LeetCode OJ - 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

分析：DFS解决，根据BST的性质 left < root < right，故递归时以root为中间点，可将整个树分为两个部分。 下面的递归过程中，枚举了分割点。

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<TreeNode *> generate(int beg, int end) {
vector<TreeNode* > ret;
if (beg > end) {
ret.push_back(NULL);
return ret;
}

for(int i = beg; i <= end; i++) {
vector<TreeNode* > leftTree = generate(beg, i - 1);
vector<TreeNode* > rightTree = generate(i + 1, end);
for(int j = 0; j < leftTree.size(); j++) {
for(int k = 0; k < rightTree.size(); k++) {
TreeNode *node = new TreeNode(i + 1);
ret.push_back(node);
node->left = leftTree[j];
node->right = rightTree[k];
}
}
}

return ret;
}

vector<TreeNode *> generateTrees(int n) {
return generate(0, n - 1);
}
};