Unique Binary Search Trees II

Given 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.

   1         3     3      2      1
           /     /      /       
     3     2     1      1   3      2
    /     /                        
   2     1         2                 3

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

ref:http://www.cnblogs.com/feiling/p/3296156.html

http://fisherlei.blogspot.com/2013/03/leetcode-unique-binary-search-trees-ii.html

划分成左右子树分别进行构造，当左右子树所有可能情况都构造完毕后，加上node即可

这里根节点可能情况为1,2,....,n

/**
 * 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 ArrayList<TreeNode> generateTrees(int n) {
        return generate(1,n);
    }
    
    public ArrayList<TreeNode> generate(int start, int end){
        // 存放所有可能的unique BST的root
        ArrayList<TreeNode> res = new ArrayList<TreeNode>();
        if(start > end){
            res.add(null);
            return res;
        }
        
        for(int  i = start; i <= end; i++){
            // 以 i 为树根的情况，分别递归得到所有的左子树和右子树的所有可能的组合
            ArrayList<TreeNode> leftSubTree = generate(start, i-1);
            ArrayList<TreeNode> rightSubTree = generate(i+1, end);
            for(int j = 0; j<leftSubTree.size(); j++){
                for(int k = 0; k<rightSubTree.size(); k++){
                    // 创建树根
                    TreeNode root = new TreeNode(i);
                    // 排列组合所有可能的unique的BST
                    root.left = leftSubTree.get(j);
                    root.right = rightSubTree.get(k);
                    res.add(root);
                }
            }
        }
        return res;
    }
}