Merge pull request #199 from gostool/leetcode0700

Leetcode0700

leetcode/0700.Search-in-a-Binary-Search-Tree/700.Search in a Binary Search Tree.go

@@ -0,0 +1,20 @@
+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)
+	}
+}

leetcode/0700.Search-in-a-Binary-Search-Tree/700.Search in a Binary Search Tree_test.go

@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question700 struct {
+	para700
+	ans700
+}
+
+// para 是参数
+type para700 struct {
+	root *TreeNode
+	val  int
+}
+
+// ans 是答案
+type ans700 struct {
+	ans *TreeNode
+}
+
+func Test_Problem700(t *testing.T) {
+
+	qs := []question700{
+		{
+			para700{&TreeNode{Val: 4,
+				Left:  &TreeNode{Val: 2, Left: &TreeNode{Val: 1, Left: nil, Right: nil}, Right: &TreeNode{Val: 3, Left: nil, Right: nil}},
+				Right: &TreeNode{Val: 7, Left: nil, Right: nil}},
+				2},
+			ans700{&TreeNode{Val: 2, Left: &TreeNode{Val: 1, Left: nil, Right: nil}, Right: &TreeNode{Val: 3, Left: nil, Right: nil}}},
+		},
+
+		{
+			para700{&TreeNode{Val: 4,
+				Left:  &TreeNode{Val: 2, Left: &TreeNode{Val: 1, Left: nil, Right: nil}, Right: &TreeNode{Val: 3, Left: nil, Right: nil}},
+				Right: &TreeNode{Val: 7, Left: nil, Right: nil}},
+				5},
+			ans700{nil},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 700------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans700, q.para700
+		fmt.Printf("【input】:%v    【output】:%v\n", p, searchBST(p.root, p.val))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0700.Search-in-a-Binary-Search-Tree/README.md

@@ -0,0 +1,61 @@
+# [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)
+	}
+}
+```