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).

M1: using extra space

traverse的時候，用hashmap存每個節點出現的次數

time: O(n), space: O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int max = 0;
    
    
 
public int[] findMode(TreeNode root) {
        if(root == null) {
            return new int[] {};
        }
        Map<Integer, Integer> map = new HashMap<>();
        traverse(root, map);
        
        List<Integer> tmp = new ArrayList<>();
        for(Map.Entry<Integer, Integer> entry : map.entrySet()) {
            if(entry.getValue() == max) {
                tmp.add(entry.getKey());
            }
        }
        
        int[] res = new int[tmp.size()];
        for(int i = 0; i < tmp.size(); i++) {
            res[i] = tmp.get(i);
        }
        return res;
    }
    
    public void traverse(TreeNode root, Map<Integer, Integer> map) {
        if(root == null) {
            return;
        }
        map.put(root.val, map.getOrDefault(root.val, 0) + 1);
        max = Math.max(max, map.get(root.val));
        traverse(root.left, map);
        traverse(root.right, map);
    }
}

M2: optimized, not using extra space