Idea is simple. Like in the version 1 question, you start from N=1 up to the desired number. for each N, you construct the list of trees, dynamically. Each N uses the trees of the smaller Ns as the left and right trees, that's effortless, well, almost.
For left trees, you can always reuse the exact same object that you have constructed for the smaller Ns, since they will always be the same. For the right trees, you will need to substitute the values in the pre-existing trees with the values you have on the right, since the structure of the trees are the same. And since you need to change the values for the right trees, you do this by constructing a new tree, otherwise you would change the copy that is shared by many trees.
class Solution: # @return a list of tree node def treeNumSubstitution(self, root,numList): if root is None: return None newRoot=TreeNode(numList[root.val-1]) if root.left: newRoot.left=self.treeNumSubstitution(root.left,numList) if root.right: newRoot.right=self.treeNumSubstitution(root.right,numList) return newRoot def generateTrees(self, n): num2Trees=dict() num2Trees=[None] for i in xrange(1,n+1): num2Trees[i]= for rootPos in xrange(1,i+1): for leftTree in num2Trees[rootPos-1]: for rightTree in num2Trees[i-rootPos]: root=TreeNode(rootPos) num2Trees[i].append(root) rightTree=self.treeNumSubstitution(rightTree,range(rootPos+1,i+1)) root.left=leftTree root.right=rightTree return num2Trees[n]