【unique-binary-search-trees-ii】

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

confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1  / \ 2   3    /   4    \     5

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

题意：输出1到n组成的所有二叉搜索树的结构

思路：

选定1-n的一个数字i作为根节点、0到i-1是左子树， i+1-n是右子树，两层循环；

class Solution{public:vector<TreeNode*> generateTrees(int n){return create(1, n);}vector<TreeNode*> create(int left, int right){vector<TreeNode*> res;if (left>right){res.push_back(NULL);return res;}for (int i=left; i<=right; i++){vector<TreeNode*> leftTree = create(left, i-1);vector<TreeNode*> rightTree = create(i+1, right);for (int j = 0; j<leftTree.size(); j++){for (int k = 0; k<rightTree.size(); k++){TreeNode* root = new TreeNode(i);root->left = leftTree[j];root->right = rightTree[k];res.push_back(root);}}}return res;}};