為什麼要用LinkedList 來建立二叉樹呢? LinkedList 是線性表,
樹是樹形的, 似乎不太合適。
其實也可以用數組完成，而且效率更高.
關鍵是我覺得你這個輸入本身就是一個二叉樹啊，
String input = "ABCDE F G";
節點編號從0到8. 層次遍歷的話:
對於節點i.
leftChild = input.charAt(2*i+1); //做子樹
rightChild = input.charAt(2*i+2);//右子樹
如果你要將帶有節點信息的樹存到LinkedList裡面, 先建立一個節點類:
class Node{
public char cValue;
public Node leftChild;
public Node rightChild;
public Node(v){
this.cValue = v;
}
}
然後遍歷input,建立各個節點對象.
LinkedList tree = new LinkedList();
for(int i=0;i< input.length;i++)
LinkedList.add(new Node(input.charAt(i)));
然後為各個節點設置左右子樹:
for(int i=0;i<input.length;i++){
((Node)tree.get(i)).leftChild = (Node)tree.get(2*i+1);
((Node)tree.get(i)).rightChild = (Node)tree.get(2*i+2);
}
這樣LinkedList 就存儲了整個二叉樹. 而第0個元素就是樹根，思路大體是這樣吧。
Ⅸ 用java實現二叉樹
我有很多個(假設10萬個)數據要保存起來，以後還需要從保存的這些數據中檢索是否存在某
個數據，（我想說出二叉樹的好處，該怎麼說呢？那就是說別人的缺點），假如存在數組中，
那麼，碰巧要找的數字位於99999那個地方，那查找的速度將很慢，因為要從第1個依次往
後取，取出來後進行比較。平衡二叉樹（構建平衡二叉樹需要先排序，我們這里就不作考慮
了）可以很好地解決這個問題，但二叉樹的遍歷（前序，中序，後序）效率要比數組低很多，
public class Node {
public int value;
public Node left;
public Node right;
public void store(intvalue)
right.value=value;
}
else
{
right.store(value);
}
}
}
public boolean find(intvalue)
{
System.out.println("happen" +this.value);
if(value ==this.value)
{
return true;
}
else if(value>this.value)
{
if(right ==null)returnfalse;
return right.find(value);
}else
{
if(left ==null)returnfalse;
return left.find(value);
}
}
public void preList()
{
System.out.print(this.value+ ",");
if(left!=null)left.preList();
if(right!=null) right.preList();
}
public void middleList()
{
if(left!=null)left.preList();
System.out.print(this.value+ ",");
if(right!=null)right.preList();
}
public void afterList()
{
if(left!=null)left.preList();
if(right!=null)right.preList();
System.out.print(this.value+ ",");
}
public static voidmain(String [] args)
{
int [] data =new int[20];
for(inti=0;i<data.length;i++)
{
data[i] = (int)(Math.random()*100)+ 1;
System.out.print(data[i] +",");
}
System.out.println();
Node root = new Node();
root.value = data[0];
for(inti=1;i<data.length;i++)
{
root.store(data[i]);
}
root.find(data[19]);
root.preList();
System.out.println();
root.middleList();
System.out.println();
root.afterList();
}
}