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：中序遍歷

思路：
Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

class Solution {
    Integer prev = null;
    int count = 1;
    int max = 0;
    public int[] findMode(TreeNode root) {
        if (root == null) return new int[0];
        
        List<Integer> list = new ArrayList<>();
        traverse(root, list);
        
        int[] res = new int[list.size()];
        for (int i = 0; i < list.size(); ++i) res[i] = list.get(i);
        return res;
    }
    
    private void traverse(TreeNode root, List<Integer> list) {
        if (root == null) return;
        traverse(root.left, list);
        if (prev != null) {
            if (root.val == prev)
                count++;
            else
                count = 1;
        }
        if (count > max) {
            max = count;
            list.clear();
            list.add(root.val);
        } else if (count == max) {
            list.add(root.val);
        }
        prev = root.val;
        traverse(root.right, list);
    }
}