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)

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List closestKValues(TreeNode root, double target, int k) {
12         List res = new ArrayList();
13         LinkedList stack = new LinkedList();
14         LinkedList queue = new LinkedList();
15         inorder(root, target, stack, queue);
16         for (int i=0; i) {
17             if (stack.isEmpty()) res.add(queue.poll());
18             else if (queue.isEmpty()) res.add(stack.pop());
19             else {
20                 int s = stack.peek();
21                 int q = queue.peek();
22                 if (Math.abs(s-target) < Math.abs(q-target)) {
23                     res.add(s);
24                     stack.pop();
25                 }
26                 else {
27                     res.add(q);
28                     queue.poll();
29                 }
30             }
31         }
32         return res;
33     }
34     
35     public void inorder(TreeNode root, double target, LinkedList stack, LinkedList queue) {
36         if (root == null) return;
37         inorder(root.left, target, stack, queue);
38         if (root.val < target) {
39             stack.push(root.val);
40         }
41         else {
42             queue.offer(root.val);
43         }
44         inorder(root.right, target, stack, queue);
45     }
46 }

 

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。

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List closestKValues(TreeNode root, double target, int k) {
12         //List res = new ArrayList();
13         LinkedList stack = new LinkedList();
14         LinkedList queue = new LinkedList();
15         
16         while (!stack.isEmpty() || root!=null) {
17             if (root != null) {
18                 stack.push(root);
19                 root = root.left;
20             }
21             else {
22                 root = stack.pop();
23                 if (queue.size() < k) {
24                     queue.offer(root.val);
25                 }
26                 else {
27                     if (Math.abs(root.val-target) < Math.abs(queue.peek()-target)) {
28                         queue.poll();
29                         queue.offer(root.val);
30                     }
31                     else break;
32                 }
33                 root = root.right;
34             }
35         }
36 
37         return (List)queue;
38     }
39     
40 }