[leetcode]272. Closest Binary Search Tree Value II二叉搜索树中最近的值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.

Example:

Input: root = [4,2,5,1,3], target = 3.714286, and k = 2

    4
   / \
  2   5
 / \
1   3

Output: [4,3]

Follow up:
Assume that the BST is balanced, could you solve it in less than O(n) runtime (where n = total nodes)?

 

题意

和之前一样，不过这次要找的是最接近的k个值。

 

Solution1：

1.  Based on BST's attributes, if we traversal BST inorder,  each node will be in acsending order

2. we choose a data structure which can help operate on both sides(LinkedList or Deque), maintaining a K size sliding window

   once there is a new item, 

   we check diff (1) next item vs target

                      (2) leftMost item vs target

 

 

code

 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 
11 /*
12 Time Complexity: O(n)
13 Space Complexity:O(k)
14 */
15 class Solution {
16     // choose a data structure which can do operations on both sides
17     LinkedList result = new LinkedList<>();
18     
19     public List closestKValues(TreeNode root, double target, int k) {
20         // corner case
21         if (root == null) {return null;}
22         // inorder traversal
23         closestKValues(root.left, target, k);
24     
25         if (result.size() < k) {
26             result.add(root.val);
27             // maintain a K size sliding window such that items are closest to the target
28         } else if(result.size() == k) {
29             if (Math.abs(result.getFirst() - target) > (Math.abs(root.val - target))) {
30                 result.removeFirst();
31                 result.addLast(root.val);
32             } 
33         }
34         // inorder traversal
35         closestKValues(root.right, target, k);
36         return result;
37     }    
38 }

 

转载于:https://www.cnblogs.com/liuliu5151/p/10767955.html