95. Unique Binary Search Trees II

Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1...n.

For example,
Given n = 3, your program should return all 5 unique BST's shown below.

大致思路如上，可以看出这也是一个可以划分成子问题求解的题目，所以考点是动态规划。 

但具体对于本题来说，采取的是自底向上的求解过程。 
1. 选出根结点后应该先分别求解该根的左右子树集合，也就是根的左子树有若干种，它们组成左子树集合，根的右子树有若干种，它们组成右子树集合。 
2. 然后将左右子树相互配对，每一个左子树都与所有右子树匹配，每一个右子树都与所有的左子树匹配。然后将两个子树插在根结点上。 
3. 最后，把根结点放入链表中。

 

class Solution {

    public List<TreeNode> generateTrees(int n) {

         if (n == 0) {

              ArrayList<TreeNode> res = new ArrayList<>();

              return res;

         }

         return generateTrees(1,n);

    }

   

    private ArrayList<TreeNode> generateTrees(int left, int right) {

        ArrayList<TreeNode> res = new ArrayList<>();

        if (left > right) {

            res.add(null);

            return res;

        }

        for (int i = left; i <= right; i++) {

            ArrayList<TreeNode> lefts = generateTrees(left, i - 1);//以i作为根节点，左子树由[1,i-1]构成

            ArrayList<TreeNode> rights = generateTrees(i + 1, right);//右子树由[i+1, n]构成

            for (int j = 0; j < lefts.size(); j++) {

                for (int k = 0; k < rights.size(); k++) {

                    TreeNode root = new TreeNode(i);

                    root.left = lefts.get(j);

                    root.right = rights.get(k);

                    res.add(root);

                }

            }

        }        

        return res;

    }

}