【LeetCode】95. 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

由于1~n是升序列，因此建起来的树天然就是BST。

递归思想，依次选择根节点，对左右子序列再分别建树。

由于左右子序列建树的结果也可能不止一种，需要考虑所有搭配情况。

vector<TreeNode *> left代表所有valid左子树。

vector<TreeNode *> right代表所有valid右子树。

/**
* 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 *> generateTrees(int n) {
return Helper(1, n);
}
vector<TreeNode *> Helper(int begin, int end)
{
vector<TreeNode *> ret;
if(begin > end)
ret.push_back(NULL);
else if(begin == end)
{
TreeNode* node = new TreeNode(begin);
ret.push_back(node);
}
else
{
for(int i = begin; i <= end; i ++)
{//root
vector<TreeNode *> left = Helper(begin, i-1);
vector<TreeNode *> right = Helper(i+1, end);
for(int l = 0; l < left.size(); l ++)
{
for(int r = 0; r < right.size(); r ++)
{
//new tree
TreeNode* root = new TreeNode(i);
root->left = left[l];
root->right = right[r];
ret.push_back(root);
}
}
}
}
return ret;
}
};