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 = Noneclass 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 = Noneclass 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 = Noneclass 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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