Coding Hashing PYTHON

Python Programming – Binary Tree in Vertical Order Hashmap based Method

An efficient solution based on hash map is discussed. We need to check the Horizontal Distances from root for all nodes and two nodes have the same

Given a binary tree, print it vertically. The following example illustrates vertical order traversal.

           1
        /    \
       2      3
      / \    / \
     4   5  6   7
             \   \
              8   9 
               
			  
The output of print this tree vertically will be:
4
2
1 5 6
3 8
7
9

binary

We have discussed a O(n2) solution in the previous post. In this post, an efficient solution based on hash map is discussed. We need to check the Horizontal Distances from root for all nodes. If two nodes have the same Horizontal Distance (HD), then they are on same vertical line. The idea of HD is simple. HD for root is 0, a right edge (edge connecting to right subtree) is considered as +1 horizontal distance and a left edge is considered as -1 horizontal distance. For example, in the above tree, HD for Node 4 is at -2, HD for Node 2 is -1, HD for 5 and 6 is 0 and HD for node 7 is +2.
We can do preorder traversal of the given Binary Tree.

While traversing the tree, we can recursively calculate HDs. We initially pass the horizontal distance as 0 for root. For left subtree, we pass the Horizontal Distance as Horizontal distance of root minus 1. For right subtree, we pass the Horizontal Distance as Horizontal Distance of root plus 1. For every HD value, we maintain a list of nodes in a hasp map. Whenever we see a node in traversal, we go to the hash map entry and add the node to the hash map using HD as a key in map.

Python Program
# Python program for printing vertical order of a given
# binary tree
 
# A binary tree node
class Node:
    # Constructor to create a new node
    def __init__(self, key):
        self.key = key
        self.left = None
        self.right = None
 
# Utility function to store vertical order in map 'm' 
# 'hd' is horizontal distance of current node from root
# 'hd' is initially passed as 0
def getVerticalOrder(root, hd, m):
 
    # Base Case
    if root is None:
        return
     
    # Store current node in map 'm'
    try:
        m[hd].append(root.key)
    except:
        m[hd] = [root.key]
     
    # Store nodes in left subtree
    getVerticalOrder(root.left, hd-1, m)
     
    # Store nodes in right subtree
    getVerticalOrder(root.right, hd+1, m)
 
# The main function to print vertical order of a binary
#tree ith given root
def printVerticalOrder(root):
     
    # Create a map and store vertical order in map using
    # function getVerticalORder()
    m = dict()
    hd = 0
    getVerticalOrder(root, hd, m)
     
    # Traverse the map and print nodes at every horizontal
    # distance (hd)
    for index, value in enumerate(sorted(m)):
        for i in m[value]:
            print i,
        print
 
 
# Driver program to test above function
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
root.right.left.right = Node(8)
root.right.right.right = Node(9)
print "Vertical order traversal is"
printVerticalOrder(root)
 
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

Output:

Vertical order traversal is
4
2
1 5 6
3 8
7
9

Time Complexity of hashing based solution can be considered as O(n) under the assumption that we have good hashing function that allows insertion and retrieval operations in O(1) time. In the above C++ implementation, map of STL is used. map in STL is typically implemented using a Self-Balancing Binary Search Tree where all operations take O(Logn) time. Therefore time complexity of above implementation is O(nLogn).

Note that the above solution may print nodes in same vertical order as they appear in tree. For example, the above program prints 12 before 9. See this for a sample run.

             1
          /     \
         2       3
        /  \    /  \
       4    5  6    7
                \  /  \
                 8 10  9 
                     \
                     11
                       \
                        12
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