Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node‘s value equals the given value. Return the subtree rooted with that node. If such node doesn‘t exist, you should return NULL.

For example,

Given the tree:
        4
       /       2   7
     /     1   3

And the value to search: 2

You should return this subtree:

      2     
     / \   
    1   3

In the example above, if we want to search the value 5, since there is no node with value 5, we should return NULL.

Note that an empty tree is represented by NULL, therefore you would see the expected output (serialized tree format) as []

給定二叉搜索樹（BST）的根節點和一個值。 你需要在BST中找到節點值等於給定值的節點。 返回以該節點為根的子樹。 如果節點不存在，則返回 NULL。

例如，

給定二叉搜索樹:

        4
       /       2   7
     /     1   3

和值: 2

你應該返回如下子樹:

      2     
     / \   
    1   3

在上述示例中，如果要找的值是 5，但因為沒有節點值為 5，我們應該返回 NULL。

 1
 
 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func searchBST(_ root: TreeNode?, _ val: Int) -> TreeNode? {
16         var root = root
17         while ((root != nil) && root!.val != val)
18         {
19             root = (root!.val > val) ? root?.left : root?.right
20         }
21         return root
22     }
23 }

168ms

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func searchBST(_ root: TreeNode?, _ val: Int) -> TreeNode? {
16         if nil == root {
17             return nil
18         } else{
19             let value = (root?.val)!
20             if value == val {
21                 return root
22             } else if value < val {
23                 return searchBST(root?.right, val)
24             } else {
25                 return searchBST(root?.left, val)
26             }
27         }
28     }
29 }

188ms

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func searchBST(_ root: TreeNode?, _ val: Int) -> TreeNode? {
16         guard root != nil else { return nil }
17         var current = root
18         while current != nil {
19             if current!.val == nil { return nil }
20             if current!.val == val { return current }
21             if val > current!.val {
22                 current = current!.right
23             } else {
24                 current = current!.left
25             }
26         }
27         return nil
28     }
29 }

[Swift]LeetCode700. 二叉搜索樹中的搜索 | Search in a Binary Search Tree