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)
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> findMode(TreeNode* root) {
vector<int> res;
int mx = 0;
unordered_map<int, int> m;
inorder(root, m, mx);
for(auto a : m)
{
if(a.second == mx)
res.push_back(a.first);
}
return res;
}
void inorder(TreeNode * node, unordered_map<int, int> &m, int &mx)
{
if(!node) return;
inorder(node->left, m, mx);
mx = max(mx, ++m[node->val]);
inorder(node->right, m, mx);
}
};