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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.
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.
题目：
这题要求将每个子树输出，思想就是对于n，以1-n为根节点，然后左右子树分别是比其小和比其大的节点构成的，对左右子树的的形态也要使用遍历，代码中使用的是foreach遍历：for(i:j)这种形式
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<TreeNode *> generateTrees(int n) { 13 if (n == 0) return generate(1, 0); 14 return generate(1, n); 15 } 16 private: 17 vector<TreeNode *> generate(int start, int end) { 18 vector<TreeNode*> subTree; 19 if (start > end) { 20 subTree.push_back(nullptr); 21 return subTree; 22 } 23 for (int k = start; k <= end; k++) { 24 vector<TreeNode*> leftSubs = generate(start, k - 1); 25 vector<TreeNode*> rightSubs = generate(k + 1, end); 26 for (auto i : leftSubs) { 27 for (auto j : rightSubs) { 28 TreeNode *node = new TreeNode(k); 29 node->left = i; 30 node->right = j; 31 subTree.push_back(node); 32 } 33 } 34 } 35 return subTree; 36 } 37 };
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原文地址：http://www.cnblogs.com/raichen/p/4956115.html
