**Difficulty: Hard**

**Frequency: N/A**

Given a non-empty binary search tree and a target value, find *k* values in the BST that are closest to the target.

**Note:**

- Given target value is a floating point.
- You may assume
*k*is always valid, that is:*k*≤ total nodes. - You are guaranteed to have only one unique set of
*k*values in the BST that are closest to the target.

**Follow up:**

Assume that the BST is balanced, could you solve it in less than *O*(*n*) runtime (where *n* = total nodes)?

**Hint:**

- Consider implement these two helper functions:
`getPredecessor(N)`

, which returns the next smaller node to N.`getSuccessor(N)`

, which returns the next larger node to N.

- Try to assume that each node has a parent pointer, it makes the problem much easier.
- Without parent pointer we just need to keep track of the path from the root to the current node using a stack.
- You would need two stacks to track the path in finding predecessor and successor node separately.

**My solution:**

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class Solution { vector<int> closestKValues(TreeNode* root, double target, int k) { priority_queue<pair<double, int>> pq; vector<int> result; dfs(root, pq, target, k); while (!pq.empty()) { result.push_back(pq.top().second); pq.pop(); } return result; } void dfs(TreeNode* node, priority_queue<pair<double, int>>& pq, double target, int k) { if (!node) return; pq.push(make_pair(abs(target - (double)node->val), node->val)); if (pq.size() > k) pq.pop(); dfs(node->left, pq, target, k); dfs(node->right, pq, target, k); } };

**Another solution:**

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**Things to learn:**