226. Invert Binary Tree
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.
226.翻转二叉树
把一棵二叉树翻转(如图)
冷知识:
这个问题启迪于Max Howell的这个推文
谷歌:虽然我们90%的工程师用你写的软件(Homebrew),但你却不能在一却白板上翻转一个二叉树,所以滚蛋吧。
(Homebrew是Mac平台上的软件包管理器,作者是Max Howell,但Max Howell被谷歌拒绝聘用,原因是算法能力不行,所以他发了这条推文。)
思路:从图上看是要把每棵树的左右子树互换,所以可以用递归快速解决问题。对于任意一个结点,先将其左子树调用当前函数完成翻转,再将其右子树调用当前函数完成翻转。然后交换左右子树,返回这个结点即可。递归的终点是判断给的结点值是否为空,如果为空说明到达叶结点的子树了,直接返回空就行了。代码中可写的稍微简练些,先记录左子树,然后让右子树完成翻转后赋给左子树,左子树完成翻转后赋给右子树。Python中可直接两个变量一起赋值,无需记录。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
struct TreeNode* invertTree(struct TreeNode* root) {
if (root == NULL) return NULL;
struct TreeNode* left = root->left;
root->left = invertTree(root->right);
root->right = invertTree(left);
return root;
}
# 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 None
root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)
return root