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95. Unique Binary Search Trees II.cpp
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95. Unique Binary Search Trees II.cpp
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//recursion
//Runtime: 36 ms, faster than 10.54% of C++ online submissions for Unique Binary Search Trees II.
//Memory Usage: 18.2 MB, less than 5.04% of C++ online submissions for Unique Binary Search Trees II.
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int n;
void generateTrees(vector<bool>& used, vector<TreeNode*>& trees){
for(int v = 1; v <= n; ++v){
if(used[v]) continue;
used[v] = true;
vector<bool> lused = used;
vector<TreeNode*> ltrees;
//elements larger than v is not usable
for(int lv = v+1; lv <= n; ++lv){
lused[lv] = true;
}
generateTrees(lused, ltrees);
vector<bool> rused = used;
vector<TreeNode*> rtrees;
for(int rv = 1; rv < v; ++rv){
rused[rv] = true;
}
generateTrees(rused, rtrees);
// cout << "left: " << ltrees.size() << endl;
// cout << "right: " << rtrees.size() << endl;
for(int li = 0; li < ltrees.size(); ++li){
for(int ri = 0; ri < rtrees.size(); ++ri){
TreeNode* root = new TreeNode(v);
root->left = ltrees[li];
root->right = rtrees[ri];
trees.push_back(root);
}
}
used[v] = false;
}
if(trees.empty()) trees.push_back(nullptr);
}
vector<TreeNode*> generateTrees(int n) {
if(n == 0) return {};
this->n = n;
//0 is for padding
vector<bool> used(n+1, false);
vector<TreeNode*> trees;
generateTrees(used, trees);
return trees;
}
};
//recursion + memorization
//note that a tree's range must be continuous, so we can use start and end to identify a tree
//Runtime: 12 ms, faster than 95.32% of C++ online submissions for Unique Binary Search Trees II.
//Memory Usage: 12.3 MB, less than 88.56% of C++ online submissions for Unique Binary Search Trees II.
class Solution {
public:
vector<vector<vector<TreeNode*>>> memo;
void generateTrees(int start, int end){
// cout << start << " - " << end << endl;
if(!memo[start][end].empty()){
return;
}
if(start > end){
memo[start][end] = {nullptr};
return;
}
vector<TreeNode*> trees;
for(int v = start; v <= end; ++v){
generateTrees(start, v-1);
vector<TreeNode*>& ltree = memo[start][v-1];
generateTrees(v+1, end);
vector<TreeNode*>&rtree = memo[v+1][end];
// cout << "left: " << ltrees.size() << endl;
// cout << "right: " << rtrees.size() << endl;
for(int li = 0; li < ltree.size(); ++li){
for(int ri = 0; ri < rtree.size(); ++ri){
TreeNode* root = new TreeNode(v);
root->left = ltree[li];
root->right = rtree[ri];
trees.push_back(root);
}
}
}
if(trees.empty()) trees.push_back(nullptr);
memo[start][end] = trees;
}
vector<TreeNode*> generateTrees(int n) {
if(n == 0) return {};
//[1,n] x [1,n] is valid range
//but 0 and n+1 are also used for convenience
memo = vector<vector<vector<TreeNode*>>>(n+2,
vector<vector<TreeNode*>>(n+2));
generateTrees(1, n);
return memo[1][n];
}
};
//bottom-up DP
//https://leetcode.com/problems/unique-binary-search-trees-ii/discuss/31493/Java-Solution-with-DP
//Runtime: 20 ms, faster than 66.68% of C++ online submissions for Unique Binary Search Trees II.
//Memory Usage: 12.8 MB, less than 83.42% of C++ online submissions for Unique Binary Search Trees II.
class Solution {
public:
TreeNode* addOffset(TreeNode* node, int offset){
/*
clone the whole tree rooted at "node",
but adding "offset" to all of its descendents
*/
if(!node) return nullptr;
TreeNode* newnode = new TreeNode(node->val+offset);
newnode->left = addOffset(node->left, offset);
newnode->right = addOffset(node->right, offset);
return newnode;
}
vector<TreeNode*> generateTrees(int n) {
if(n == 0) return {};
vector<vector<TreeNode*>> dp(n+1);
dp[0] = {nullptr};
//count: size of tree
for(int count = 1; count <= n; ++count){
//val: root's value
for(int val = 1; val <= count; ++val){
// cout << count << ", " << val << endl;
for(TreeNode* ltree : dp[val-1]){
for(TreeNode* rtree : dp[count-val]){
TreeNode* node = new TreeNode(val);
node->left = ltree;
node->right = addOffset(rtree, val);
dp[count].push_back(node);
}
}
}
}
return dp[n];
}
};