题目链接
https://leetcode.com/problems/invert-binary-tree/
题目原文
Invert 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.
题目翻译
翻转二叉树。
花絮：
Max Howell 在 tweet 中说：
谷歌：虽然我们的工程师中有90%的人在用你写的软件，但你居然不能再白板上写个翻转二叉树的代码，滚吧。
思路方法
思路一
（DFS）递归算法，每次递归交换当前节点的左右子树，同时对左右子树做同样的处理。
代码一
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def invertTree(self, root):
"""
:type root: TreeNode
:rtype: TreeNode
"""
if root == None:
return root
tmp = root.left
root.left = self.invertTree(root.right)
root.right = self.invertTree(tmp)
return root
既然这（可能）是一道谷歌的面试题，我们要尝试写出不同的代码来应付可能的题目限制，比如不准用递归，那我们用栈好了。。。
代码二
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def invertTree(self, root):
"""
:type root: TreeNode
:rtype: TreeNode
"""
if root == None:
return root
stack = [root]
while len(stack) != 0:
node = stack.pop()
node.left, node.right = node.right, node.left
if node.left:
stack.append(node.left)
if node.right:
stack.append(node.right)
return root
def invertTree(self, root):
#stack初始化
stack = [root]
while stack:
#循环做如下动作：pop会删除list末尾的元素，做操作，新的元素也加在list末尾
node = stack.pop()
if node:
node.left, node.right = node.right, node.left
stack += node.left, node.right #向stack中加入待处理节点
return root
思路二
（BFS）除了上面的用栈的解法，用队列也可以解决该问题。先将根节点入队，交换左右节点并将非空的节点加入队列，再将根节点出队，这样循环下去即可。
代码
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def invertTree(self, root):
"""
:type root: TreeNode
:rtype: TreeNode
"""
if root == None:
return root
q = [root]
while len(q) != 0:
q[0].left, q[0].right = q[0].right, q[0].left
if q[0].left:
q.append(q[0].left)
if q[0].right:
q.append(q[0].right)
del q[0]
return root
原文：
http://blog.csdn.net/coder_orz/article/details/51383933