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Find Mode in Binary Search Tree
Category	Difficulty	Likes	Dislikes
algorithms	Easy (56.06%)	3878	782
Tags
tree

Companies
google

Given the root of a binary search tree (BST) with duplicates, return all the mode(s) (i.e., the most frequently occurred element) in it.

If the tree has more than one mode, return them in any order.

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.
 

Example 1:


Input: root = [1,null,2,2]
Output: [2]
Example 2:

Input: root = [0]
Output: [0]
 

Constraints:

The number of nodes in the tree is in the range [1, 104].
-105 <= Node.val <= 105
 

Follow up: Could you do that without using any extra space? (Assume that the implicit stack space incurred due to recursion does not count).

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class Solution1 {
    //1. tree to Hashmap <node.val, frequent> hashmap
    //2. sort hashmap by frequent, get max 
    //3.traverse hashmap to collect node.vals by max
    HashMap<Integer, Integer> val2frequent = new HashMap<>();
    public int[] findMode(TreeNode root) {
 
        List<Integer> result = new ArrayList<>();
        inorderTraverse(root);
        
        List<Integer>frequentList = new ArrayList<>();
        for(Integer frequent :val2frequent.values()){
            frequentList.add(frequent);
        }
        Integer max = frequentList.stream().max(Integer::compareTo).get();
        for(Map.Entry<Integer, Integer> entry: val2frequent.entrySet()){
            if(entry.getValue() == max){
                result.add(entry.getKey());
            }
        }
        return result.stream().mapToInt(Integer::intValue).toArray(); 
    }

    public void inorderTraverse(TreeNode root){
        if(null == root){
            return;
        }
        inorderTraverse(root.left);
        val2frequent.put(root.val, val2frequent.getOrDefault(root.val,0)+1);
        inorderTraverse(root.right);
    }
}

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class Solution {
    // inorder traverse tree, and count frequent
    TreeNode pre = null;
    int maxfrequent = 0;
    int currentfrequent = 0;
    List<Integer> result = new ArrayList<>();
    public int[] findMode(TreeNode root) {
        inorder(root);
        return result.stream().mapToInt(Integer::intValue).toArray(); 
    }

    public void inorder(TreeNode root){
        if(null == root){
            return;
        }
        inorder(root.left);

        //第一个节点
        if(pre == null ){
            currentfrequent = 1;
        } else if(pre.val == root.val) {
            //与前一个节点值相同
            currentfrequent++;
        } else {
            //与前一个节点值不相同的新节点
            currentfrequent = 1;
        } 
        if(currentfrequent > maxfrequent){    
            //更新
            result.clear();
            result.add(root.val);
            maxfrequent = currentfrequent;
        } else if(currentfrequent == maxfrequent){
            //添加
            result.add(root.val);
        }
        //更新pre节点
        pre = root;
        inorder(root.right);
    }
}