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LeetCode-Solutions/Python/invert-binary-tree.py at master · tehsints/LeetCode-Solutions · GitHub
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invert-binary-tree.py
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LeetCode-Solutions
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Python
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invert-binary-tree.py
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# Time: O(n)
# Space: O(h)
import
collections
# BFS solution.
class
Queue
(
object
):
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
(
object
):
# @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
(
object
):
# @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
(
object
):
# @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
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