Leetcode 95. Unique Binary Search Tree II

由于BST的性质，所以右子树或者左子树中Node的值是连续的： 左子树 = [1, i -1], root = i, 右子树 = [i + 1, n]。使用一个递归函数构造这个BST。其中返回值应该是所有的Unique BST的root node。

 1 def generateTrees(self, n):
 2         """
 3         :type n: int
 4         :rtype: List[TreeNode]
 5         """
 6         if n == 0:
 7             return []
 8         return self.buildTree(1, n)
 9
10     def buildTree(self, start, end):
11         result = []
12         if start > end:
13             result.append(None)
14             return result
15
16         for i in range(start, end + 1):
17             leftChild = self.buildTree(start, i - 1)
18             rightChild = self.buildTree(i+1, end)
19
20             for l in leftChild:
21                 for r in rightChild:
22                     node = TreeNode(i)
23                     node.left = l
24                     node.right = r
25                     result.append(node)
26
27         return result