new_contents: # Time:  O(n)
  # Space: O(1)
  
  # 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).
  
  # Definition for a binary tree node.
  # class TreeNode(object):
  #     def __init__(self, x):
  #         self.val = x
  #         self.left = None
  #         self.right = None
  
  class Solution(object):
      def findMode(self, root):
          """
          :type root: TreeNode
          :rtype: List[int]
          """
          def inorder(root, prev, cnt, max_cnt, result):
              if not root:
                  return prev, cnt, max_cnt
      
              prev, cnt, max_cnt = inorder(root.left, prev, cnt, max_cnt, result)
              if prev:
                  if root.val == prev.val:
                      cnt += 1
                  else:
                      cnt = 1
              if cnt > max_cnt:
                  max_cnt = cnt;
                  del result[:]
                  result.append(root.val)
              elif cnt == max_cnt:
                  result.append(root.val)
              return inorder(root.right, root, cnt, max_cnt, result)
          
          if not root:
              return []
          result = []
          inorder(root, None, 1, 0, result);
          return result
  
subject: Create find-mode-in-binary-search-tree.py