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)?

Solution：two Stack

思路： in order 正反two pass, 分别put至stack，再从stacks求得结果
Time Complexity: O(N) Space Complexity: O(N)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    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
    private 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);
    }
}