[Swift]LeetCode272. 最近的二分搜索树的值 II $ Closest Binary Search Tree Value II

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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:
　　i. getPredecessor(N), which returns the next smaller node to N.
　　ii. 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.

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给定一个非空的二进制搜索树和一个目标值，在BST中查找离目标最近的k值。

注：

给定的目标值是一个浮点。

您能够假定k始终有效，即：k≤总节点。

保证BST中只有一组最接近目标的惟一k值。

跟进：

假设BST是平衡的，您能够在小于O（n）的运行时（其中n=总节点）中解决它吗？

提示：

一、考虑实现这两个助手函数：

　　i.getPrevistory（n），它将下一个较小的节点返回到n。

　　ii.getsuccessor（n），它将下一个较大的节点返回到n。

二、尝试假设每一个节点都有一个父指针，这会使问题变得更容易。

三、若是没有父指针，咱们只须要使用堆栈跟踪从根到当前节点的路径。

四、在单独查找前置节点和后续节点时，须要两个堆栈来跟踪路径。

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Solution

 1 public class TreeNode {
 2     public var val: Int
 3     public var left: TreeNode?
 4     public var right: TreeNode?
 5     public init(_ val: Int) {
 6         self.val = val
 7         self.left = nil
 8         self.right = nil
 9     }
10 }
11 
12 class Solution {
13     func closestKValues(_ root: TreeNode?,_ target:Double,_ k:Int) -> [Int] {
14         var res:[Int] = [Int]()
15         inorder(root, target, k, &res)
16         return res
17     }
18     
19     func inorder(_ root: TreeNode?,_ target:Double,_ k:Int,_ res:inout [Int])
20     {
21         if root == nil {return}
22         inorder(root?.left, target, k, &res)
23         if res.count < k
24         {
25             res.append(root!.val)
26         }
27         else if abs(Double(root!.val) - target) < abs(Double(res[0]) - target)
28         {
29             res.removeFirst()
30             res.append(root!.val)
31         }
32         else
33         {
34             return
35         }
36         inorder(root?.right, target, k, &res)
37     }
38 }