【LeetCode】【Python题解】Single Number & Maximum Depth of Binary Tree

标签：leetcode   python   递归   面试   遍历   

今天做了三道LeetCode上的简单题目，每道题都是用c++和Python两种语言写的，因为c++版的代码网上比较多，所以就只分享一下Python的代码吧，刚学完Python的基本语法，做做LeetCode的题目还是不错的，对以后找工作面试也有帮助！

刚开始就从AC率最高的入手吧！

1.Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

开始纠结于线性时间复杂度和不要额外的存储空间，思路是遍历一遍数组，想象成翻硬币，碰到一次就翻面，再碰到就翻回来了，剩下的那个仍是背面的就是只出现一次的。但是这种思路就势必要有额外的存储空间。后来看了别人的思路，原来全部异或就可以了，因为A XOR A =0，0 XOR any = any，简单多了。

class Solution:
    # @param A, a list of integer
    # @return an integer
    def singleNumber(self, A):
        result = 0
        for i in A:
            result = result ^ i
        return result

2.Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

递归寻找左子树和右子树的深度，取二者较大的，再加上根的深度1，即为答案。

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param root, a tree node
    # @return an integer
    def maxDepth(self, root):
        if root is None:
            return 0
        else:
            return max(self.maxDepth(root.left),self.maxDepth(root.right))+1

【LeetCode】【Python题解】Single Number & Maximum Depth of Binary Tree

标签：leetcode   python   递归   面试   遍历   

原文地址：http://blog.csdn.net/monkeyduck/article/details/37991201