Version I - hash map + traversal

in this way, there's no difference between each traversal.

public class Solution {
    private int maxOccur;

    public int[] findMode(TreeNode root) {
        Map<Integer, Integer> map = new HashMap<>();

        List<Integer> list = new ArrayList<>();

        inOrderTraversal(root, map);

        for (int key : map.keySet()) {
            if (map.get(key) == maxOccur) {
                list.add(key);
            }
        }

        int[] result = new int[list.size()];
        for (int i = 0; i < list.size(); ++i) result[i] = list.get(i);

        return result;
    }

    private void inOrderTraversal(TreeNode node, Map<Integer, Integer> map) {
        //exit
        if (node == null) return;

        map.put(node.val, map.getOrDefault(node.val, 0) + 1);
        maxOccur = Math.max(maxOccur, map.get(node.val));

        inOrderTraversal(node.left, map);
        inOrderTraversal(node.right, map);

    }
}

Version II - what makes BST unique? - in order traversal

do inorder traversal

since the traversal should be an non-descending order (or equal), then, check the value with previous node, if equal, then count++, else reset count

public class Solution {

    private TreeNode prevNode;
    private int count = 1;
    private int max;

    public int[] findMode(TreeNode root) {

        List<Integer> list = new ArrayList<>();

        inOrderTraversal(root, list);

        int[] result = new int[list.size()];

        for (int i = 0; i < result.length; ++i) result[i] = list.get(i);

        return result;
    }

    private void inOrderTraversal(TreeNode node, List<Integer> list) {
        if (node == null) return;

        inOrderTraversal(node.left, list);

        //compare with prev
        if (prevNode != null) {
            if (prevNode.val == node.val) {
                count++;
            } else {
                count = 1;
            }
        }

        if (count > max) {
            max = count;
            list.clear();
            list.add(node.val);
        } else if (count == max) {
            list.add(node.val);
        }

        prevNode = node;

        inOrderTraversal(node.right, list);
    }
}