Given a binary search tree (BST) with duplicates, find all themode(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:

//O(n) space: hashmap

//O(1) space: in-order traversal, 记录之前的node的val,如果当前root.val与它相等，count++，否则count=1。然后count于max比较，若大于max，则清空list，然后加入root.val。

    Integer prev;
    int max = 0;
    int count = 1;
    public int[] findMode(TreeNode root) {
        if(root == null){
            return new int[0];
        }
        List<Integer> list = new ArrayList<Integer>();
        traverse(root, list);
        int[] ans = new int[list.size()];
        for(int i = 0; i < ans.length; i++){
            ans[i] = list.get(i);
        }
        return ans;
    }
    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);
    }