Leetcode 树 Unique Binary Search TreesII

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Unique Binary Search Trees II

 Total Accepted: 7349 Total
Submissions: 27648

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.

题意：给定数字n,生成以数字1到n为节点的二叉查找树

思路：dfs暴力枚举 

以i为根的树的左右树分别是[1,i-1]和[i+1,n]

递归函数：

vector<TreeNode *> generateTree(int begin, int end)

表示生成以[begin,end]的值为节点的二叉查找树

复杂度：不懂分析 

相关题目：Unique Binary Search Trees

/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<TreeNode *> generateTrees(int begin, int end){
vector<TreeNode *> results;
if(begin > end){
results.push_back(NULL);
return results;
}

for(int i = begin; i <= end; i++){
vector<TreeNode *> left_trees = generateTrees(begin, i - 1);
vector<TreeNode *> right_trees = generateTrees(i + 1, end);
for(int l = 0; l < left_trees.size(); l++){
for(int r = 0; r < right_trees.size(); r++){
TreeNode *head = new TreeNode(i);
head->left = left_trees[l];
head->right = right_trees[r];
results.push_back(head);
}
}
}

}

vector<TreeNode *> generateTrees(int n){
return generateTrees(1, n);
}
};