Leetcode 494 python 解题报告

AC代码：

class Solution(object):    def findMode(self, root):        """        :type root: TreeNode        :rtype: List[int]        """        self.ans = []        self.max = 1        if root == None:            return []        r = root        self.tmp = root.val+1        self.num = 0        while r.left:            r = r.left            self.tmp = r.val+1        self.findnum(root)        return self.ans    def findnum(self, node):        if node.left != None:            self.findnum(node.left)        if node.val == self.tmp:            self.num += 1        else:            self.num = 1        self.tmp = node.val        if self.num > self.max:                self.ans = [node.val]                self.max = self.num        elif self.num == self.max:            if node.val not in self.ans:                self.ans.append(node.val)        if node.right != None:            self.findnum(node.right)
解题思路：为了不使用额外的存储空间，需要按照递增的顺序遍历BST，中序遍历可满足遍历的顺序递增，增加三个变量，tmp，max及num，tmp记录的是上次访问的数，num记录的是连续相等数，max记录的是最大的num。将num与max比较，若相等且ans中没有该node，则将node加入ans中。若num大于max，则将ans清零，且加入该node，max换为num。