我在遍历树时遇到了问题，并且输出如下。
如果树是这样的图像。[http://www.java-forums.org/attachments/advanced-java/3355d1332821031t-traversing-binary-tree-root-each-branch-binarytree.png][1]
输出：
- A、A1、A2、B1、B2
- A、A1、B1、A2、B2
- A、A1、B1、B2、A2
- A, B1, A1, A2, B2
- A、B1、A1、B2、A2
- A、B1、B2、A1、A2
我知道这和前序遍历非常相似，但是前序在节点分裂为左右节点时不会再次输出父节点。有什么建议么？
这是我的代码，但我卡在打印输出上。
public class BinaryTreeTest {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
int countA = 0;
int countB = 0;
ArrayList listA = new ArrayList();
ArrayList listB = new ArrayList();
listA.add("A1");
listA.add("A2");
listA.add("A3");
listB.add("B1");
listB.add("B2");
listB.add("B3");
//listB.add("B1");
Node root = new Node("START");
constructTree(root, countA, countB, listA, listB);
//printInOrder(root);
//printFromRoot(root);
}
public static class Node{
private Node left;
private Node right;
private String value;
public Node(String value){
this.value = value;
}
}
public static void constructTree(Node node, int countA, int countB, ArrayList listA, ArrayList listB){
if(countA < listA.size()){
if(node.left == null){
System.out.println("There is no left node. CountA is " + countA);
System.out.println("Created new node with value: " + listA.get(countA).toString() + " with parent, "
+ node.value);
System.out.println();
node.left = new Node(listA.get(countA).toString());
constructTree(node.left, countA+1, countB, listA, listB);
}else{
System.out.println("There is a left node. CountA + 1 is " + countA+1);
constructTree(node.left, countA+1, countB, listA, listB);
}
}
if(countB < listB.size()){
if(node.right == null){
System.out.println("There is no right node. CountB is " + countB);
System.out.println("Created new node with value: " + listB.get(countB).toString() + " with parent, "
+ node.value);
System.out.println();
node.right = new Node(listB.get(countB).toString());
constructTree(node.right, countA, countB+1, listA, listB);
}else{
System.out.println("There is a right node. CountB + 1 is " + countB+1);
constructTree(node.right, countA, countB+1, listA, listB);
}
}
}