【LeetCode】95. Unique Binary Search Trees II

题目

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.

思路

此题最好想的思路是递归，当确定根节点是 i 时，左子树由 1..(i−1) 组成， 右子树由 (i+1)..n 构成。但是要注意在递归的时候要返回vector，保存的是在start to end的区间内所有子树。
还有一种思路就是动态规划，用一个二维数组保存从s到e的所有子树的vector。amap[s][e] 保存的是一个数组，表示s到e之间所有子树的根节点。在动态规划遍历的时候，先计算s和e相等的情况，再依次计算s和e差1，差2的情况。

代码

递归

vector<TreeNode*> dfs (int s, int e){
        vector<TreeNode*> v;
        if (s>e){
            v.push_back(NULL);
            return v;
        }
        if (s==e){
            v.push_back(new TreeNode(s));
            return v;
        }
        for (int i=s;i<=e;++i){
            vector<TreeNode*> left = dfs(s,i-1);
            vector<TreeNode*> right = dfs(i+1,e);
            for (TreeNode* lnode:left){
                for (TreeNode* rnode:right){
                    TreeNode* root = new TreeNode(i);
                    root -> left = lnode;
                    root ->right = rnode;
                    v.push_back(root);
                }
            }
        }
        return v;
    }
    vector<TreeNode*> generateTrees(int n) {
        if (!n) {vector<TreeNode*> v;return v;}
        return dfs(1,n);
    }

动归

vector<TreeNode*> generateTrees(int n) {
        if (!n) {vector<TreeNode*> v;return v;}
        vector<TreeNode*> vec1;
        vec1.push_back(NULL);
        vector<vector<TreeNode*> > vec2(n+2,vec1);
        vector<vector<vector<TreeNode*> > > amap(n+2,vec2);
        for (int gap=0;gap<n;++gap){
            for (int s=1;s<=n-gap;++s){
                int e = s+gap;
                vector<TreeNode*> v;
                for (int i=s;i<=e;++i){
                    for (auto lnode:amap[s][i-1]){
                        for (auto rnode:amap[i+1][e]){
                            TreeNode* root = new TreeNode(i);
                            root->left=lnode;
                            root->right=rnode;
                            v.push_back(root);
                        }
                    }
                }
                amap[s][e]=v;
            }
        }
        return amap[1][n];
    }