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.
Example:
Input: root = [4,2,5,1,3], target = 3.714286, and k = 2

    4
   / \
  2   5
 / \
1   3

Output: [4,3]
Follow up:
Assume that the BST is balanced, could you solve it in less than O(n) runtime (where n = total nodes)?









The idea is to compare the predecessors and successors of the closest node to the target, we can use two stacks to track the predecessors and successors, then like what we do in merge sort, we compare and pick the closest one to the target and put it to the result list.

As we know, inorder traversal gives us sorted predecessors, whereas reverse-inorder traversal gives us sorted successors.

We can use iterative inorder traversal rather than recursion, but to keep the code clean, here is the recursion version.

public List<Integer> closestKValues(TreeNode root, double target, int k) {
  List<Integer> res = new ArrayList<>();

  Stack<Integer> s1 = new Stack<>(); // predecessors
  Stack<Integer> s2 = new Stack<>(); // successors

  inorder(root, target, false, s1);
  inorder(root, target, true, s2);
  
  while (k-- > 0) {
    if (s1.isEmpty())
      res.add(s2.pop());
    else if (s2.isEmpty())
      res.add(s1.pop());
    else if (Math.abs(s1.peek() - target) < Math.abs(s2.peek() - target))
      res.add(s1.pop());
    else
      res.add(s2.pop());
  }
  
  return res;
}

// inorder traversal
void inorder(TreeNode root, double target, boolean reverse, Stack<Integer> stack) {
  if (root == null) return;

  inorder(reverse ? root.right : root.left, target, reverse, stack);
  // early terminate, no need to traverse the whole tree
  if ((reverse && root.val <= target) || (!reverse && root.val > target)) return;
  // track the value of current node
  stack.push(root.val);
  inorder(reverse ? root.left : root.right, target, reverse, stack);
}