Question

我遇到了遍历树并输出如下的问题。

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

我知道这与preorder遍历非常相似，但是当节点分裂成左右节点时，preorder不再输出父节点。有什么建议吗？

这是我的代码，但我仍然坚持打印出来。

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);
        }
    }
}

Answer 1

您想要做的是使用深度优先算法遍历树。

你会在互联网上找到很多例子。取决于你如何制作你的树。如果你已经加载了一个对象树，你可以使用递归算法从左到右传递每个孩子或使用访问者模式。

首先，看看http://en.wikipedia.org/wiki/Tree_traversal#Depth-first

Java中的遍历树

1 个答案: