Unique Binary Search Trees II
描述
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
见前面一题。
代码
- Java
- C++
// Unique Binary Search Trees II
// 时间复杂度TODO，空间复杂度TODO
public class Solution {
public List<TreeNode> generateTrees(int n) {
if (n == 0) return new ArrayList<>();
return generate(1, n);
}
private static List<TreeNode > generate(int start, int end) {
List<TreeNode> subTree = new ArrayList<>();
if (start > end) {
subTree.add(null);
return subTree;
}
for (int k = start; k <= end; k++) {
List<TreeNode> leftSubs = generate(start, k - 1);
List<TreeNode> rightSubs = generate(k + 1, end);
for (TreeNode i : leftSubs) {
for (TreeNode j : rightSubs) {
TreeNode node = new TreeNode(k);
node.left = i;
node.right = j;
subTree.add(node);
}
}
}
return subTree;
}
}
// Unique Binary Search Trees II
// 时间复杂度TODO，空间复杂度TODO
class Solution {
public:
vector<TreeNode *> generateTrees(int n) {
if (n == 0) return vector<TreeNode*>();
return generate(1, n);
}
private:
vector<TreeNode *> generate(int start, int end) {
vector<TreeNode*> subTree;
if (start > end) {
subTree.push_back(nullptr);
return subTree;
}
for (int k = start; k <= end; k++) {
vector<TreeNode*> leftSubs = generate(start, k - 1);
vector<TreeNode*> rightSubs = generate(k + 1, end);
for (auto i : leftSubs) {
for (auto j : rightSubs) {
TreeNode *node = new TreeNode(k);
node->left = i;
node->right = j;
subTree.push_back(node);
}
}
}
return subTree;
}
};