Find mode in Binary Search Tree

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: 题目要求O(1)的空间复杂度，所以我们需要先得到有多少个modes，再申请空间。所产生的stack space不计入空间复杂度，因此可用递归。

/**
 * 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 currentModes = 0;
    private int currentValue = 0;
    private int currentCount = 0;
    private int modes[];
    private int maxCount = 0;
    
    public int[] findMode (TreeNode root) {
        helper(root);
        modes = new int[currentModes];
        currentModes = 0;
        currentCount = 0;
        helper(root);
        return modes;
    }
    
    private void helper (TreeNode root) {
        if (root == null) return;
        
        helper(root.left);
        
        if (root.val != currentValue) {
            currentCount = 1;
            currentValue = root.val;
        } else {
            currentCount++;
        }
        
        if (currentCount > maxCount) {
            maxCount = currentCount;
            currentModes = 1;
        } else if (currentCount == maxCount) {
            if (modes != null)
                modes[currentModes] = root.val;
            currentModes++;
        }
        
        helper(root.right);
    }
}