[leetcode]Unique Binary Search Trees II

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 

算法思路：

分治法：求出左子树的list和右子树的list，然后循环两个子树，分别插入root的左右子树。

代码如下：

/**
 * 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) {
        return buildTree(1,n);
    }
    private List<TreeNode> buildTree(int begin,int end){
        List<TreeNode> res = new ArrayList<TreeNode>();
        if(begin > end){
            res.add(null);
            return res;
        }
        List<TreeNode> left = new ArrayList<TreeNode>();
        List<TreeNode> right = new ArrayList<TreeNode>();
        for(int i = begin; i <= end; i++){
　　　　　　　　　//TreeNode node = new TreeNode(i); 本来这句写在这里，结果发现错鸟，大家可以脑补一下原因
            left = buildTree(begin, i - 1);
            right = buildTree(i + 1, end);
            for(int j = 0; j < left.size(); j++){
                for(int k = 0; k < right.size(); k++){
                    TreeNode node = new TreeNode(i);
                    node.left = left.get(j);
                    node.right = right.get(k);
                    res.add(node);
                }
            }
        }
        return res;
    }
}

参考：

http://www.cnblogs.com/cheapcrook/archive/2013/01/29/2880903.html