Printing BFS (Binary Tree) in Level Order with Specific Formatting

Just build one level at a time, e.g.:

class Node(object):  def __init__(self, value, left=None, right=None):    self.value = value    self.left = left    self.right = rightdef traverse(rootnode):  thislevel = [rootnode]  while thislevel:    nextlevel = list()    for n in thislevel:      print n.value,      if n.left: nextlevel.append(n.left)      if n.right: nextlevel.append(n.right)    print    thislevel = nextlevelt = Node(1, Node(2, Node(4, Node(7))), Node(3, Node(5), Node(6)))traverse(t)

Edit: if you're keen to get a small saving in maximum consumed "auxiliary" memory (never having simultaneously all this level and the next level in such "auxiliary" memory), you can of course use collection.deque instead of list, and consume the current level as you go (via popleft) instead of simply looping. The idea of creating one level at a time (as you consume --or iterate on-- the previous one) remains intact -- when you do need to distinguish levels, it's just more direct than using a single big deque plus auxiliary information (such as depth, or number of nodes remaining in a given level).

However, a list that is only appended to (and looped on, rather than "consumed") is quite a bit more efficient than a deque (and if you're after C++ solutions, quite similarly, a std::vector using just push_back for building it, and a loop for then using it, is more efficient than a std::deque). Since all the producing happens first, then all the iteration (or consuming), an interesting alternative if memory is tightly constrained might be to use a list anyway to represent each level, then .reverse it before you start consuming it (with .pop calls) -- I don't have large trees around to check by measurement, but I suspect that this approach would still be faster (and actually less memory-consuming) than deque (assuming that the underlying implementation of list [[or std::vector]] actually does recycle memory after a few calls to pop [[or pop_back]] -- and with the same assumption for deque, of course;-).

Sounds like breadth-first traversal to me.

Breadth-first traversal is implemented with a queue. Here, simply insert in the queue a special token that indicate that a newline must be printed. Each time the token is found, print a newline and re-insert the token in the queue (at the end -- that's the definition of a queue).

Start the algorithm with a queue containing the root followed by the special newline token.

My solution is similar to Alex Martelli's, but I separate traversal of the data structure from processing the data structure. I put the meat of the code into iterLayers to keep printByLayer short and sweet.

from collections import dequeclass Node:    def __init__(self, val, lc=None, rc=None):        self.val = val        self.lc = lc        self.rc = rc    def iterLayers(self):        q = deque()        q.append(self)        def layerIterator(layerSize):            for i in xrange(layerSize):                n = q.popleft()                if n.lc: q.append(n.lc)                if n.rc: q.append(n.rc)                yield n.val        while (q):            yield layerIterator(len(q))    def printByLayer(self):        for layer in self.iterLayers():            print ' '.join([str(v) for v in layer])root = Node(1, Node(2, Node(4, Node(7))), Node(3, Node(5), Node(6)))root.printByLayer()

which prints the following when run:

12 34 5 67