打印一边的二叉树

Question

如何在其侧面打印二叉树,以便输出如下所示？

   __/a
__/  \b
  \   _/c
   \_/ \d
     \e

_{(更漂亮的ascii-art欢迎)}

这里有一些代码不太合适:

def print_tree(tree):
    def emit(node,prefix):
        if "sequence" in node:
            print "%s%s"%(prefix[:-1],node["name"])
        else:
            emit(node["left"],"%s_/ "%prefix.replace("/ "," /")[:-1].replace("_"," "))
            emit(node["right"],"%s \\ "%prefix.replace("\\ "," \\")[:-1])
    emit(tree,"")    

哪个输出:

      _/hg19
    _/ \rheMac2
  _/ \mm9
  /\_/bosTau4
  /  \_/canFam2
_/     \pteVam1
 \_/loxAfr3
   \dasNov2

范围蔓延:如果您可以传入一个函数来返回字符串以打印任何节点,那将是非常好的; 通过这种方式,我有时也可以打印有关非离开节点的信息.因此,节点是否有任何要打印的内容由作为参数传入的函数控制.

这是JSON中的一些测试数据:

{
    "left": {
        "left": {
            "left": {
                "left": {
                    "name": "hg19", 
                    "sequence": 0
                }, 
                "right": {
                    "name": "rheMac2", 
                    "sequence": 1
                }
            }, 
            "right": {
                "name": "mm9", 
                "sequence": 2
            }
        }, 
        "right": {
            "left": {
                "name": "bosTau4", 
                "sequence": 3
            }, 
            "right": {
                "left": {
                    "name": "canFam2", 
                    "sequence": 4
                }, 
                "right": {
                    "name": "pteVam1", 
                    "sequence": 5
                }
            }
        }
    }, 
    "right": {
        "left": {
            "name": "loxAfr3", 
            "sequence": 6
        }, 
        "right": {
            "name": "dasNov2", 
            "sequence": 7
        }
    }
}

Answer 1

这里有一些代码实现了其他地方描述的通用递归方法.树的内部表示是子节点的字符串(叶子)或元组(对).节点的中间"片段"的内部表示是元组(above, below, lines),其中above和below是根的上方和下方的行数,并且lines是每个部分行上的迭代器(左侧没有空格).

#!/usr/local/bin/python3.3

from itertools import chain
from random import randint


def leaf(t):
    return isinstance(t, str)

def random(n):
    def extend(t):
        if leaf(t):
            return (t+'l', t+'r')
        else:
            l, r = t
            if randint(0, 1): return (l, extend(r))
            else: return (extend(l), r)
    t = ''
    for _ in range(n-1): t = extend(t)
    return t

def format(t):
    def pad(prefix, spaces, previous):
        return prefix + (' ' * spaces) + previous
    def merge(l, r):
        l_above, l_below, l_lines = l
        r_above, r_below, r_lines = r
        gap = r_below + l_above
        gap_above = l_above
        gap_below = gap - gap_above
        def lines():
            for (i, line) in enumerate(chain(r_lines, l_lines)):
                if i < r_above:
                    yield ' ' + line
                elif i - r_above < gap_above:
                    dash = '_' if i - r_above == gap_above - 1 else ' '
                    if i < r_above + r_below:
                        yield pad(dash + '/', 2 * (i - r_above), line)
                    else:
                        yield pad(dash + '/', 2 * gap_below - 1, line)
                elif i - r_above - gap_above < gap_below:
                    if i < r_above + r_below:
                        yield pad(' \\', 2 * gap_above - 1, line)
                    else:
                        spaces = 2 * (r_above + gap_above + gap_below - i - 1)
                        yield pad(' \\', spaces, line)
                else:
                    yield ' ' + line
        return (r_above + gap_above, gap_below + l_below, lines())
    def descend(left, t):
        if leaf(t):
            if left:
                return (1, 0, [t])
            else:
                return (0, 1, [t])
        else:
            l, r = t
            return merge(descend(True, l), descend(False, r))
    def flatten(t):
        above, below, lines = t
        for (i, line) in enumerate(lines):
            if i < above: yield (' ' * (above - i - 1)) + line
            else: yield (' ' * (i - above)) + line
    return '\n'.join(flatten(descend(True, t)))


if __name__ == '__main__':
    for n in range(1,20,3):
        tree = random(n)
        print(format(tree))

这是一些示例输出:

          _/rrrr
        _/ \_/rrrlr
       / \   \rrrll
     _/   \_/rrlr
    / \     \rrll
   /   \   _/rlrr
  /     \_/ \rlrl
_/        \_/rllr
 \          \_/rlllr
  \           \rllll
   \        _/lrrr
    \     _/ \lrrl
     \   / \_/lrlr
      \_/    \lrll
        \   _/llrr
         \_/ \llrl
           \_/lllr
             \_/llllr
               \lllll

还有一点非对称的,也许是为什么我不会在左边用空格填充行直到结束(通过flatten).如果下半部分被填充在左侧,则一些上臂将穿过衬垫区域.

               _/rrrrr
             _/ \rrrrl
           _/ \rrrl
         _/ \_/rrlr
        / \   \rrll
       /   \_/rlr
      /      \rll
     /        /lrrr
    /       _/  _/lrrlrr
   /       / \_/ \lrrlrl
  /       /    \lrrll
_/      _/     _/lrlrrr
 \     / \   _/ \lrlrrl
  \   /   \_/ \lrlrl
   \_/      \lrll
     \      _/llrrr
      \   _/ \llrrl
       \_/ \llrl
         \lll

这是"明显的"递归算法 - 魔鬼在细节中.没有"_"这是最容易编写的,这使得逻辑稍微复杂一些.

也许唯一的"见解"是gap_above = l_above- 这就是说正确的"手臂"具有左子树右侧的长度(你需要阅读几次).它使事情相对平衡.请参阅上面的非对称示例.

更详细地理解事物的一种好方法是修改pad例程以取代字符而不是' '为每个调用赋予不同的字符.然后你可以确切地看到哪个逻辑生成了哪个空格.这是你使用A. B,C和D从上到下调用​​填充(当空间量为零时显然没有字符):

             _/rrrr
            / \rrrl
          _/B _/rrlrr
         / \_/ \rrlrl
        /AA  \rrll
      _/BBB  _/rlrrr
     / \DD _/ \rlrrl
    /AA \_/ \_/rlrlr
   /AAAA  \C  \rlrll
  /AAAAAA  \_/rllr
_/AAAAAAAA   \rlll
 \DDDDDDDD   _/lrrrr
  \DDDDDD  _/ \lrrrl
   \DDDD  / \lrrl
    \DD _/B _/lrlrr
     \_/ \_/ \lrlrl
       \C  \lrll
        \_/llr
          \lll

还有更多的解释在这里(虽然树是非常略有不同).