DFS Traversal of a Tree Without using Recursion in Python

This is a Python program to perform depth-first search on a binary tree without using recursion.

Problem Description

The program creates a binary tree and presents a menu to the user to perform operations on the tree including depth-first search.

Problem Solution

1. Create a class Stack to implement a stack.
2. The class Stack will have methods is_empty, push and pop.
3. Create a class BinaryTree with instance variables key, left and right.
4. Define methods set_root, insert_left, insert_right, preorder_depth_first and search.
5. The method set_root takes a key as argument and sets the variable key equal to it.
6. The methods insert_left and insert_right insert a node as the left and right child respectively.
7. The method search returns a node with a specified key.
8. The method preorder_depth_first displays a preorder DFS traversal of the tree.
9. The preorder traversal is implemented using a stack to avoid recursion.

Program/Source Code

Here is the source code of a Python program to perform depth-first search on a binary tree without using recursion. The program output is shown below.

class BinaryTree:
    def __init__(self, key=None):
        self.key = key
        self.left = None
        self.right = None
 
    def set_root(self, key):
        self.key = key
 
    def insert_left(self, new_node):
        self.left = new_node
 
    def insert_right(self, new_node):
        self.right = new_node
 
    def search(self, key):
        if self.key == key:
            return self
        if self.left is not None:
            temp =  self.left.search(key)
            if temp is not None:
                return temp
        if self.right is not None:
            temp =  self.right.search(key)
            return temp
        return None
 
    def preorder_depth_first(self):
        s = Stack()
        s.push(self)
        while (not s.is_empty()):
            node = s.pop()
            print(node.key, end=' ')
            if node.right is not None:
                s.push(node.right)
            if node.left is not None:
                s.push(node.left)
 
 
class Stack:
    def __init__(self):
        self.items = []
 
    def is_empty(self):
        return self.items == []
 
    def push(self, data):
        self.items.append(data)
 
    def pop(self):
        return self.items.pop()
 
 
btree = BinaryTree()
 
print('Menu (this assumes no duplicate keys)')
print('insert <data> at root')
print('insert <data> left of <data>')
print('insert <data> right of <data>')
print('dfs')
print('quit')
 
while True:
    do = input('What would you like to do? ').split()
 
    operation = do[0].strip().lower()
    if operation == 'insert':
        data = int(do[1])
        new_node = BinaryTree(data)
        suboperation = do[2].strip().lower() 
        if suboperation == 'at':
                btree = new_node
        else:
            position = do[4].strip().lower()
            key = int(position)
            ref_node = None
            if btree is not None:
                ref_node = btree.search(key)
            if ref_node is None:
                print('No such key.')
                continue
            if suboperation == 'left':
                ref_node.insert_left(new_node)
            elif suboperation == 'right':
                ref_node.insert_right(new_node)
 
    elif operation == 'dfs':
        print('pre-order dfs traversal: ', end='')
        if btree is not None:
            btree.preorder_depth_first()
        print()
 
    elif operation == 'quit':
        break
Program Explanation

1. A variable is created to store the binary tree.
2. The user is presented with a menu to perform operations on the tree.
3. The corresponding methods are called to perform each operation.
4. The method preorder_depth_first is called to display a DFS pre-order traversal of the tree.

advertisement
advertisement
Runtime Test Cases
Case 1:
Menu (this assumes no duplicate keys)
insert <data> at root
insert <data> left of <data>
insert <data> right of <data>
dfs
quit
What would you like to do? insert 1 at root
What would you like to do? insert 2 left of 1
What would you like to do? insert 3 right of 1
What would you like to do? insert 4 right of 2
What would you like to do? insert 5 left of 4
What would you like to do? dfs
pre-order dfs traversal: 1 2 4 5 3 
What would you like to do? quit
 
Case 2:
Menu (this assumes no duplicate keys)
insert <data> at root
insert <data> left of <data>
insert <data> right of <data>
dfs
quit
What would you like to do? insert 3 at root
What would you like to do? insert 6 left of 3
What would you like to do? insert 7 right of 3
What would you like to do? insert 8 left of 7
What would you like to do? insert 10 right of 7
What would you like to do? dfs
pre-order dfs traversal: 3 6 7 8 10 
What would you like to do? quit

Sanfoundry Global Education & Learning Series – Python Programs.

To practice all Python programs, here is complete set of 150+ Python Problems and Solutions.

Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

If you find any mistake above, kindly email to [email protected]

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.