Leetcode: 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.
2. Try to assume that each node has a parent pointer, it makes the problem much easier.
3. Without parent pointer we just need to keep track of the path from the root to the current node using a stack.
4. You would need two stacks to track the path in finding predecessor and successor node separately.

 用一个栈一个队列，用recursion写inorder Traversal, Time: O(N+K), Space: O(N)

/**
 * 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<Integer>();
        LinkedList<Integer> stack = new LinkedList<Integer>();
        LinkedList<Integer> queue = new LinkedList<Integer>();
        inorder(root, target, stack, queue);
        for (int i=0; i<k; i++) {
            if (stack.isEmpty()) res.add(queue.poll());
            else if (queue.isEmpty()) res.add(stack.pop());
            else {
                int s = stack.peek();
                int q = queue.peek();
                if (Math.abs(s-target) < Math.abs(q-target)) {
                    res.add(s);
                    stack.pop();
                }
                else {
                    res.add(q);
                    queue.poll();
                }
            }
        }
        return res;
    }
    
    public void inorder(TreeNode root, double target, LinkedList<Integer> stack, LinkedList<Integer> queue) {
        if (root == null) return;
        inorder(root.left, target, stack, queue);
        if (root.val < target) {
            stack.push(root.val);
        }
        else {
            queue.offer(root.val);
        }
        inorder(root.right, target, stack, queue);
    }
}

Better Idea: 只用一个大小为K的队列的方法：Time O(N), Space O(K)

前K个数直接加入队列，之后每来一个新的数（较大的数），如果该数和目标的差，相比于队头的数离目标的差来说，更小，则将队头拿出来，将新数加入队列。如果该数的差更大，则直接退出并返回这个队列，因为后面的数更大，差值也只会更大。

/**
 * 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<Integer>();
        LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
        LinkedList<Integer> queue = new LinkedList<Integer>();
        
        while (!stack.isEmpty() || root!=null) {
            if (root != null) {
                stack.push(root);
                root = root.left;
            }
            else {
                root = stack.pop();
                if (queue.size() < k) {
                    queue.offer(root.val);
                }
                else {
                    if (Math.abs(root.val-target) < Math.abs(queue.peek()-target)) {
                        queue.poll();
                        queue.offer(root.val);
                    }
                    else break;
                }
                root = root.right;
            }
        }

        return (List<Integer>)queue;
    }
    
}