Leetcode 95. Unique Binary Search Trees II

原题：
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

解决方法：

采用dfs还是比较容易求解的，基本思路是：由于是BST，所以左节点肯定小于根节点和左节点，这样就可以根据跟节点的值，将数组分成左右两部分，由于左右节点也要求是BST，这变成同样的子问题了，只需要再次调用dfs函数即可。最后将左节点列表、根节点、右节点列表做一个组合，即是所有的解。


代码：

void dfs(vector<TreeNode*>& res, int l, int r){
if ( l == r){
TreeNode* node = new TreeNode(l);
res.push_back(node);
return;
}

for(int i = l; i <= r; i++){
vector<TreeNode*> left, right;
dfs(left, l, i-1);
dfs(right, i+1, r);

if (left.size() < 1){
for(int k = 0; k < right.size(); k++){
TreeNode* node = new TreeNode(i);
node->right = right[k];
res.push_back(node);
}
}
if (right.size() < 1){
for(int j = 0; j < left.size(); j++){
TreeNode* node = new TreeNode(i);
node->left = left[j];
res.push_back(node);
}
}

for(int j = 0; j < left.size(); j++){
for(int k = 0; k < right.size(); k++){
TreeNode* node = new TreeNode(i);
node->left = left[j];
node->right = right[k];
res.push_back(node);
}
}
}
}

vector<TreeNode*> generateTrees(int n) {
vector<TreeNode*> res;
dfs(res, 1, n);
return res;
}