272. Closest Binary Search Tree Value II

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:

1. Consider implement these two helper functions:
   1. getPredecessor(N), which returns the next smaller node to N.
   2. getSuccessor(N), which returns the next larger node to N.
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 用inorder来做 ， bst inorder是有序的。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 //use inorder it will be a increase array
public class Solution {
    public List<Integer> closestKValues(TreeNode root, double target, int k) {
        List<Integer> res = new LinkedList<Integer>();
        getClosestKValues(res, root, target, k);
        return res;
    }
    public void getClosestKValues(List<Integer> res, TreeNode root, double target, int k){
        if(root == null) return;
        getClosestKValues(res, root.left, target, k);
        if(res.size() == k){
            if(Math.abs(res.get(0) - target) > Math.abs(root.val - target)){
                res.remove(0);
                res.add(root.val);
            }
            else
                return;
        }
        else
            res.add(root.val);
        getClosestKValues(res, root.right, target, k);
    }
}