272. Closest Binary Search Tree Value II

Description

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 thanO(n) runtime (where n= total nodes)?

Discussion

Method 1

O(n) sliding window

Method 2

O(n) 2 stack / deque (LeetCode)

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);
}