We'll use the first tree of the treebank_chunk corpus as our example. The obvious first step is to join all the words in the tree with a space:

>>> from nltk.corpus import treebank_chunk
>>> tree = treebank_chunk.chunked_sents()[0]
>>> ' '.join([w for w, t in tree.leaves()])
'Pierre Vinken , 61 years old , will join the board as a nonexecutive director Nov. 29 .'

But as you can see, the punctuation isn't quite right. The commas and period are treated as individual words, and so get the surrounding spaces as well. But we can fix this using regular expression substitution. This is implemented in the chunk_tree_to_sent() function found in transforms.py:

import re
punct_re = re.compile(r's([,.;?])')

def chunk_tree_to_sent(tree, concat=' '):
  s = concat.join([w for w, t in tree.leaves()])
  return re.sub(punct_re, r'g<1>', s)

Using chunk_tree_to_sent() results in a cleaner sentence, with no space before each punctuation mark:

>>> from transforms import chunk_tree_to_sent
>>> chunk_tree_to_sent(tree)
'Pierre Vinken, 61 years old, will join the board as a nonexecutive director Nov. 29.'

To correct the extra spaces in front of the punctuation, we create a regular expression, punct_re, that will match a space followed by any of the known punctuation characters. We have to escape both '.' and '?' with a '' since they are special characters. The punctuation is surrounded by parentheses so we can use the matched group for substitution.

Once we have our regular expression, we define chunk_tree_to_sent(), whose first step is to join the words by a concatenation character that defaults to a space. Then, we can call re.sub() to replace all the punctuation matches with just the punctuation group. This eliminates the space in front of the punctuation characters, resulting in a more correct string.

We can simplify this function a little using nltk.tag.untag() to get words from the tree's leaves, instead of using our own list comprehension:

import nltk.tag, re
punct_re = re.compile(r's([,.;?])')

def chunk_tree_to_sent(tree, concat=' '):
  s = concat.join(nltk.tag.untag(tree.leaves()))
  return re.sub(punct_re, r'g<1>', s)

The nltk.tag.untag() function is covered at the end of the Default tagging recipe in Chapter 4, Part-of-speech Tagging.