Leetcode 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)?

1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public int val;
 5  *     public TreeNode left;
 6  *     public TreeNode right;
 7  *     public TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public IList<int> ClosestKValues(TreeNode root, double target, int k) {
12         var result = new List<int>();
13         
14         Stack<int> s1 = new Stack<int>(), s2 = new Stack<int>();
15         Inorder(root, target, true, s1);
16         Inorder(root, target, false, s2);
17         
18         for (int i = 0; i < k; i++)
19         {
20             if (s1.Count == 0)
21             {
22                 result.Add(s2.Pop());
23             }
24             else if (s2.Count == 0)
25             {
26                 result.Add(s1.Pop());
27             }
28             else
29             {
30                 double d1 = Math.Abs((double)s1.Peek() - target);
31                 double d2 = Math.Abs((double)s2.Peek() - target);
32                 
33                 if (d1 < d2)
34                 {
35                     result.Add(s1.Pop());
36                 }
37                 else
38                 {
39                     result.Add(s2.Pop());
40                 }
41             }
42         }
43         
44         return result;
45     }
46     
47     
48     private void Inorder(TreeNode node, double target, bool reverse, Stack<int> stack)
49     {
50         if (node == null) return;
51         
52         Inorder(reverse ? node.right : node.left, target, reverse, stack);
53         
54         if (reverse ? (double)node.val <= target : (double)node.val > target)
55         {
56             return;
57         }
58         
59         stack.Push(node.val);
60         
61         Inorder(reverse ? node.left : node.right, target, reverse, stack);
62     }
63 }      

轉載于:https://www.cnblogs.com/liangmou/p/8071212.html