我如何在Java中打印一个二叉树,这样输出就像:
4
/ \
2 5
我的节点:
public class Node<A extends Comparable> {
Node<A> left, right;
A data;
public Node(A data){
this.data = data;
}
}
我如何在Java中打印一个二叉树,这样输出就像:
4
/ \
2 5
我的节点:
public class Node<A extends Comparable> {
Node<A> left, right;
A data;
public Node(A data){
this.data = data;
}
}
当前回答
private StringBuilder prettyPrint(Node root, int currentHeight, int totalHeight) {
StringBuilder sb = new StringBuilder();
int spaces = getSpaceCount(totalHeight-currentHeight + 1);
if(root == null) {
//create a 'spatial' block and return it
String row = String.format("%"+(2*spaces+1)+"s%n", "");
//now repeat this row space+1 times
String block = new String(new char[spaces+1]).replace("\0", row);
return new StringBuilder(block);
}
if(currentHeight==totalHeight) return new StringBuilder(root.data+"");
int slashes = getSlashCount(totalHeight-currentHeight +1);
sb.append(String.format("%"+(spaces+1)+"s%"+spaces+"s", root.data+"", ""));
sb.append("\n");
//now print / and \
// but make sure that left and right exists
char leftSlash = root.left == null? ' ':'/';
char rightSlash = root.right==null? ' ':'\\';
int spaceInBetween = 1;
for(int i=0, space = spaces-1; i<slashes; i++, space --, spaceInBetween+=2) {
for(int j=0; j<space; j++) sb.append(" ");
sb.append(leftSlash);
for(int j=0; j<spaceInBetween; j++) sb.append(" ");
sb.append(rightSlash+"");
for(int j=0; j<space; j++) sb.append(" ");
sb.append("\n");
}
//sb.append("\n");
//now get string representations of left and right subtrees
StringBuilder leftTree = prettyPrint(root.left, currentHeight+1, totalHeight);
StringBuilder rightTree = prettyPrint(root.right, currentHeight+1, totalHeight);
// now line by line print the trees side by side
Scanner leftScanner = new Scanner(leftTree.toString());
Scanner rightScanner = new Scanner(rightTree.toString());
// spaceInBetween+=1;
while(leftScanner.hasNextLine()) {
if(currentHeight==totalHeight-1) {
sb.append(String.format("%-2s %2s", leftScanner.nextLine(), rightScanner.nextLine()));
sb.append("\n");
spaceInBetween-=2;
}
else {
sb.append(leftScanner.nextLine());
sb.append(" ");
sb.append(rightScanner.nextLine()+"\n");
}
}
return sb;
}
private int getSpaceCount(int height) {
return (int) (3*Math.pow(2, height-2)-1);
}
private int getSlashCount(int height) {
if(height <= 3) return height -1;
return (int) (3*Math.pow(2, height-3)-1);
}
https://github.com/murtraja/java-binary-tree-printer
只适用于1到2位整数(我懒得让它通用)
其他回答
一个Scala解决方案,改编自Vasya Novikov的答案,专门用于二叉树:
/** An immutable Binary Tree. */
case class BTree[T](value: T, left: Option[BTree[T]], right: Option[BTree[T]]) {
/* Adapted from: http://stackoverflow.com/a/8948691/643684 */
def pretty: String = {
def work(tree: BTree[T], prefix: String, isTail: Boolean): String = {
val (line, bar) = if (isTail) ("└── ", " ") else ("├── ", "│")
val curr = s"${prefix}${line}${tree.value}"
val rights = tree.right match {
