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

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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 [], not null.

---

 給定二叉搜索樹（BST）的根節點和一個值。 你須要在BST中找到節點值等於給定值的節點。 返回以該節點爲根的子樹。 若是節點不存在，則返回 NULL。

例如，

給定二叉搜索樹:

        4
       / \
      2   7
     / \
    1   3

和值: 2

你應該返回以下子樹:

      2     
     / \   
    1   3

在上述示例中，若是要找的值是 5，但由於沒有節點值爲 5，咱們應該返回 NULL。

---

Runtime: 168 ms

Memory Usage: 19.6 MB

 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 }