leetcode: Unique Binary Search Trees II

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递归求解....对于BST来说，比根节点小的点构成了根的左子树，大的点构成了右子树.....循环每个点作为根节点，递归查找到所有可能的左子树和右子树的组合即可。

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; left = null; right = null; } * } */public class Solution {        public List<TreeNode> generateTrees(int n) {        // if(n<1)        // {        //     List<TreeNode> res = new ArrayList<TreeNode>();        //     TreeNode r = null;        //     res.add(r);        //     return res;        // }                return fun(1,n);    }    List<TreeNode> fun(int left,int right)    {        List<TreeNode> res = new ArrayList<TreeNode>();        if(right<left)        {            res.add(null);            return res;        }                for(int i=left;i<=right;i++)        {            //TreeNode root = new TreeNode(i);            List<TreeNode> leftList = fun(left,i-1);            List<TreeNode> rightList = fun(i+1,right);            for(int j=0;j<leftList.size();j++)            {                for(int k=0;k<rightList.size();k++)                {                    TreeNode root = new TreeNode(i);                    res.add(root);                    root.left = leftList.get(j);                    root.right = rightList.get(k);                }            }        }        return res;    }}

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