case None => s"${prefix}${bar} ├── ∅"
case Some(r) => work(r, s"${prefix}${bar} ", false)
}
val lefts = tree.left match {
case None => s"${prefix}${bar} └── ∅"
case Some(l) => work(l, s"${prefix}${bar} ", true)
}
s"${curr}\n${rights}\n${lefts}"
}
work(this, "", true)
}
}
按行打印[大]树。
输出的例子:
z
├── c
│ ├── a
│ └── b
├── d
├── e
│ └── asdf
└── f
代码:
public class TreeNode {
final String name;
final List<TreeNode> children;
public TreeNode(String name, List<TreeNode> children) {
this.name = name;
this.children = children;
}
public String toString() {
StringBuilder buffer = new StringBuilder(50);
print(buffer, "", "");
return buffer.toString();
}
private void print(StringBuilder buffer, String prefix, String childrenPrefix) {
buffer.append(prefix);
buffer.append(name);
buffer.append('\n');
for (Iterator<TreeNode> it = children.iterator(); it.hasNext();) {
TreeNode next = it.next();
if (it.hasNext()) {
next.print(buffer, childrenPrefix + "├── ", childrenPrefix + "│ ");
} else {
next.print(buffer, childrenPrefix + "└── ", childrenPrefix + " ");
}
}
}
}
附注:这个答案并不完全关注“二叉”树——相反,它打印了各种类型的树。解决方案的灵感来自linux中的“树”命令。
试试这个:
public static void print(int[] minHeap, int minWidth) {
int size = minHeap.length;
int level = log2(size);
int maxLength = (int) Math.pow(2, level) * minWidth;
int currentLevel = -1 ;
int width = maxLength;
for (int i = 0; i < size; i++) {
if (log2(i + 1) > currentLevel) {
currentLevel++;
System.out.println();
width = maxLength / (int) Math.pow(2, currentLevel);
}
System.out.print(StringUtils.center(String.valueOf(minHeap[i]), width));
}
System.out.println();
}
private static int log2(int n) {
return (int) (Math.log(n) / Math.log(2));
}
这段代码片段的思想是用maxLength(即底线的长度)除以每一行的元素数量来得到块宽度。然后把元素放在每个块的中间。
参数minWidth表示底部行中块的长度。
用一张图片来说明想法并展示结果。
https://github.com/AharonSambol/PrettyPrintTreeJava
我知道我迟到了。但是我做了这个解决方案,不仅适用于简单的树,也适用于更复杂的树(如多行字符串)
示例输出:
与垂直表示相比,水平表示有点复杂。垂直打印只是简单的RNL(右->节点->左或镜像的顺序)遍历,以便先打印右子树,然后打印左子树。
def printFullTree(root, delim=' ', idnt=[], left=None):
if root:
idnt.append(delim)
x, y = setDelims(left)
printFullTree(root.right, x, idnt, False)
indent2(root.val, idnt)
printFullTree(root.left, y, idnt, True)
idnt.pop()
def setDelims(left):
x = ' '; y='|'
return (y,x) if (left == True) else (x,y) if (left == False) else (x,x)
def indent2(x, idnt, width=6):
for delim in idnt:
print(delim + ' '*(width-1), end='')
print('|->', x)
output:
|-> 15
|-> 14
| |-> 13
|-> 12
| | |-> 11
| |-> 10
| |-> 9
|-> 8
| |-> 7
| |-> 6
| | |-> 4
|-> 3
| |-> 2
|-> 1
|-> 0
在水平表示中,显示由TreeMap的HashMap或HashMap<Integer, TreeMap<Integer, Object>> xy构建;其中HashMap包含节点的y轴/level_no作为Key, TreeMap作为value。Treemap内部保存同一级别的所有节点,按它们的x轴值排序,作为键,从最左端开始-ve,根=0,最右端=+ve。
如果使用自平衡树/Treap,则使用HashMap使算法在每个级别的O(1)查找中工作,并在O(logn)中使用TreeMap排序。
不过,在这样做的时候,不要忘记为空子存储占位符,例如' '/空格,这样树看起来就像预期的那样。
现在唯一剩下的就是计算水平节点的距离,这可以用一些数学计算来完成,
计算树的宽度和高度。 一旦完成,在显示节点时,根据计算的宽度,高度和倾斜信息(如果有的话),以最佳距离呈现它们。