以ASCII显示树

这是一个涵盖您所需要的大部分解决方案。

像任何树算法一样，递归树的子级，并在每个节点上合并结果。诀窍是：

display()

aaaaaaaaaaaaaaaaaa

大部分矩形将为空白。仅返回文本的矩形可轻松组合结果。我们将使用以下两个辅助函数，一个用于测量块的宽度，另一个用于将块水平组合成更大的块：

def block_width(block):    try:        return block.index('n')    except ValueError:        return len(block)def stack_str_blocks(blocks):    """Takes a list of multiline strings, and stacks them horizontally.    For example, given 'aaanaaa' and 'bbbbnbbbb', it returns    'aaa bbbbnaaa bbbb'.  As in:    'aaa  +  'bbbb   =  'aaa bbbb     aaa'     bbbb'      aaa bbbb'    Each block must be rectangular (all lines are the same length), but blocks    can be different sizes.    """    builder = []    block_lens = [block_width(bl) for bl in blocks]    split_blocks = [bl.split('n') for bl in blocks]    for line_list in itertools.izip_longest(*split_blocks, fillvalue=None):        for i, line in enumerate(line_list): if line is None:     builder.append(' ' * block_lens[i]) else:     builder.append(line) if i != len(line_list) - 1:     builder.append(' ')  # Padding        builder.append('n')    return ''.join(builder[:-1])

看到这是怎么回事？子级返回一个显示自己及其后代的矩形，每个节点会将这些矩形组合成一个包含自身的较大矩形。其余代码仅呈现破折号和加号：

class Node:    def display(self): # Here        if not self.children: return self.name        child_strs = [child.display() for child in self.children]        child_widths = [block_width(s) for s in child_strs]        # How wide is this block?        display_width = max(len(self.name),         sum(child_widths) + len(child_widths) - 1)        # Determines midpoints of child blocks        child_midpoints = []        child_end = 0        for width in child_widths: child_midpoints.append(child_end + (width // 2)) child_end += width + 1        # Builds up the brace, using the child midpoints        brace_builder = []        for i in xrange(display_width): if i < child_midpoints[0] or i > child_midpoints[-1]:     brace_builder.append(' ') elif i in child_midpoints:     brace_builder.append('+') else:     brace_builder.append('-')        brace = ''.join(brace_builder)        name_str = '{:^{}}'.format(self.name, display_width)        below = stack_str_blocks(child_strs)        return name_str + 'n' + brace + 'n' + below    # SNIP (your other methods)

而且我们要参加比赛了！

       a       +-+-+---------------------------+    b e f     g    +     +-+-------------------------+  c     h i   k  +       + +-+-+-+-------------+-------------+-+------+     d       j l m p r  s  O P      Q      + +   +-+-+-+---------+  +-----+     n q   t u w x         y  R     S     +       +     +-------+-------+       +---+---+ o       v     z       A       M       T   U   Z      +-+-+-+-+-+-+ ++   +      B D E H I K L NV   a      +   +   +    +-+-+ +      C   F   J    W X Y b          +   G

（诸如“将^置于父母以下的要求”留给读者练习）