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.

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

Tree类题目。

1,...,n都可以作为root，确定一个root后，左边的即为左子树，右边的即为右子树。 可通过recursion来同样处理左右子树。 这是divide-and-conquer的思想。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<TreeNode> generateTrees(int n) {
        if(n == 0) return new ArrayList<TreeNode>();
        return generateSubtrees(1, n);
    }
    
    private List<TreeNode> generateSubtrees(int start, int end) {
        List<TreeNode> res = new ArrayList<TreeNode>();
        if(start > end) {
            res.add(null);
            return res;
        }
        for(int i = start; i <= end; i++) {
            List<TreeNode> leftSubtrees = generateSubtrees(start, i-1);
            List<TreeNode> rightSubtrees = generateSubtrees(i+1, end);
            for(TreeNode left : leftSubtrees) {
                for(TreeNode right: rightSubtrees) {
                    TreeNode root = new TreeNode(i);
                    root.left = left;
                    root.right = right;
                    res.add(root);
                }
            }
        }
        return res;
    }
}