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=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()
{
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...
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())
}
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=0i<input.lengthi++)
LinkedList.add(new Node(input.charAt(i)))
然后为各个节点设置左右子树:
for(int i=0i<input.lengthi++){
((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。
