You are given the root of a binary search tree (BST) and an integer val.

Find the node in the BST that the node's value equals val and return the subtree rooted with that node. If such a node does not exist, return null.

Example 1:

Input: root = [4,2,7,1,3], val = 2 Output: [2,1,3] 

Example 2:

Input: root = [4,2,7,1,3], val = 5 Output: [] 

Constraints:

• The number of nodes in the tree is in the range [1, 5000].
• 1 <= Node.val <= 10000000
• root is a binary search tree.
• 1 <= val <= 10000000

给定二叉搜索树（BST）的根节点和一个值。 你需要在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。 如果节点不存在，则返回 NULL。

• 根据二叉搜索树的性质(根节点的值大于左子树所有节点的值，小于右子树所有节点的值),进行递归求解

package leetcode import ( "github.com/halfrost/LeetCode-Go/structures" ) // TreeNode definetypeTreeNode= structures.TreeNode/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */funcsearchBST(root*TreeNode, valint) *TreeNode { ifroot==nil { returnnil } ifroot.Val==val { returnroot } elseifroot.Val<val { returnsearchBST(root.Right, val) } else { returnsearchBST(root.Left, val) } }