LeetCode OJ ：Unique Binary Search Trees II（唯一二叉搜索树）

题目如下所示：返回的结果是一个Node的Vector：

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
　　
树节点的定义是下面这样的

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<TreeNode*> generateTrees(int n) {
return createNode(1, n);
}

vector<TreeNode*> createNode(int start, int end)
{
vector<TreeNode*> result;
if(start > end){
result.push_back(NULL);
return result;
}
for(int i = start; i <= end; ++i){
vector<TreeNode*> leftNode = createNode(start, i - 1);
vector<TreeNode*> rightNode = createNode(i + 1, end);
for(int j = 0; j < leftNode.size(); ++j){
for(int k = 0; k < rightNode.size(); ++k){
TreeNode * tmpNode = new TreeNode(i);
tmpNode->left = leftNode[j];　　
tmpNode->right = rightNode[k];
result.push_back(tmpNode);
}
}
}
return result;
}
};

这一题实际上更另外一个叫做different ways to add parentheses的题目比较相似，这个详见上一篇博文。

java版本的代码如下：

/**
* 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)//如果不加上这一条，当为0的时候会返回[[]]，不知道为什么，很奇怪
return new ArrayList<TreeNode>();
return createTree(1, n);
}

public List<TreeNode> createTree(int start, int end){
ArrayList<TreeNode> result = new ArrayList<TreeNode>();
if(start > end){
result.add(null);
return result;
}
for(int i = start; i <= end; ++i){
for(TreeNode leftNode : createTree(start, i - 1)){
for(TreeNode rightNode : createTree(i + 1, end)){
TreeNode node = new TreeNode(i);
node.left = leftNode;
node.right = rightNode;
result.add(node);
}
}
}
return result;
}
}