A good way into this one is to ask: what do we need to remember so we never redo work blindly? For 95. Unique Binary Search Trees II, the solution in this repository is mainly a dynamic programming solution.
Guide
What?
The first job is to translate the English prompt into state, transition, and stopping conditions. Instead of trying to be clever immediately, read the code as a sequence of questions:
- What state are we keeping?
- How do we move from one state to the next?
- When do we know the answer is already determined?
For this file, the main tools are: dynamic programming, two pointers.
The notes already sitting in the source point us in the right direction:
- recursion
Guide
When?
This is the kind of solution you want when the problem has structure hiding inside a messy-looking input. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are generateTrees, used, addOffset.
Guide
Why?
The win comes from making each line carry responsibility: store the useful state, discard the rest, keep moving.
- The two-dimensional vector is the memory of the solution: grid state, DP state, or adjacency shape.
- The final return is not magic; it is the invariant after the loops or recursion have finished doing their accounting.
Guide
How?
Walk through the solution in this order:
- Read the setup variables first.
- Follow the main loop or recursive helper next.
- Watch where invalid states get skipped.
- Check which value survives to the return statement.
The most important competitive-programming habit here is to trust the invariant. Once the invariant is right, the loops become much less scary.
Guide
Complexity
- Time: O(n) to O(n log n), depending on the dominant loop or data structure operation
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
solution.cpp
01//recursion
02//Runtime: 36 ms, faster than 10.54% of C++ online submissions for Unique Binary Search Trees II.
03//Memory Usage: 18.2 MB, less than 5.04% of C++ online submissions for Unique Binary Search Trees II.
04/**
05 * Definition for a binary tree node.
06 * struct TreeNode {
07 * int val;
08 * TreeNode *left;
09 * TreeNode *right;
10 * TreeNode() : val(0), left(nullptr), right(nullptr) {}
11 * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
12 * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
13 * };
14 */
15class Solution {
16public:
17 int n;
18
19 void generateTrees(vector<bool>& used, vector<TreeNode*>& trees){
20 for(int v = 1; v <= n; ++v){
21 if(used[v]) continue;
22 used[v] = true;
23
24 vector<bool> lused = used;
25 vector<TreeNode*> ltrees;
26 //elements larger than v is not usable
27 for(int lv = v+1; lv <= n; ++lv){
28 lused[lv] = true;
29 }
30 generateTrees(lused, ltrees);
31
32 vector<bool> rused = used;
33 vector<TreeNode*> rtrees;
34 for(int rv = 1; rv < v; ++rv){
35 rused[rv] = true;
36 }
37 generateTrees(rused, rtrees);
38
39 // cout << "left: " << ltrees.size() << endl;
40 // cout << "right: " << rtrees.size() << endl;
41
42 for(int li = 0; li < ltrees.size(); ++li){
43 for(int ri = 0; ri < rtrees.size(); ++ri){
44 TreeNode* root = new TreeNode(v);
45 root->left = ltrees[li];
46 root->right = rtrees[ri];
47 trees.push_back(root);
48 }
49 }
50
51 used[v] = false;
52 }
53 if(trees.empty()) trees.push_back(nullptr);
54 }
55
56 vector<TreeNode*> generateTrees(int n) {
57 if(n == 0) return {};
58
59 this->n = n;
60 //0 is for padding
61 vector<bool> used(n+1, false);
62 vector<TreeNode*> trees;
63
64 generateTrees(used, trees);
65
66 return trees;
67 }
68};
69
70//recursion + memorization
71//note that a tree's range must be continuous, so we can use start and end to identify a tree
72//Runtime: 12 ms, faster than 95.32% of C++ online submissions for Unique Binary Search Trees II.
73//Memory Usage: 12.3 MB, less than 88.56% of C++ online submissions for Unique Binary Search Trees II.
74class Solution {
75public:
76 vector<vector<vector<TreeNode*>>> memo;
77
78 void generateTrees(int start, int end){
79 // cout << start << " - " << end << endl;
80
81 if(!memo[start][end].empty()){
82 return;
83 }
84
85 if(start > end){
86 memo[start][end] = {nullptr};
87 return;
88 }
89
90 vector<TreeNode*> trees;
91 for(int v = start; v <= end; ++v){
92 generateTrees(start, v-1);
93 vector<TreeNode*>& ltree = memo[start][v-1];
94
95 generateTrees(v+1, end);
96 vector<TreeNode*>&rtree = memo[v+1][end];
97
98 // cout << "left: " << ltrees.size() << endl;
99 // cout << "right: " << rtrees.size() << endl;
100
101 for(int li = 0; li < ltree.size(); ++li){
102 for(int ri = 0; ri < rtree.size(); ++ri){
103 TreeNode* root = new TreeNode(v);
104 root->left = ltree[li];
105 root->right = rtree[ri];
106 trees.push_back(root);
107 }
108 }
109 }
110 if(trees.empty()) trees.push_back(nullptr);
111 memo[start][end] = trees;
112 }
113
114 vector<TreeNode*> generateTrees(int n) {
115 if(n == 0) return {};
116 //[1,n] x [1,n] is valid range
117 //but 0 and n+1 are also used for convenience
118 memo = vector<vector<vector<TreeNode*>>>(n+2,
119 vector<vector<TreeNode*>>(n+2));
120 generateTrees(1, n);
121 return memo[1][n];
122 }
123};
124
125//bottom-up DP
126//https://leetcode.com/problems/unique-binary-search-trees-ii/discuss/31493/Java-Solution-with-DP
127//Runtime: 20 ms, faster than 66.68% of C++ online submissions for Unique Binary Search Trees II.
128//Memory Usage: 12.8 MB, less than 83.42% of C++ online submissions for Unique Binary Search Trees II.
129class Solution {
130public:
131 TreeNode* addOffset(TreeNode* node, int offset){
132 /*
133 clone the whole tree rooted at "node",
134 but adding "offset" to all of its descendents
135 */
136 if(!node) return nullptr;
137 TreeNode* newnode = new TreeNode(node->val+offset);
138
139 newnode->left = addOffset(node->left, offset);
140 newnode->right = addOffset(node->right, offset);
141
142 return newnode;
143 }
144
145 vector<TreeNode*> generateTrees(int n) {
146 if(n == 0) return {};
147
148 vector<vector<TreeNode*>> dp(n+1);
149
150 dp[0] = {nullptr};
151
152 //count: size of tree
153 for(int count = 1; count <= n; ++count){
154 //val: root's value
155 for(int val = 1; val <= count; ++val){
156 // cout << count << ", " << val << endl;
157 for(TreeNode* ltree : dp[val-1]){
158 for(TreeNode* rtree : dp[count-val]){
159 TreeNode* node = new TreeNode(val);
160 node->left = ltree;
161 node->right = addOffset(rtree, val);
162 dp[count].push_back(node);
163 }
164 }
165 }
166 }
167
168 return dp[n];
169 }
170};Cost
Complexity
Time
O(n) to O(n log n), depending on the dominant loop or data structure operation
Dominated by the main traversal, recursion, or data-structure operations in the code.
Space
O(n) in the usual case for auxiliary containers or recursion
Auxiliary state plus the answer structure where the problem requires one.