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# https://leetcode.com/problems/invert-binary-tree/

# convert a binary tree.

#

# 4

# / \

# 2 7

# / \ / \

# 1 3 6 9

# to

# 4

# / \

# 7 2

# / \ / \

# 9 6 3 1

#

# Trivia:

# This problem was inspired by this original tweet by Max Howell:

# Google: 90% of our engineers use the software you wrote (Homebrew),

# but you can’t invert a binary tree on a whiteboard so fuck off.

# Definition for a binary tree node.

# class TreeNode(object):

# def __init__(self, x):

# self.val = x

# self.left = None

# self.right = None

# The solution to this problem involves using a Queue

# Step1 : Create a Queue with enqueue , dequeue and isempty functionality. These will come handy while

# inverting the binary tree

class Queue():

def __init__(self):

self.items = []

def enqueue(self,key):

self.key = key

self.items.insert(0,key)

def dequeue(self):

return self.items.pop()

def empty(self):

return self.items

def isempty(self):

return len(self.items)==0

def print_q(self):

return self.items

class Solution(object):

def invertTree(self, root):

"""

:type root: TreeNode

:rtype: TreeNode

"""

# Step 2 : if the tree is not an empty tree , enqueue the root into the Queue

# Step 3 : dequeue an element from the Queue , swap its left and right children

# Step 4 : enqueue these children that you just swapped in the Queue , as long as they are not None

# Step 5 : dequeue the next node from the Queue and swap its children .. Keep going.

if root is not None:

my_queue = Queue()

my_queue.enqueue(root)

while not my_queue.isempty():

node = my_queue.dequeue()

node.left , node.right = node.right , node.left

if node.left is not None:

my_queue.enqueue(node.left)

if node.right is not None:

my_queue.enqueue(node.right)

return root