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.

問題不復雜，就是一個遍歷加map的問題

基本做法如下

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int max = 0;
    public int[] findMode(TreeNode root) {
        if(root == null){
            return new int[0];
        }
        Map<Integer, Integer> map = new HashMap<>();
        helper(map, root);
        
        List<Integer> list = new LinkedList<>();
        for(int key: map.keySet()){
            if(map.get(key) == max) list.add(key);
        }
        
        int[] res = new int[list.size()];
        for(int i = 0; i<res.length; i++) res[i] = list.get(i);
        return res;
    }
    
    
    private void helper(Map<Integer, Integer> map, TreeNode root){
        if(root == null){
            return;
        }
        if(map.containsKey(root.val)){
            map.put(root.val, map.get(root.val) + 1);
        }else{
            map.put(root.val, 1);
        }
        max = Math.max(max, map.get(root.val));
        helper(map, root.right);
        helper(map, root.left);
    }
}
 

大大的做法