AC代碼：

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)
解題思路：為了不使用額外的存儲(chǔ)空間，需要按照遞增的順序遍歷BST，中序遍歷可滿足遍歷的順序遞增，增加三個(gè)變量，tmp，max及num，tmp記錄的是上次訪問的數(shù)，num記錄的是連續(xù)相等數(shù)，max記錄的是最大的num。將num與max比較，若相等且ans中沒有該node，則將node加入ans中。若num大于max，則將ans清零，且加入該node，max換為num。