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题目：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
这道题是Unique Binary Search Trees I的后续，可以先看看http://www.cnblogs.com/laihaiteng/p/3789279.html这个，思路上有相似的地方
个人思路：
1、还是从1-n逐个遍历，作为BST的根节点，左半部分作为根节点的左子树，右半部分作为根节点的右子树
2、递归函数返回从start到end的节点所能产生的不同的BST的集合，则把不同的左BST与右BST做一个乘积（或者说一一匹配），就是此时某个i（1-n中的某个数）作为根节点的BST的集合
代码：
1 #include <stddef.h> 2 #include <vector> 3 4 using namespace std; 5 6 struct TreeNode 7 { 8 int val; 9 TreeNode *left; 10 TreeNode *right; 11 TreeNode(int x) : val(x), left(NULL), right(NULL) {} 12 }; 13 14 class Solution 15 { 16 public: 17 vector<TreeNode *> generateTrees(int n) 18 { 19 vector<TreeNode *> result; 20 21 if (n < 1) 22 { 23 TreeNode *root = NULL; 24 result.push_back(root); 25 return result; 26 } 27 28 result = generatePartTrees(1, n); 29 30 return result; 31 } 32 private: 33 vector<TreeNode *> generatePartTrees(int start, int end) 34 { 35 vector<TreeNode *> part; 36 37 for (int i = start; i <= end; ++i) 38 { 39 vector<TreeNode *> leftTrees = generatePartTrees(start, i - 1); 40 vector<TreeNode *> rightTrees = generatePartTrees(i + 1, end); 41 //左右子树非空 42 if (!leftTrees.empty() && !rightTrees.empty()) 43 { 44 for (int x = 0; x < leftTrees.size(); ++x) 45 { 46 for (int y = 0; y < rightTrees.size(); ++y) 47 { 48 TreeNode *root = new TreeNode(i); 49 TreeNode *left = leftTrees[x]; 50 root->left = left; 51 TreeNode *right = rightTrees[y]; 52 root->right = right; 53 part.push_back(root); 54 } 55 } 56 } 57 //左子树非空，右子树为空 58 if (!leftTrees.empty() && rightTrees.empty()) 59 { 60 for (int x = 0; x < leftTrees.size(); ++x) 61 { 62 TreeNode *root = new TreeNode(i); 63 TreeNode *left = leftTrees[x]; 64 root->left = left; 65 root->right = NULL; 66 part.push_back(root); 67 } 68 } 69 //左子树为空，右子树非空 70 if (leftTrees.empty() && !rightTrees.empty()) 71 { 72 for (int y = 0; y < rightTrees.size(); ++y) 73 { 74 TreeNode *root = new TreeNode(i); 75 TreeNode *right = rightTrees[y]; 76 root->left = NULL; 77 root->right = right; 78 part.push_back(root); 79 } 80 } 81 //左右子树均为空 82 if (leftTrees.empty() && rightTrees.empty()) 83 { 84 TreeNode *root = new TreeNode(i); 85 root->left = NULL; 86 root->right = NULL; 87 part.push_back(root); 88 } 89 } 90 91 return part; 92 } 93 };
网上看了一下，大致都是这个思路，但发现有更好的写法，链接：http://www.cnblogs.com/remlostime/archive/2012/11/19/2777841.html
精简了一下代码：
1 #include <stddef.h> 2 #include <vector> 3 4 using namespace std; 5 6 struct TreeNode 7 { 8 int val; 9 TreeNode *left; 10 TreeNode *right; 11 TreeNode(int x) : val(x), left(NULL), right(NULL) {} 12 }; 13 14 class Solution 15 { 16 public: 17 vector<TreeNode *> generateTrees(int n) 18 { 19 return generatePartTrees(1, n); 20 } 21 private: 22 vector<TreeNode *> generatePartTrees(int start, int end) 23 { 24 vector<TreeNode *> part; 25 26 if (start > end) 27 { 28 part.push_back(NULL); 29 return part; 30 } 31 32 for (int i = start; i <= end; ++i) 33 { 34 vector<TreeNode *> leftTrees = generatePartTrees(start, i - 1); 35 vector<TreeNode *> rightTrees = generatePartTrees(i + 1, end); 36 for (int x = 0; x < leftTrees.size(); ++x) 37 { 38 for (int y = 0; y < rightTrees.size(); ++y) 39 { 40 TreeNode *root = new TreeNode(i); 41 TreeNode *left = leftTrees[x]; 42 root->left = left; 43 TreeNode *right = rightTrees[y]; 44 root->right = right; 45 part.push_back(root); 46 } 47 } 48 } 49 50 return part; 51 } 52 };
leetcode - Unique Binary Search Trees II,布布扣,bubuko.com
leetcode - Unique Binary Search Trees II
标签：style class blog code http tar
原文地址：http://www.cnblogs.com/laihaiteng/p/3801607.html
