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用java怎么构造一个二叉树呢？
java构造二叉树，可以通过链表来构造，如下代码：
public class BinTree {
public final static int MAX=40;
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
private Object data; //数据元数
private BinTree left,right; //指向左,右孩子结点的链
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=0;rear=1;
while(frontrear)
{
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 == nullright == 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 = leftHeightrightHeight?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(leftLevel0rightLevel0)
return -1;
levelInTree = leftLevelrightLevel?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()
{
if(this == null)
return;
//若本节点是叶节点，则将其移走
if(left==nullright == 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...
String 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());
}
hashmap链表大于多少后成为红黑树
java8不是用红黑树来管理hashmap，而是在hash值相同的情况下（且重复数量大于8），用红黑树来管理数据。 红黑树相当于排序数据。可以自动的使用二分法进行定位。性能较高。
一般情况下，hash值做的比较好的话基本上用不到红黑树。
java中没有指针，怎么构建一个链表树
java中要用到链表结构的话有LinkedList、LinkedHashSet和LinkedHashMap等集合可供使用，这些集合的底层都是由链表实现的