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website/content/ChapterFour/0095.Unique-Binary-Search-Trees-II.md

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+# [95. Unique Binary Search Trees II](https://leetcode.com/problems/unique-binary-search-trees-ii/)
+
+
+## 题目
+
+Given an integer *n*, generate all structurally unique **BST's** (binary search trees) that store values 1 ... *n*.
+
+**Example**:
+
+    Input: 3
+    Output:
+    [
+      [1,null,3,2],
+      [3,2,null,1],
+      [3,1,null,null,2],
+      [2,1,3],
+      [1,null,2,null,3]
+    ]
+    Explanation:
+    The above output corresponds to the 5 unique BST's shown below:
+    
+       1         3     3      2      1
+        \       /     /      / \      \
+         3     2     1      1   3      2
+        /     /       \                 \
+       2     1         2                 3
+
+
+## 题目大意
+
+给定一个整数 n，生成所有由 1 ... n 为节点所组成的二叉搜索树。
+
+## 解题思路
+
+- 输出 1~n 元素组成的 BST 所有解。这一题递归求解即可。外层循环遍历 1~n 所有结点，作为根结点，内层双层递归分别求出左子树和右子树。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func generateTrees(n int) []*TreeNode {
+	if n == 0 {
+		return []*TreeNode{}
+	}
+	return generateBSTree(1, n)
+}
+
+func generateBSTree(start, end int) []*TreeNode {
+	tree := []*TreeNode{}
+	if start > end {
+		tree = append(tree, nil)
+		return tree
+	}
+	left, right := []*TreeNode{}, []*TreeNode{}
+	for i := start; i <= end; i++ {
+		left = generateBSTree(start, i-1)
+		right = generateBSTree(i+1, end)
+		for _, l := range left {
+			for _, r := range right {
+				root := &TreeNode{Val: i, Left: l, Right: r}
+				tree = append(tree, root)
+			}
+		}
+	}
+	return tree
+}
+
+```