个数据，（我想说出二叉树的好处，该怎么说呢？那就是说别人的缺点），假如存在数组中，
那么，碰巧要找的数字位于99999那个地方，那查找的速度将很慢，因为要从第1个依次往
后取，取出来后进行比较。平衡二叉树（构建平衡二叉树需要先排序，我们这里就不作考虑
了）可以很好地解决这个问题，但二叉树的遍历（前序，中序，后序）效率要比数组低很多，
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=0i<data.lengthi++)
{
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=1i<data.lengthi++)
{
root.store(data[i])
}
root.find(data[19])
root.preList()
System.out.println()
root.middleList()
System.out.println()
root.afterList()
}
}
//******************************************************************************************************////*****本程序包括简单的二叉树类的实现和前序,中序,后序,层次遍历二叉树算法,*******//
//******以及确定二叉树的高度,制定对象在树中的所处层次以及将树中的左右***********//
//******孩子节点对换位置,返回叶子节点个数删除叶子节点,并输出所删除的叶子节点**//
//*******************************CopyRight By phoenix*******************************************//
//************************************Jan 12,2008*************************************************//
//****************************************************************************************************//
public class BinTree {
public final static int MAX=40
private Object data//数据元数
private BinTree left,right//指向左,右孩子结点的链
BinTree []elements = new BinTree[MAX]//层次遍历时保存各个节点
int front//层次遍历时队首
int rear//层次遍历时队尾
public BinTree()
{
}
public BinTree(Object data)
{ //构造有值结点
this.data = data
left = right = null
}
public BinTree(Object data,BinTree left,BinTree right)
{ //构造有值结点
this.data = data
this.left = left
this.right = right
}
public String toString()
{
return data.toString()
}//前序遍历二叉树
public static void preOrder(BinTree parent){
if(parent == null)
return
System.out.print(parent.data+" ")
preOrder(parent.left)
preOrder(parent.right)
}//中序遍历二叉树
public void inOrder(BinTree parent){
if(parent == null)
return
inOrder(parent.left)
System.out.print(parent.data+" ")
inOrder(parent.right)
}//后序遍历二叉树
public void postOrder(BinTree parent){
if(parent == null)
return
postOrder(parent.left)
postOrder(parent.right)
System.out.print(parent.data+" ")
}// 层次遍历二叉树
public void LayerOrder(BinTree parent)
{
elements[0]=parent
front=0rear=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}
}
}//返回树的叶节点个数
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())
}//结果返回树的高度
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
}
//如果对象不在树中,结果返回-1否则结果返回该对象在树中所处的层次,规定根节点为第一层
public int level(Object object)
{
int levelInTree
if(this == null)
return -1
if(object == data)
return 1//规定根节点为第一层
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
}
//将树中的每个节点的孩子对换位置
public void reflect()
{
if(this == null)
return
if(left != null)
left.reflect()
if(right != null)
right.reflect()
BinTree temp = left
left = right
right = temp
}// 将树中的所有节点移走,并输出移走的节点
public void defoliate()
{
String innerNode = ""
if(this == null)
return
//若本节点是叶节点，则将其移走
if(left==null&&right == null)
{
System.out.print(this + " ")
data = null
return
}
//移走左子树若其存在
if(left!=null){
left.defoliate()
left = null
}
//移走本节点，放在中间表示中跟移走...
innerNode += this + " "
data = null
//移走右子树若其存在
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("前序遍历二叉树结果: ")
tree.preOrder(tree)
System.out.println()
System.out.println("中序遍历二叉树结果: ")
tree.inOrder(tree)
System.out.println()
System.out.println("后序遍历二叉树结果: ")
tree.postOrder(tree)
System.out.println()
System.out.println("层次遍历二叉树结果: ")
tree.LayerOrder(tree)
System.out.println()
System.out.println("F所在的层次: "+tree.level("F"))
System.out.println("这棵二叉树的高度: "+tree.height())
System.out.println("--------------------------------------")
tree.reflect()
System.out.println("交换每个节点的孩子节点后......")
System.out.println("前序遍历二叉树结果: ")
tree.preOrder(tree)
System.out.println()
System.out.println("中序遍历二叉树结果: ")
tree.inOrder(tree)
System.out.println()
System.out.println("后序遍历二叉树结果: ")
tree.postOrder(tree)
System.out.println()
System.out.println("层次遍历二叉树结果: ")
tree.LayerOrder(tree)
System.out.println()
System.out.println("F所在的层次: "+tree.level("F"))
System.out.println("这棵二叉树的高度: "+tree.height())
}
