Mirror image of a binary tree

Question

Suppose there is a tree:

             1
            / \\
           2   3
              / \\
             4   5

Then the mirror image will

Answer 1

Here's a non-recursive way of doing it in python using queues. The Queue class may be initialised like this:

class Queue(object):
    def __init__(self):
        self.items = []

    def enqueue(self, item):
        self.items.insert(0, item)

    def dequeue(self):
        if not self.is_empty():
            return self.items.pop()

    def is_empty(self):
        return len(self.items) == 0

    def peek(self):
        if not self.is_empty():
            return self.items[-1]

    def __len__(self):
        return self.size()

    def size(self):
        return len(self.items)

The class which represents a node:

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

And here's the method along with traversal example which does the task:

class BinaryTree(object):
    def __init__(self, root):
        self.root = Node(root)

    def inorder(self, start, traversal):
        if start:
            traversal = self.inorder(start.left, traversal)
            traversal += f"{start.val} "
            traversal = self.inorder(start.right, traversal)
        return traversal


def mirror_tree_iterative(root):
    if root is None:
        return

    q = Queue()
    q.enqueue(root)

    while not q.is_empty():
        curr = q.peek()
        q.dequeue()
        curr.left, curr.right = curr.right, curr.left
        if curr.left:
            q.enqueue(curr.left)
        if curr.right:
            q.enqueue(curr.right)        

tree = BinaryTree(1)
tree.root.left = Node(2)
tree.root.right = Node(3)
tree.root.left.left = Node(4)
tree.root.left.right = Node(5)
tree.root.right.left = Node(6)

print(tree.inorder(tree.root, ''))
mirror_tree_iterative(tree.root)
print(tree.inorder(tree.root, ''))