Unique Binary Search Trees
Given n, how many structurally unique BST's (binary search trees) that store values 1...n?
For example, Given n = 3, there are a total of 5 unique BST's.
1 3 3 2 1
/ / /
3 2 1 1 3 2
/ /
2 1 2 3
思路： f(n) = Σi=1n f(n-i)*f(i-1), 其中 f(0) = f(1) = 1; 利用动归记下之前的 f(2)~f(n-1)即可。
class Solution {
public:
int numTrees(int n) {
vector<int> f(n+1, 0);
f[0] = f[1] = 1;
for(int v = 2; v <= n; ++v)
for(int pos = 1; pos <= v; ++pos)
f[v] += f[pos-1] * f[v-pos];
return f[n];
}
};
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
思路：分别以 1~n 为根节点，左右子树根的集合数量相乘,递归，依次得出结果。
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
vector<TreeNode *> generateTreesCore(int start, int end) {
vector<TreeNode *> vec;
if(start > end) { vec.push_back(NULL); return vec; }
for(int cur = start; cur <= end; ++cur) {
vector<TreeNode *> left = generateTreesCore(start, cur-1);
vector<TreeNode *> right = generateTreesCore(cur+1, end);
for(size_t i = 0; i < left.size(); ++i) {
for(size_t j = 0; j < right.size(); ++j) {
TreeNode *root = new TreeNode(cur);
root->left = left[i];
root->right = right[j];
vec.push_back(root);
}
}
}
return vec;
}
class Solution {
public:
vector<TreeNode *> generateTrees(int n) {
return generateTreesCore(1, n);
}
};