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272. Closest Binary Search Tree Value II 👍

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  vector<int> closestKValues(TreeNode* root, double target, int k) {
    deque<int> dq;

    inorder(root, dq);

    while (dq.size() > k)
      if (abs(dq.front() - target) > abs(dq.back() - target))
        dq.pop_front();
      else
        dq.pop_back();

    return {dq.begin(), dq.end()};
  }

 private:
  void inorder(TreeNode* root, deque<int>& dq) {
    if (root == nullptr)
      return;

    inorder(root->left, dq);
    dq.push_back(root->val);
    inorder(root->right, dq);
  }
};

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class Solution {
  public List<Integer> closestKValues(TreeNode root, double target, int k) {
    Deque<Integer> dq = new ArrayDeque<>();

    inorder(root, dq);

    while (dq.size() > k)
      if (Math.abs(dq.peekFirst() - target) > Math.abs(dq.peekLast() - target))
        dq.pollFirst();
      else
        dq.pollLast();

    return new ArrayList<>(dq);
  }

  private void inorder(TreeNode root, Deque<Integer> dq) {
    if (root == null)
      return;

    inorder(root.left, dq);
    dq.offerLast(root.val);
    inorder(root.right, dq);
  }
}

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class Solution:
  def closestKValues(self, root: Optional[TreeNode], target: float, k: int) -> List[int]:
    dq = collections.deque()

    def inorder(root: Optional[TreeNode]) -> None:
      if not root:
        return

      inorder(root.left)
      dq.append(root.val)
      inorder(root.right)

    inorder(root)

    while len(dq) > k:
      if abs(dq[0] - target) > abs(dq[-1] - target):
        dq.popleft()
      else:
        dq.pop()

    return list(dq)