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)?

links : 94 Binary Tree Inorder Traversal

 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
public class Solution {
    public List<Integer> closestKValues(TreeNode root, double target, int k) {
        Queue<Integer> list = new LinkedList<>();

        Stack<TreeNode> stack = new Stack<>();
        TreeNode node = root;
        while(!stack.isEmpty() || node != null){
            while(node != null){
                node = node.left;

            node = stack.pop();

            if(list.size() < k){
                if(Math.abs(list.peek() - target) > Math.abs(node.val - target)){

            node = node.right;

        return (List<Integer>) list;

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