标签：style blog color 使用 io strong for ar
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
思路：使用递归来求解该问题。GenerateSub( start, end )生成[start,end]区间的所有二叉树。为了求解GenerateSub( start, end )，依次取i=start,start+1,...,end为根节点，则，其所有左子树由GenerateSub( start, i-1 )得到，所有右子树由GenerateSub( i+1, end )得到。
1 class Solution { 2 public: 3 vector<TreeNode*> generateTrees( int n ) { 4 return GenerateSub( 1, n ); 5 } 6 private: 7 vector<TreeNode*> GenerateSub( int start, int end ) { 8 vector<TreeNode*> roots; 9 if( start > end ) { roots.push_back( 0 ); } 10 for( int i = start; i <= end; ++i ) { 11 vector<TreeNode*> left = GenerateSub( start, i-1 ); 12 vector<TreeNode*> right = GenerateSub( i+1, end ); 13 for( auto left_iter = left.begin(); left_iter != left.end(); ++left_iter ) { 14 for( auto right_iter = right.begin(); right_iter != right.end(); ++right_iter ) { 15 TreeNode *treeNode = new TreeNode( i ); 16 treeNode->left = *left_iter; 17 treeNode->right = *right_iter; 18 roots.push_back( treeNode ); 19 } 20 } 21 } 22 return roots; 23 } 24 };
Unique Binary Search Trees II,布布扣,bubuko.com
标签：style blog color 使用 io strong for ar
原文地址：http://www.cnblogs.com/moderate-fish/p/3918338.html
