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Given a binary search tree (BST) with duplicates, find all the mode(s) (the most frequently occurred element) in the given BST.

Assume a BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than or equal to the node‘s key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node‘s key.
• Both the left and right subtrees must also be binary search trees.

For example:
Given BST [1,null,2,2],

   1
         2
    /
   2

return [2].

Note: If a tree has more than one mode, you can return them in any order.

Follow up: Could you do that without using any extra space? (Assume that the implicit stack space incurred due to recursion does not count).

Solution: 1. use dictionary to store and count the frequency here we use recursive way

 1  if root is None:
 2             return []
 3         dic = {}
 4         
 5         def averageOfLevelsHelper(root):
 6             
 7             if root is not None:
 8                 # print("root: ", root.val)
 

 9                 if root.val not in dic:
10                     dic[root.val] = 1
11                 else:
12                     dic[root.val] += 1
13             if root and root.left is not None:
14                 averageOfLevelsHelper(root.left)
15             if root and root.right is not None:
16                 averageOfLevelsHelper(root.right)
17         averageOfLevelsHelper(root)
18         resLst = []
19         
20         maxV = max([v for v in dic.values()])
21         for k, v in dic.items():
22             if v == maxV:
23                 resLst.append(k)
24         return resLst

2. o(1) space use inorder traversal. We find After the traversal, the node values is sorted in nondescending order maxCnt to record current maximum times among node values use currCnt to record the current node visiting use preVal to record the node value which visited last before traversal use indexAnswer the record the index position of the result list to fetch the final answer. so we have to judge the preVal and current Val, and get the repeated number to update the maxCnt and indexAnswer. Refer to the following code:

 1       if root is None:
 2             return []
 3         st = []
 4         p = TreeNode(-1*2**32)
 5         p.right = root
 6         st.append(p)
 7         top = st[0]
 8         
 9         maxCnt = 1
10         currCnt = 1
11         preVal = -1
12         resLst = []
13         indexAnswer = 0      #indexAnswer in resLst
14         i = 0                #the index of visiting list
15         while (st):
16             top = st.pop()
17             if top.val != preVal:
18                 currCnt = 1
19                 if top.val != -1*2**32 and currCnt >= maxCnt:
20                     resLst.append(top.val)
21                     #print ("resLst: ", resLst)
22                 
23             else:
24                 currCnt += 1
25                 if currCnt > maxCnt:
26                     maxCnt = currCnt
27                     indexAnswer = len(resLst)
28                     resLst.append(top.val)
29                 elif currCnt == maxCnt:
30                     resLst.append(top.val)
31 
32             preVal = top.val
33             if top.right:
34                 tmp = top.right
35                 while (tmp):
36                     st.append(tmp)
37                     tmp = tmp.left
38         
39         return resLst[indexAnswer::]
40         

[Leetcode] Binary tree -- 501. Find Mode in Binary Search Tree