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Java-二进制家谱树-Can';找不到节点_Java_Tree_Traversal_Preorder - Fatal编程技术网

Java-二进制家谱树-Can';找不到节点

Java-二进制家谱树-Can';找不到节点,java,tree,traversal,preorder,Java,Tree,Traversal,Preorder,我正在做一项作业,要求我输入并显示一棵族谱,首先将它转换为一棵二叉树——孩子在左边,兄弟姐妹在右边。 我了解树,遍历树,以及如何使用前、中、后顺序方法搜索某些节点 我已经编写了插入新节点、查找节点和打印整个树的代码,但是我的findNode方法无法正常工作。我需要它使用pre-order搜索树,并返回它要查找的节点。目前,递归方法一直到左下角(最低的子节点)和右下角(最低的同级节点),但它从不返回到我调用的原始节点,因此破坏了递归 以下是my FamilyTree和主要类的代码: public

我正在做一项作业,要求我输入并显示一棵族谱,首先将它转换为一棵二叉树——孩子在左边,兄弟姐妹在右边。 我了解树,遍历树,以及如何使用前、中、后顺序方法搜索某些节点

我已经编写了插入新节点、查找节点和打印整个树的代码,但是我的findNode方法无法正常工作。我需要它使用pre-order搜索树,并返回它要查找的节点。目前,递归方法一直到左下角(最低的子节点)和右下角(最低的同级节点),但它从不返回到我调用的原始节点,因此破坏了递归

以下是my FamilyTree和主要类的代码:

public class FamilyTree
{
Node root;

// Initialize tree
public FamilyTree()
{
    root = null;
}


// --------------------------------------------------------------
// This method inserts a new family member into the tree.
// It takes two parameters - the parent who the new node should
// be inserted under, and the name of the new member being added.
// --------------------------------------------------------------

public void insertNode(String par, String name)
{
    Node parent, current;
    Node newMember = new Node(name);

    // If tree is empty, then the new member becomes the root
    if(root == null)
        root = newMember;


    // If adding a sibling to the root, insert to root's right child
    else if(par == "")
    {
        // Check if root's sibling is empty
        if(root.rightChild == null)
            root.rightChild = newMember;


        // Traverse root's siblings until end, then insert at end
        else
        {
            current = root;

            while(current.rightChild != null)
                current = current.rightChild;

            current.rightChild = newMember;
        }
    }

    else
    {
        // Find the parent where we will insert under
        parent = findNode(par, root);

        System.out.println("parent is = " + parent);
        System.out.println("newMember is = " + newMember + "\n");

        // If that parent doesn't exist, print error msg
        if (parent == null)
            System.out.println("Parent doesn't exist");


        // If parent does exist, but has no left child,
        // then the new member becomes left child
        else if(parent.leftChild == null)   
            parent.leftChild = newMember;


        // If parent already has a left child, then traverse
        // to the end of it's left children and insert node
        else
        {
            current = parent.leftChild;

            while(current.rightChild != null)
                current = current.rightChild;               

            current.rightChild = newMember;
        }   
    }
}


// ------------------------------------------------------------
// This method recursively finds a node in the family tree,
// given the name of the node to look for, and the tree.
// It is run pre-order, and, if found, returns the node
// ------------------------------------------------------------

public Node findNode(String name, Node localTree)
{
    Node current = localTree;

    // Visit the node
    if(current.name == name)
        return current;

    // Pre-order - go left
    if(current.leftChild != null)
    {
        System.out.println("going left to " + current.leftChild);
        return findNode(name, current.leftChild);
    }

    // Pre-order - go right
    if(current.rightChild != null)
    {
        System.out.println("going right to " + current.rightChild);
        return findNode(name, current.rightChild);
    }


    return null;
}

// ------------------------------------------------------------
// This method prints the family tree, given a parent name
// and a tree to print from. It will attempt to find the parent
// node with the given name, then print the entire tree 
// (all children and grandchildren) from that point.
// ------------------------------------------------------------

public void printTree(String par, Node localTree)
{
    Node parent, current;

    // Find the parent to start printing from
    parent = findNode(par, root);
    System.out.println("parent= " + parent);

    // If parent doesn't exist, print error msg
    if (parent == null)
        System.out.println(par + " doesn't exist.");

    else
    {
        current = localTree;

        System.out.println(current);

        if(current.leftChild != null)
            printTree(par, current.leftChild);

        else if(current.rightChild != null)
            printTree(par, current.rightChild);
    }

}

public class Node
{
    Node leftChild, rightChild;
    String name;

    public Node(String n)
    {
        leftChild = null;
        rightChild = null;
        name = n;
    }

    public String toString()
    {
        return name;
    }
}
}


}

问题在于您从搜索中过早返回的

public Node findNode(String name, Node localTree)
{
...
    // Pre-order - go left
    if(current.leftChild != null)
    {
        System.out.println("going left to " + current.leftChild);
        return findNode(name, current.leftChild); // <===== HERE!
    }
...
}

另外,在Java中,永远不要使用
==
来比较字符串。它不会正常工作。对于字符串,始终使用
String.equals(…)
。例如,请参见上面的代码,并且。

您不应该
current.name==name
,或者
par==”
。对于
String
s,始终使用
equals()!哇!首先感谢您的快速回复!!!第二,这非常有效!我没有意识到return语句导致它崩溃。我认为递归总是“返回函数(param a,param b-1)”等等。
public Node findNode(String name, Node localTree)
{
...
    // Pre-order - go left
    if(current.leftChild != null)
    {
        System.out.println("going left to " + current.leftChild);
        return findNode(name, current.leftChild); // <===== HERE!
    }
...
}
public Node findNode(String name, Node localTree)
{
    Node current = localTree;

    // Visit the node
    if(current.name.equals(name))
        return current;

    // Pre-order - go left
    if(current.leftChild != null)
    {
        System.out.println("going left to " + current.leftChild);
        Node nodeFound = findNode(name, current.leftChild);
        if ( nodeFound != null ) {
          // Only return from findNode if we have already found what we're looking for.
          return nodeFound;
        }
    }

    // Pre-order - go right
    if(current.rightChild != null)
    {
        System.out.println("going right to " + current.rightChild);
        return findNode(name, current.rightChild);
    }

    return null;
}