我有一个项目是“从tree.java程序开始（代码清单8.1）并修改它以创建二进制文件 来自用户输入的一串字母（如A，B等）的树。每 字母将显示在自己的节点中。构造树以便所有节点 包含字母的是叶子。父节点可以包含一些非字母 符号像+。确保每个父节点都有两个子节点。 不要担心这棵树是不平衡的。“这本书给了我们一个如何开始的提示。 “开始的一种方法是制作一系列树木。 （一组未连接的树木 被称为森林。）取用户输入的每个字母并将其放在一个节点中。采取 这些节点中的每一个都将它放在树中，它将成为树的根。现在把所有 这些单节点树在数组中。首先在根目录下创建一个带有+的新树 和两个单节点树作为它的子节点。然后继续添加单节点树 从数组到这个更大的树。如果它是一棵不平衡的树，请不要担心。“
import java.io.*;
import java.util.*;
class Node
{
public String iData; // data item (key)
public Node leftChild; // this node’s left child
public Node rightChild; // this node’s right child
public void displayNode() // display ourself
{
System.out.print('{');
System.out.print(iData);
System.out.print("} ");
}
} // end class Node
class Tree
{
private Node root; // first node of tree
public void setNode(Node newNode)
{root = newNode;}
public Node getNode()
{return root;}
// -------------------------------------------------------------
public Tree() // constructor
{ root = null; } // no nodes in tree yet
// -------------------------------------------------------------
public void traverse(int traverseType)
{
switch(traverseType)
{
case 1: System.out.print("nPreorder traversal: ");
preOrder(root);
break;
case 2: System.out.print("nInorder traversal: ");
inOrder(root);
break;
case 3: System.out.print("nPostorder traversal: ");
postOrder(root);
break;
}
System.out.println();
}
private void preOrder(Node localRoot)
{
if(localRoot != null)
{
System.out.print(localRoot.iData + " ");
preOrder(localRoot.leftChild);
preOrder(localRoot.rightChild);
}
}
// -------------------------------------------------------------
private void inOrder(Node localRoot)
{
if(localRoot != null)
{
inOrder(localRoot.leftChild);
System.out.print(localRoot.iData + " ");
inOrder(localRoot.rightChild);
}
}
// -------------------------------------------------------------
private void postOrder(Node localRoot)
{
if(localRoot != null)
{
postOrder(localRoot.leftChild);
postOrder(localRoot.rightChild);
System.out.print(localRoot.iData + " ");
}
}
// -------------------------------------------------------------
public void displayTree()
{
Stack globalStack = new Stack();
globalStack.push(root);
int nBlanks = 32;
boolean isRowEmpty = false;
System.out.println(
"......................................................");
while(isRowEmpty==false)
{
Stack localStack = new Stack();
isRowEmpty = true;
for(int j=0; j<nBlanks; j++)
System.out.print(' ');
while(globalStack.isEmpty()==false)
{
Node temp = (Node)globalStack.pop();
if(temp != null)
{
System.out.print(temp.iData);
localStack.push(temp.leftChild);
localStack.push(temp.rightChild);
if(temp.leftChild != null ||
temp.rightChild != null)
isRowEmpty = false;
}
else
{
System.out.print("--");
localStack.push(null);
localStack.push(null);
}
for(int j=0; j<nBlanks*2-2; j++)
System.out.print(' ');
} // end while globalStack not empty
System.out.println();
nBlanks /= 2;
while(localStack.isEmpty()==false)
globalStack.push( localStack.pop() );
} // end while isRowEmpty is false
System.out.println(
"......................................................");
} // end displayTree()
// -------------------------------------------------------------
}
public class Leaves
{
//function used to enter the single node trees into a larger tree
public static void enterLetters(Node localRoot, Tree[] nodeTree, int i)
{
if(localRoot != null && i == nodeTree.length)
{
if(nodeTree.length == i - 1)
{
localRoot.leftChild = nodeTree[i].getNode();
localRoot.rightChild = nodeTree[i + 1].getNode();
enterLetters(localRoot.leftChild, nodeTree, i + 1);
}
else
{
Node plusNode = new Node();
plusNode.iData = "+";
localRoot.leftChild = plusNode;
localRoot.rightChild = nodeTree[i].getNode();
enterLetters(localRoot.leftChild, nodeTree, i + 1);
}
}
}
public static void main(String[] args)
{
Tree[] forest = new Tree[10];
Scanner sc = new Scanner(System.in);
for(int i = 0; i < 10; i++)
{
String letter;
forest[i] = new Tree();
System.out.println("Enter a letter: ");
letter = sc.nextLine();
Node newNode = new Node();
newNode.iData = letter;
forest[i].setNode(newNode);
}
Tree letterTree = new Tree();
Node firstNode = new Node();
firstNode.iData = "+";
letterTree.setNode(firstNode);
enterLetters(letterTree.getNode(), forest, 0);
letterTree.displayTree();
}
}
我的问题是尝试将单个节点树的数组放入更大的树中。我尝试制作一个递归函数，但是当我显示较大的树时，它只显示第一个节点，就好像函数enterLeaves从未完成它的工作。
答案 0 :(得分：0)
这可能不正确：
public static void enterLetters(Node localRoot, Tree[] nodeTree, int i) {
if (localRoot != null && i == nodeTree.length) {
