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。

 def generateTrees(self, n):
         """
         :type n: int
         :rtype: List[TreeNode]
         """
         if n == 0:
             return []
         return self.buildTree(1, n)

     def buildTree(self, start, end):
         result = []
         if start > end:
             result.append(None)
             return result

         for i in range(start, end + 1):
             leftChild = self.buildTree(start, i - 1)
             rightChild = self.buildTree(i+1, end)

             for l in leftChild:
                 for r in rightChild:
                     node = TreeNode(i)
                     node.left = l
                     node.right = r
                     result.append(node)

         return result