[Swift]LeetCode700. 二叉搜索树中的搜索 | 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.

---

 给定二叉搜索树（BST）的根节点和一个值。 你须要在BST中找到节点值等于给定值的节点。 返回以该节点为根的子树。 若是节点不存在，则返回 NULL。

例如，

给定二叉搜索树:

        4
       / \
      2   7
     / \
    1   3

和值: 2

你应该返回以下子树:

      2     
     / \   
    1   3

在上述示例中，若是要找的值是 5，但由于没有节点值为 5，咱们应该返回 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 }