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) {
unordered_map<int, int> map;
vector<int> vi;
findMode(root, map);
int max = 0;
for (auto it = map.begin(); it != map.end(); it++)
{
if (it->second > max) {
max = it->second;
}
}
for (auto it = map.begin(); it != map.end(); it++)
{
if (it->second == max) {
vi.push_back(it->first);
}
}
return vi;
}
void findMode(TreeNode* root, unordered_map<int, int>& map) {
if (root == nullptr) {
return;
}
map[root->val]++;
if (root->left) {
findMode(root->left, map);
}
if (root->right) {
findMode(root->right, map);
}
}
};