lintCode 入门题 python版解答

献给我的人生第二春，下面开始问题与答案

1. 矩阵面积
   

class Rectangle():    '''     * Define a constructor which expects two parameters width and height here.    '''    width = 0.1    height= 0.1    # write your code here    def __init__(self, width,height):        self.width = width        self.height = height        '''     * Define a public method `getArea` which can calculate the area of the     * rectangle and return.    '''    # write your code here    def getArea(self):        return self.width*self.height

2.二叉树遍历

二叉树原理点击打开链接

其它解答http://www.cnblogs.com/bozhou/p/LintCode.html

　　LintCode已实现NodeTree类，传入的参数实际为一个NodeTree类型的二叉树，没有找到方法接触源码，只能通过网页一步一步调试，分析出结构：

其中，Node节点类拥有三个属性

　　left：当前节点的左节点

　　right：当前节点的右节点

　　val：当前节点的值

class Solution:    """    @param: root: the root of tree    @return: the max node    """    maxVal = -9999    node = None    # write your code here    def maxNode(self, root):        if root == None:            return None        self.max(root)        return self.node        def max(self,node):        if node == None:            return None        if node.val > self.maxVal:            self.node = node            self.maxVal = node.val        self.max(node.left)        self.max(node.right)

3 ,整数排序

class Solution:    """    @param: A: an integer array    @return:     """    def sortIntegers(self, A):        # write your code here        for i in range(len(A)-1):            for j in range(len(A)-i-1):                if A[j]>A[j+1]:                    B=A[j]                    A[j]=A[j+1]                    A[j+1]=B        return A