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).

public class Solution {
    Map<Integer, Integer> map; 
    int max = 0;
    public int[] findMode(TreeNode root) {
        if(root==null) return new int[0]; 
        this.map = new HashMap<>(); 
        
        inorder(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 inorder(TreeNode node){
        if(node.left!=null) inorder(node.left);
        map.put(node.val, map.getOrDefault(node.val, 0)+1);
        max = Math.max(max, map.get(node.val));
        if(node.right!=null) inorder(node.right); 
    }
}