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:

这是打印树的一个非常简单的解决方案。它不是那么漂亮，但它真的很简单:

enum { kWidth = 6 };
void PrintSpace(int n)
{
  for (int i = 0; i < n; ++i)
    printf(" ");
}

void PrintTree(struct Node * root, int level)
{
  if (!root) return;
  PrintTree(root->right, level + 1);
  PrintSpace(level * kWidth);
  printf("%d", root->data);
  PrintTree(root->left, level + 1);
}

样例输出:

      106
            105
104
            103
                  102
                        101
      100

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)
  }
}

你的树每一层需要两倍的距离:

       a
      / \
     /   \
    /     \
   /       \
   b       c
  / \     / \
 /   \   /   \
 d   e   f   g
/ \ / \ / \ / \
h i j k l m n o

你可以将你的树保存在一个数组的数组中，每个数组对应一个深度:

[[a],[b,c],[d,e,f,g],[h,i,j,k,l,m,n,o]]

如果你的树没有满，你需要在数组中包含空值:

       a
      / \
     /   \
    /     \
   /       \
   b       c
  / \     / \
 /   \   /   \
 d   e   f   g
/ \   \ / \   \
h i   k l m   o
[[a],[b,c],[d,e,f,g],[h,i, ,k,l,m, ,o]]

然后你可以遍历数组来打印你的树，根据深度打印第一个元素之前和元素之间的空格，根据下一层数组中对应的元素是否被填充打印行。 如果您的值可以超过一个字符长，您需要在创建数组表示时找到最长的值，并相应地乘以所有宽度和行数。

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);

    }
}

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。