【Leetcode】【python】Hamming Distance, Merge Two Binary Trees

Hamming Distance

题目大意

两个整数的汉明距离是指其二进制不相等的位的个数。
给定两个整数x和y，计算汉明距离。
注意：
0 ≤ x, y < 2^31.

解题思路

异或运算

代码

class Solution(object):    def hammingDistance(self, x, y):        """        :type x: int        :type y: int        :rtype: int        """        return bin(x ^ y).count('1')

我提交的

class Solution(object):    def hammingDistance(self, x, y):        """        :type x: int        :type y: int        :rtype: int        """        x_bin = list(bin(x))        y_bin = list(bin(y))        x_bin.reverse()        y_bin.reverse()        x_judge = []        y_judge = []        for x in x_bin:            if x == 'b':                break                x_judge.append(x)        for y in y_bin:            if y == 'b':                break                y_judge.append(y)        list_distance = abs(len(x_judge) - len(y_judge))        if len(x_judge) < len(y_judge):            for i in range(list_distance):                x_judge.append('0')        else:            for i in range(list_distance):                y_judge.append('0')        distance = 0        for i in range(len(x_judge)):            if x_judge[i] != y_judge[i]:                distance += 1        return distance 

总结

以后谁和我说二进制异或运算我和谁急

Merge Two Binary Trees

题目大意

给两个二叉树想要合并，有一些结点会重叠而有一些不会，现在想把重叠的结点值变为两者值相加，不重叠的直接用，构建出新的树

解题思路

考察的就是二叉树的遍历，遍历每个结点然后如果重叠（两个二叉树结点都不为空）新结点值便为两者和，不重叠（只有一个结点为空）新结点值为不为空的值，全为空到达底部返跳出。按照这个逻辑进行迭代

联想：二叉树遍历方式有深度优先和广度优先，深度（纵向）优先在Python中一般使用列表，广度优先（横向）一般使用迭代

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 mergeTrees(self, t1, t2):        """        :type t1: TreeNode        :type t2: TreeNode        :rtype: TreeNode        """        if t1 is None and t2 is None:            return        if t1 is None:            return t2        if t2 is None:            return t1        t1.val += t2.val        t1.right = self.mergeTrees(t1.right, t2.right)        t1.left = self.mergeTrees(t1.left, t2.left)        return t1

总结

此题做时没有理解这个TreeNode类是怎么用的，看了答案才明白。
后来尝试了：

t1.right.val>> 2t1.left.left.val>> 5

该答案应该是广度优先的迭代方法。