标签：style blog color io strong for ar div
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
分析：这道题与unique binary search tree I不同在于我们要生成所以的二叉查找树并返回树的根。同样我们用递归的方法，如果是空树我们加入一个NULL指针，这个对于简化代码是有帮助的。
如果在空树的情况是我们不加入NULL,而是让<TreeNode *> res为空，那么在通过左右两个子树组成新树时，代码会繁琐很多。因为我们必须要处理左子树、右子树是否为空总共四种情况。
1 class Solution { 2 public: 3 vector<TreeNode *> generateTrees(int n) { 4 vector<TreeNode *> res; 5 res = generateBST(1,n); 6 return res; 7 } 8 vector<TreeNode *> generateBST(int left, int right){ 9 vector<TreeNode *> res; 10 if(left > right){ 11 res.push_back(NULL); 12 return res; 13 } 14 for(int i = left; i <= right; i++){ 15 vector<TreeNode *> lefts = generateBST(left, i-1); 16 vector<TreeNode *> rights = generateBST(i+1,right); 17 for(auto k:lefts) 18 for(auto j:rights){ 19 TreeNode * root = new TreeNode(i); 20 root->left = k; 21 root->right = j; 22 res.push_back(root); 23 } 24 } 25 return res; 26 } 27 };
Unique Binary Search Tree II,布布扣,bubuko.com
标签：style blog color io strong for ar div
原文地址：http://www.cnblogs.com/Kai-Xing/p/3912983.html
