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+# [700. Search in a Binary Search Tree](https://leetcode-cn.com/problems/search-in-a-binary-search-tree/)
+
+## 题目
+
+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**:
+
+![https://assets.leetcode.com/uploads/2021/01/12/tree1.jpg](https://assets.leetcode.com/uploads/2021/01/12/tree1.jpg)
+
+    Input: root = [4,2,7,1,3], val = 2
+    Output: [2,1,3]
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2021/01/12/tree2.jpg](https://assets.leetcode.com/uploads/2021/01/12/tree2.jpg)
+
+    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
+
+## 题目大意
+
+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。
+
+## 解题思路
+
+- 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
+
+## 代码
+
+```go
+package leetcode
+
+type TreeNode struct {
+	Val int
+	Left *TreeNode
+	Right *TreeNode
+}
+
+func searchBST(root *TreeNode, val int) *TreeNode {
+	if root == nil {
+		return nil
+	}
+	if root.Val == val {
+		return root
+	} else if root.Val < val {
+		return searchBST(root.Right, val)
+	} else {
+		return searchBST(root.Left, val)
+	}
+}
+```
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