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

Question

To begin with, this question is not a dup of this one, but builds on it.

Taking the tree in that question as an example,

    1 
   / \\
  2   3
 /            


        

Answer 1

why not keep sentinal in queue and check when all the nodes in current level are processed.

public void printLevel(Node n) {
    Queue<Integer> q = new ArrayBlockingQueue<Integer>();
    Node sentinal = new Node(-1);
    q.put(n);
    q.put(sentinal);
    while(q.size() > 0) {
        n = q.poll();
        System.out.println(n.value + " "); 
        if (n == sentinal && q.size() > 0) {
           q.put(sentinal); //push at the end again for next level
           System.out.println();
        }
        if (q.left != null) q.put(n.left);
        if (q.right != null) q.put(n.right);
    }
}

Answer 2

A simple Version based on Bread First Search, This code is applicable for graphs in general and can be used for binary trees as well.

def printBfsLevels(graph,start):
  queue=[start]
  path=[]
  currLevel=1
  levelMembers=1
  height=[(0,start)]
  childCount=0
  print queue
  while queue:
    visNode=queue.pop(0)
    if visNode not in path:
      if  levelMembers==0:
        levelMembers=childCount
        childCount=0
        currLevel=currLevel+1
      queue=queue+graph.get(visNode,[])
      if levelMembers > 0:
        levelMembers=levelMembers-1
        for node in graph.get(visNode,[]):
          childCount=childCount+1
          height.append((currLevel,node))
      path=path+[visNode]

  prevLevel=None

  for v,k in sorted(height):
        if prevLevel!=v:
          if prevLevel!=None:
            print "\n"
        prevLevel=v
        print k,
  return height

g={1: [2, 3,6], 2: [4, 5], 3: [6, 7],4:[8,9,13]}
printBfsLevels(g,1)


>>> 
[1]
1 

2 3 6 

4 5 6 7 

8 9 13
>>> 

Another version based on Recursion, which is specific to binary tree

class BinTree:
  "Node in a binary tree"
  def __init__(self,val,leftChild=None,rightChild=None,root=None):
    self.val=val
    self.leftChild=leftChild
    self.rightChild=rightChild
    self.root=root
    if not leftChild and not rightChild:
      self.isExternal=True

  def getChildren(self,node):
    children=[]
    if node.isExternal:
      return []
    if node.leftChild:
      children.append(node.leftChild)
    if node.rightChild:
      children.append(node.rightChild)
    return children

  @staticmethod
  def createTree(tupleList):
    "Creates a Binary tree Object from a given Tuple List"
    Nodes={}
    root=None
    for item in treeRep:
      if not root:
        root=BinTree(item[0])
        root.isExternal=False
        Nodes[item[0]]=root
        root.root=root
        root.leftChild=BinTree(item[1],root=root)
        Nodes[item[1]]=root.leftChild
        root.rightChild=BinTree(item[2],root=root)
        Nodes[item[2]]=root.rightChild
      else:
        CurrentParent=Nodes[item[0]]
        CurrentParent.isExternal=False
        CurrentParent.leftChild=BinTree(item[1],root=root)
        Nodes[item[1]]=CurrentParent.leftChild
        CurrentParent.rightChild=BinTree(item[2],root=root)
        Nodes[item[2]]=CurrentParent.rightChild
    root.nodeDict=Nodes
    return root

  def printBfsLevels(self,levels=None):
    if levels==None:
      levels=[self]
    nextLevel=[]
    for node in levels:
      print node.val,
    for node in levels:
      nextLevel.extend(node.getChildren(node))
    print '\n'
    if nextLevel:
      node.printBfsLevels(nextLevel)  


##       1
##     2     3
##   4   5  6  7
##  8

treeRep = [(1,2,3),(2,4,5),(3,6,7),(4,8,None)]
tree= BinTree.createTree(treeRep)
tree.printBfsLevels()

>>> 
1 

2 3 

4 5 6 7 

8 None 

Answer 3

class TNode:
  def __init__(self, data, left=None, right=None):
    self.data = data
    self.left = left
    self.right = right

class BST:
  def __init__(self, root):
    self._root = root

  def bfs(self):
    list = [self._root]
    while len(list) > 0:
        print [e.data for e in list]
        list = [e.left for e in list if e.left] + \
               [e.right for e in list if e.right]
bst = BST(TNode(1, TNode(2, TNode(4), TNode(5)), TNode(3, TNode(6), TNode(7))))
bst.bfs()

Answer 4

A version that doesn't require extra storage:

std::deque<Node> bfs;
bfs.push_back(start);
int nodesInThisLayer = 1;
int nodesInNextLayer = 0;
while (!bfs.empty()) {
    Node front = bfs.front();
    bfs.pop_front();
    for (/*iterate over front's children*/) {
        ++nodesInNextLayer;
        nodes.push_back(child);
    }
    std::cout << node.value;
    if (0 == --nodesInThisLayer) {
        std::cout << std::endl;
        nodesInThisLayer = nodesInNextLayer; 
        nodesInNextLayer = 0;
    } else {
        std::cout << " ";
    }
}

P.S. sorry for the C++ code, I'm not very fluent in Python yet.

Answer 5

This is mostly the same code as provided by Alex Martelli except this is modified for python 3.

class Node(object):
  def __init__(self, value, left=None, right=None):
    self.value = value
    self.left = left
    self.right = right

def traverse(rootnode):
  thislevel = [rootnode]
  while thislevel:
    nextlevel = list()
    for n in thislevel:
      print (n.value,' ', end=''),
      if n.left: nextlevel.append(n.left)
      if n.right: nextlevel.append(n.right)
    print(" ")
    thislevel = nextlevel

t = Node(1, Node(2, Node(4, Node(7))), Node(3, Node(5), Node(6)))

traverse(t)

Answer 6

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 deque

class 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:

1
2 3
4 5 6
7