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:

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).
因為這道題左子節點可能等于根節點和右子節點，宜使用中序遍歷求解。此外，這道題限制空間復雜度O(1)，不少sulotion采用hashmap或者list，如果遇到1，2，3，4...n-1，n，n，n的情況，空間復雜度會變成O(n)。所以應該先得到共有幾個modes，再申請空間。代碼如下：
/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    PRivate int max = 0;    private int currval = 0;    private int currcount = 0;    private int currmodes = 0;    private int[] modes;        public int[] findMode(TreeNode root) {        helper(root);        modes = new int[currmodes];        currcount = 0;        currmodes = 0;        helper(root);        return modes;    }        public void helper(TreeNode root) {        if (root == null) {            return;        }        helper(root.left);        if (root.val != currval) {            currval = root.val;            currcount = 1;        } else {            currcount ++;        }        if (currcount > max) {            max = currcount;            currmodes = 1;        }else if (currcount == max) {            if (modes != null) {                modes[currmodes] = root.val;            }            currmodes ++;        }        helper(root.right);    }}