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。

public static class Node<T extends Comparable<T>> {
    T value;
    Node<T> left, right;

    public void insertToTree(T v) {
        if (value == null) {
            value = v;
            return;
        }
        if (v.compareTo(value) < 0) {
            if (left == null) {
                left = new Node<T>();
            }
            left.insertToTree(v);
        } else {
            if (right == null) {
                right = new Node<T>();
            }
            right.insertToTree(v);
        }
    }

    public void printTree(OutputStreamWriter out) throws IOException {
        if (right != null) {
            right.printTree(out, true, "");
        }
        printNodeValue(out);
        if (left != null) {
            left.printTree(out, false, "");
        }
    }
    private void printNodeValue(OutputStreamWriter out) throws IOException {
        if (value == null) {
            out.write("<null>");
        } else {
            out.write(value.toString());
        }
        out.write('\n');
    }
    // use string and not stringbuffer on purpose as we need to change the indent at each recursion
    private void printTree(OutputStreamWriter out, boolean isRight, String indent) throws IOException {
        if (right != null) {
            right.printTree(out, true, indent + (isRight ? "        " : " |      "));
        }
        out.write(indent);
        if (isRight) {
            out.write(" /");
        } else {
            out.write(" \\");
        }
        out.write("----- ");
        printNodeValue(out);
        if (left != null) {
            left.printTree(out, false, indent + (isRight ? " |      " : "        "));
        }
    }

}

将打印:

                 /----- 20
                 |       \----- 15
         /----- 14
         |       \----- 13
 /----- 12
 |       |       /----- 11
 |       \----- 10
 |               \----- 9
8
 |               /----- 7
 |       /----- 6
 |       |       \----- 5
 \----- 4
         |       /----- 3
         \----- 2
                 \----- 1

对于输入

8 4 12 2 6 10 14 1 3 5 7 9 11 13 20 15

这是@anurag回答的一个变体——看到额外的|让我很烦

using map...
{
Map<Integer,String> m = new LinkedHashMap<>();

         tn.printNodeWithLvl(node,l,m);

        for(Entry<Integer, String> map :m.entrySet()) {
            System.out.println(map.getValue());
        }
then....method


   private  void printNodeWithLvl(Node node,int l,Map<Integer,String> m) {
       if(node==null) {
           return;
       }
      if(m.containsKey(l)) {
          m.put(l, new StringBuilder(m.get(l)).append(node.value).toString());
      }else {
          m.put(l, node.value+"");
      }
      l++;
      printNodeWithLvl( node.left,l,m);
      printNodeWithLvl(node.right,l,m);

    }
}

下面是可视化树的另一种方法:将节点保存为xml文件，然后让浏览器显示层次结构:

class treeNode{
    int key;
    treeNode left;
    treeNode right;

    public treeNode(int key){
        this.key = key;
        left = right = null;
    }

    public void printNode(StringBuilder output, String dir){
        output.append("<node key='" + key + "' dir='" + dir + "'>");
        if(left != null)
            left.printNode(output, "l");
        if(right != null)
            right.printNode(output, "r");
        output.append("</node>");
    }
}

class tree{
    private treeNode treeRoot;

    public tree(int key){
        treeRoot = new treeNode(key);
    }

    public void insert(int key){
        insert(treeRoot, key);
    }

    private treeNode insert(treeNode root, int key){
        if(root == null){
            treeNode child = new treeNode(key);
            return child;
        }

        if(key < root.key)
            root.left = insert(root.left, key);
        else if(key > root.key)
            root.right = insert(root.right, key);

        return root;
    }

    public void saveTreeAsXml(){
        StringBuilder strOutput = new StringBuilder();
        strOutput.append("<?xml version=\"1.0\" encoding=\"UTF-8\"?>");
        treeRoot.printNode(strOutput, "root");
        try {
            PrintWriter writer = new PrintWriter("C:/tree.xml", "UTF-8");
            writer.write(strOutput.toString());
            writer.close();
        }
        catch (FileNotFoundException e){

        }
        catch(UnsupportedEncodingException e){

        }
    }
}

下面是测试它的代码:

    tree t = new tree(1);
    t.insert(10);
    t.insert(5);
    t.insert(4);
    t.insert(20);
    t.insert(40);
    t.insert(30);
    t.insert(80);
    t.insert(60);
    t.insert(50);

    t.saveTreeAsXml();

输出如下所示:

根据VasyaNovikov的回答。改进了一些Java魔术:泛型和函数接口。

/**
 * Print a tree structure in a pretty ASCII fromat.
 * @param prefix Currnet previx. Use "" in initial call!
 * @param node The current node. Pass the root node of your tree in initial call.
 * @param getChildrenFunc A {@link Function} that returns the children of a given node.
 * @param isTail Is node the last of its sibblings. Use true in initial call. (This is needed for pretty printing.)
 * @param <T> The type of your nodes. Anything that has a toString can be used.
 */
private <T> void printTreeRec(String prefix, T node, Function<T, List<T>> getChildrenFunc, boolean isTail) {
    String nodeName = node.toString();
    String nodeConnection = isTail ? "└── " : "├── ";
    log.debug(prefix + nodeConnection + nodeName);
    List<T> children = getChildrenFunc.apply(node);
    for (int i = 0; i < children.size(); i++) {
        String newPrefix = prefix + (isTail ? "    " : "│   ");
        printTreeRec(newPrefix, children.get(i), getChildrenFunc, i == children.size()-1);
    }
}

初始调用示例:

Function<ChecksumModel, List<ChecksumModel>> getChildrenFunc = node -> getChildrenOf(node)
printTreeRec("", rootNode, getChildrenFunc, true);

将输出如下内容

└── rootNode
    ├── childNode1
    ├── childNode2
    │   ├── childNode2.1
    │   ├── childNode2.2
    │   └── childNode2.3
    ├── childNode3
    └── childNode4

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。