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)?

一刷
題解：
我們用兩個(gè)棧磨镶，全部采用inorder訪問樹溃蔫。
一個(gè)棧從小到大，存儲(chǔ)<= target的節(jié)點(diǎn)
一個(gè)棧由大到小琳猫，存儲(chǔ)>target的節(jié)點(diǎn)伟叛。
然后比較兩個(gè)棧頭的值，存儲(chǔ)距離較小的點(diǎn)脐嫂，一直進(jìn)行k次统刮。

/**
 * 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) {
        List<Integer> res = new ArrayList<>();
        
        Stack<Integer> s1 = new Stack<>();//predecessors
        Stack<Integer> s2 = new Stack<>();//successors
        if(root == null) return res;
        
        inorder(root, target, false, s1);//ascending, and store the node with value lower than or equal to target
        inorder(root, target, true, s2);//descending
        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;
    }
    
    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);
    }
}

二刷
題解：思路同上。但是我們這次用兩個(gè)array和index來模擬上述行為账千。
inorder traverse,
如果root.val小于target, append在smaller尾部
否則侥蒙，append在bigger尾部
所以smaller尾部和bigger的頭部更滿足要求。

/**
 * 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) {
        List<Integer> smaller = new ArrayList<>();
        List<Integer> bigger = new ArrayList<>();
        //inorder
        inorder(root, target, smaller, bigger);
        int sIndex = smaller.size()-1;
        int bIndex = 0;
        List<Integer> res = new ArrayList<>(k);
        int index = 0;
        while(res.size()<k){
            if(sIndex<0){
                res.add(bigger.get(bIndex++));
            }else if(bIndex == bigger.size()){
                res.add(smaller.get(sIndex--));
            }else{
                if(Math.abs(smaller.get(sIndex) - target) < Math.abs(bigger.get(bIndex) - target)){
                    res.add(smaller.get(sIndex--));
                }else res.add(bigger.get(bIndex++));
            }
        }
        return res;
    }
    
    private void inorder(TreeNode root, double target, List<Integer> smaller, List<Integer> bigger){
        if(root == null) return;
        inorder(root.left, target, smaller, bigger);
        if(root.val<target) smaller.add(root.val);
        else bigger.add(root.val);
        inorder(root.right, target, smaller, bigger);
    }
}