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invert-binary-tree.py 1.96 KB
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Allen Liu 提交于 2018-10-13 13:24 +08:00 . update
# 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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