Find Mode in Binary Search Tree：带重复元素的搜索二叉树Morris遍历查找频率最大的元素

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).

思路：

因为可能有多个个数相同的元素且频率最大，所以搜索两遍：第一遍判断有几个元素是个数最大的（1或者n），方便最后返回结果；第二遍把个数最大的元素保存下来。

由于空间复杂度有限制，所以采用Morris  Traversal（线索二叉树）遍历二叉树，这样不适用额外的空间（普通递归或非递归遍历使用栈O（n））。

空间复杂度:O（1）

时间复杂度:O（n）。（因为Morris Traversal时间复杂度是O（n））

Morris  Traversal 可以看这篇：http://www.cnblogs.com/AnnieKim/archive/2013/06/15/MorrisTraversal.html

写的听明白的，也不复杂