我已经构建了一个二叉搜索树。存储在树中的原始数据类型是整数。我尝试将它存储在2-D char数组中，然后将其打印出来，如下图所示（数字表示行号和列号，我不需要打印它，忽略“ - ”符号，我只使用它表示确切的位置）
-----0---1---2---3---4---5---6---7---8---9---10---11---12---13---14---15---16
0---------------------------------------12
1--------------------------------/-------------------\
2----------------------8--------------------------------------14
3-----------------/----------\ -----------------------------------------\
4-------------5----------------9-------------------------------------------34
5--------/-------------------------------------------------------------/-------------\
6---2---------------------------------------------------------24------------------------35
第1行是第2行，“/”是位置[1] [6]，“\”是在位置[1] [10]等等。它们也在两个数字之间的中间位置
以下是我的代码
public class MainClass {
public static void main(String[] args) {
//level represents row number;
// start indicates the column I am going to
//store number in, and end is a fixed column number
// BinarySearchTree is a BinaryTree type instance,
// I already story integers on it and follow with the format
// of binary search trees, and I did tested it.
int level=0; int start=0; int end=80;
BinaryTree.plot(BinarySearchTree, level, start, end);
}
private static class BinaryTree {
private BinaryNode root;
static char[][] offset = new char [10][20];
public BinaryTree(){
root = null;
}
public BinaryTree(Object x){
root = new BinaryNode(x);
}
public boolean isEmpty(){
return root == null;
}
public Object getRootobj() throws BinaryTreeException{
if(root == null)
throw new BinaryTreeException("Empty Tree");
else
return root.element;
}
public BinaryTree getLeft() throws BinaryTreeException{
if(root == null)
throw new BinaryTreeException("Empty Tree");
else {
BinaryTree t = new BinaryTree();
t.root = root.left;
return t;
}
}
public BinaryTree getRight() throws BinaryTreeException{
if(root == null)
throw new BinaryTreeException("Empty Tree");
else {
BinaryTree t = new BinaryTree();
t.root = root.right;
return t;
}
}
public static void plot(BinaryTree t, int level, int start, int end){
if(!t.isEmpty()){
plot(t.getLeft(), level+2, start/2, end/2);
String string = Integer.toString((Integer)t.getRootobj());
for(char c: string.toCharArray())
offset[level][start++]=c;
if(!(t.getLeft().isEmpty()))
offset[++level][start/4*3] = '/';
if(!(t.getRight().isEmpty()))
offset[++level][((start+end)/2+start)/2] = '\\';
plot(t.getRight(), level+2, end/2, end);
}
for(int i = 0; i<10; i++){
for(int j= 0; j<20; j++)
System.out.print(offset[i][j]);
}
}
}
private static class BinaryNode {
Object element;
BinaryNode left,right;
BinaryNode() {
this(0);
}
BinaryNode(Object e) {
this(e, null, null);
}
BinaryNode(Object e, BinaryNode ln, BinaryNode m){
element=e;
left=ln;
right=m;
}
}
}
问题：我用来存储和打印出binarysearchtree的方法图无效，导致java.lang.ArrayIndexOutOfBoundsException：
任何人都可以看看它。感谢您的帮助。
答案 0 :(得分：1)
您的固定大小的char-Array无法处理动态大小的BinaryTree。仅对于您给出的示例，您需要每行超过20个字符！这就是你的例外来源。
但为了让您了解另一种方法 - 尽管需要一段时间，但是对您的代码进行了以下添加：
首先，我向BinaryNode类添加了一个方法：
int getDepth() {
int subTreeDepth;
if (left == null && right == null) {
subTreeDepth = 0;
} else if (left == null) {
subTreeDepth = right.getDepth();
} else if (right == null) {
subTreeDepth = left.getDepth();
} else {
subTreeDepth = Math.max(left.getDepth(), right.getDepth());
}
return 1 + subTreeDepth;
}
其次，我删除了你的固定char-Array并替换了BinaryTree中的整个绘图算法（我无法绕过所有相关的数组索引操作）：
public void plot() {
if (root == null) {
throw new BinaryTreeException("Empty Tree");
}
int lineCount = 2 * root.getDepth() - 1;
StringBuilder[] lines = new StringBuilder[lineCount];
for (int lineIndex = 0; lineIndex < lineCount; lineIndex++) {
lines[lineIndex] = new StringBuilder();
}
// get the right most node (which contains the largest element value)
BinaryNode rightMostNode = root;
while (rightMostNode.right != null) {
rightMostNode = rightMostNode.right;
}
// check how many characters we have to reserve for a single node element
int maxElementLength = String.valueOf(rightMostNode.element).length();
plot(root, 0, 0, maxElementLength, lines);
for (StringBuilder singleLine : lines) {
System.out.println(singleLine.toString());
}
}
private void plot(BinaryNode subTreeRoot, int offset, int lineIndex, int elementLength, StringBuilder[] lines) {
int actualOffset;
if (subTreeRoot.left == null) {
actualOffset = offset;
} else {
actualOffset = offset + (int) Math.pow(2, subTreeRoot.left.getDepth() - 1) * elementLength;
}
StringBuilder currentLine = lines[lineIndex];
String elementValue = String.valueOf(subTreeRoot.element);
for (int lineFillIndex = currentLine.length() + elementValue.length() / 2; lineFillIndex < actualOffset; lineFillIndex++) {
currentLine.append(' ');
}
currentLine.append(elementValue);
if (subTreeRoot.left != null) {
// draw connection to left sub tree
int connectPosition = (actualOffset - offset) * 3 / 4 + offset;
StringBuilder connectLine = lines[lineIndex + 1];
for (int lineFillIndex = connectLine.length(); lineFillIndex < connectPosition; lineFillIndex++) {
connectLine.append(' ');
}
connectLine.append('/');
// insert the left part of the next value line
plot(subTreeRoot.left, offset, lineIndex + 2, elementLength, lines);
}
if (subTreeRoot.right != null) {
// draw connection to right sub tree
int connectPosition = actualOffset + elementLength - elementValue.length() / 2;
if (subTreeRoot.right.left != null) {
connectPosition += (int) Math.pow(2, subTreeRoot.right.left.getDepth() - 1) * elementLength / 2;
}
StringBuilder connectLine = lines[lineIndex + 1];
for (int lineFillIndex = connectLine.length(); lineFillIndex < connectPosition; lineFillIndex++) {
connectLine.append(' ');
}
connectLine.append('\\');
// insert the right part of the next value line
plot(subTreeRoot.right, actualOffset + elementLength, lineIndex + 2, elementLength, lines);
}
}
对于与您的树类似的树，您包含在您的问题中：
BinaryTree binarySearchTree = new BinaryTree(
new BinaryNode(12,
new BinaryNode(8,
new BinaryNode(5,
new BinaryNode(3),
null),
new BinaryNode(9)),
new BinaryNode(14,
null,
new BinaryNode(34,
new BinaryNode(24),
new BinaryNode(35)))));
binarySearchTree.plot();
我得到以下输出：
12
/ \
8 14
/ \ \
5 9 34
/ / \
3 24 35