226. Invert Binary Tree [easy] (Python)

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

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

PS: 新手刷LeetCode，新手写博客，写错了或者写的不清楚还请帮忙指出，谢谢！
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