if (nodeTree.length == i - 1) {
localRoot.leftChild = nodeTree[i].getNode();
localRoot.rightChild = nodeTree[i + 1].getNode();
enterLetters(localRoot.leftChild, nodeTree, i + 1);
} else {
Node plusNode = new Node();
plusNode.iData = "+";
localRoot.leftChild = plusNode;
localRoot.rightChild = nodeTree[i].getNode();
enterLetters(localRoot.leftChild, nodeTree, i + 1);
}
}
}
输入此方法时：localRoot！= null，i == 0，nodeTree.length == 10
所以if语句失败了。我猜if语句应该是：
if (localRoot != null && i < nodeTree.length)
另外，我很确定你的第二个if语句也是错误的;我相信它应该是。
if (nodeTree.length-2 == i) {
localRoot.leftChild = nodeTree[i].getNode();
localRoot.rightChild = nodeTree[i + 1].getNode();
return;
}
而不是：
if (nodeTree.length == i - 1) {
localRoot.leftChild = nodeTree[i].getNode();
localRoot.rightChild = nodeTree[i + 1].getNode();
enterLetters(localRoot.leftChild, nodeTree, i + 1);
}
当你有两个要处理的节点（nodeTree.length-2 == i）时你想要停止，并且在你这样做之后你应该返回而不是输入剩余的字母。
答案 1 :(得分：0)
以下是我提出的有效方法：
Node.java
/** Represents a node in a binary tree data structure */
public class Node {
private char letter;
private Node leftChild;
private Node rightChild;
public Node(char letter) {
this.letter = letter;
}
public void setRightChild(Node rightChild) {
this.rightChild = rightChild;
}
public Node getRightChild() {
return rightChild;
}
public void setLeftChild(Node leftChild) {
this.leftChild = leftChild;
}
public Node getLeftChild() {
return leftChild;
}
/** Returns a String representation of this node. */
@Override
public String toString() {
return "" + letter;
}
}
Tree.java
import java.util.Stack;
/**
* A binary tree
*/
public class Tree {
private Node root;
public void setRoot(Node root) {
this.root = root;
}
public void addToLeft(Node node) {
root.setLeftChild(node);
}
public void addToRight(Node node) {
root.setRightChild(node);
}
public Node getRoot() {
return root;
}
public void displayTree() {
Stack<Node> globalStack = new Stack<>();
globalStack.push(root);
int nBlanks = 32;
boolean isRowEmpty = false;
System.out.println(
"......................................................");
while (!isRowEmpty) {
Stack<Node> localStack = new Stack<>();
isRowEmpty = true;
for (int j = 0; j < nBlanks; j++)
System.out.print(' ');
while (!globalStack.isEmpty()) {
Node temp = (Node) globalStack.pop();
if (temp != null) {
System.out.print(temp);
localStack.push(temp.getLeftChild());
localStack.push(temp.getRightChild());
if (temp.getLeftChild() != null ||
temp.getRightChild() != null)
isRowEmpty = false;
} else {
System.out.print("--");
localStack.push(null);
localStack.push(null);
}
for (int j = 0; j < nBlanks * 2 - 2; j++)
System.out.print(' ');
} // end while globalStack not empty
System.out.println();
nBlanks /= 2;
while (!localStack.isEmpty())
globalStack.push(localStack.pop());
} // end while isRowEmpty is false
System.out.println(
"......................................................");
}
}
Forest.java
/**
* A collection of OneNodeTrees combined together in one tree
*/
public class Forest {
private Tree[] forest;
private int forestIndex;
public Forest(int numTrees) {
forest = new Tree[numTrees];
forestIndex = 0;
}
public boolean add(Tree tree) {
if(forestIndex < forest.length) {
forest[forestIndex++] = tree;
return true;
} else {
return false;
}
}
public Tree createMainTree() {
Tree firstTree = new Tree();
firstTree.setRoot(new Node('+'));
firstTree.addToLeft(forest[0].getRoot());
firstTree.addToRight(forest[1].getRoot());
forest[1] = firstTree;
int mainTreeIndex = 0;
Tree tempTree;
for(mainTreeIndex = 2; mainTreeIndex < forest.length; mainTreeIndex++) {
tempTree = new Tree();
tempTree.setRoot(new Node('+'));
tempTree.addToLeft(forest[mainTreeIndex - 1].getRoot());
tempTree.addToRight(forest[mainTreeIndex].getRoot());
forest[mainTreeIndex] = tempTree;
}
return forest[mainTreeIndex - 1];
}
public static void main(String[] args) {
final int numberOfTrees = 6;
Forest forest = new Forest(numberOfTrees);
// Add characters starting from A which has ASCII value 65
int charLimit = 65 + numberOfTrees;
for(int i = 65; i < charLimit; i++) {
// Make new node.
Node newNode = new Node((char) i);
// Add that node to Tree as a root.
Tree newTree = new Tree();
newTree.setRoot(newNode);
// And add that one-node tree to forest(array)
forest.add(newTree);
}
Tree mainTree = forest.createMainTree();
mainTree.displayTree();
}
}
答案 2 :(得分：-1)
if(localRoot != null && i == nodeTree.length -1)
如果你没有从节点树长度中减去一个，你将在最终父节点下有一个重复的孩子，有2个孩子