96. 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
S: DP
第n个元素是所有元素最大的,其他n-1个元素只可能出现在它的左子树或者父节点(n为最右节点)
int numTrees(int n) {
if(n == 0) return 0;
vector<int> result(n+1, 0);
result[0] = result[1] = 1;
for(int i = 2; i <= n; ++i) {
for(int j = 0; j < i; ++j) {
result[i] += result[j]*result[i-1-j];
}
}
return result[n];
}
95. Unique Binary Search Trees II
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
S: recursion
对于任意数字i作为root,左子树一定全部小于root(1到i-1),右子树全部大于root(i+i到n)
class Solution {
public:
vector<TreeNode*> generateTrees(int n) {
if(n == 0) return vector<TreeNode*> {};
return genTreeHelper(1, n);
}
vector<TreeNode*> genTreeHelper(int start, int end) {
if(start > end) return vector<TreeNode*>{NULL};//这里不能返回空数组
vector<TreeNode*> result;
for(int i = start; i <= end; ++i) {
vector<TreeNode*> leftNodes = genTreeHelper(start, i-1);
vector<TreeNode*> rightNodes = genTreeHelper(i+1, end);
for(auto left: leftNodes) {
for(auto right: rightNodes) {
TreeNode* node = new TreeNode(i);
node->left = left;
node->right = right;
result.push_back(node);
}
}
}
return result;
}
};