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# Time: O(n)

# Space: O(h)

#

# Invert a binary tree.

#

# 4

# / \

# 2 7

# / \ / \

# 1 3 6 9

# to

# 4

# / \

# 7 2

# / \ / \

# 9 6 3 1

#

# Time: O(n)

# Space: O(w), w is the max number of the nodes of the levels.

import collections

# BFS solution.

class Queue:

def __init__(self):

self.data = collections.deque()

def push(self, x):

self.data.append(x)

def peek(self):

return self.data[0]

def pop(self):

return self.data.popleft()

def size(self):

return len(self.data)

def empty(self):

return len(self.data) == 0

# Definition for a binary tree node.

# class TreeNode:

# def __init__(self, x):

# self.val = x

# self.left = None

# self.right = None

class Solution:

# @param {TreeNode} root

# @return {TreeNode}

def invertTree(self, root):

if root is not None:

nodes = Queue()

nodes.push(root)

while not nodes.empty():

node = nodes.pop()

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

if node.left is not None:

nodes.push(node.left)

if node.right is not None:

nodes.push(node.right)

return root

# Time: O(n)

# Space: O(h)

# Stack solution.

class Solution2:

# @param {TreeNode} root

# @return {TreeNode}

def invertTree(self, root):

if root is not None:

nodes = []

nodes.append(root)

while nodes:

node = nodes.pop()

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

if node.left is not None:

nodes.append(node.left)

if node.right is not None:

nodes.append(node.right)

return root

# Time: O(n)

# Space: O(h)

# DFS, Recursive solution.

class Solution3:

# @param {TreeNode} root

# @return {TreeNode}

def invertTree(self, root):

if root is not None:

root.left, root.right = self.invertTree(root.right), \

self.invertTree(root.left)

return root