【Unique Binary Search Trees II】cpp
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
代码:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<TreeNode*> generateTrees(int n) {
return Solution::generateBST(, n);
}
static vector<TreeNode*> generateBST(int min, int max)
{
vector<TreeNode*> ret;
if ( min>max ) { ret.push_back(NULL); return ret; }
for ( int i = min; i<=max; ++i )
{
vector<TreeNode*> left = Solution::generateBST(min, i-);
vector<TreeNode*> right = Solution::generateBST(i+,max);
for ( size_t l = ; l < left.size(); ++l )
{
for ( size_t r = ; r < right.size(); ++r )
{
TreeNode *root = new TreeNode(i);
root->left = left[l];
root->right = right[r];
ret.push_back(root);
}
}
}
return ret;
}
};
tips:
直接学习大神的代码
http://bangbingsyb.blogspot.sg/2014/11/leetcode-unique-binary-search-trees-i-ii.html
一开始一直有一个疑问,如果min==max的时候(即只有一个元素的时候)能构造一个新的节点返回么?
肯定是可以的。因为这时left返回的是含有一个NULL的vector,right返回的是含有一个NULL的vector;两个vector的长度都是1,因此可以构造出这个新的点。
===================================================
第二次过这道题,没啥可说的,再学一遍前人的代码。没啥可说的,递归确实很漂亮。
这里的精髓在于,即使是begin>end这种情况,也返回一个长度为1的NULL点;这样做的好处是即使只有一个点传入了,l.size() 和r.szie()也都是1(虽然里面都是NULL);这样代码就很简洁了。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<TreeNode*> generateTrees(int n)
{
return Solution::generate(, n);
}
static vector<TreeNode*> generate(int begin, int end)
{
vector<TreeNode* > ret;
if ( begin>end )
{
ret.push_back(NULL);
return ret;
}
for ( int i=begin; i<=end; ++i )
{
vector<TreeNode*> l = Solution::generate(begin, i-);
vector<TreeNode*> r = Solution::generate(i+, end);
for ( int j=; j<l.size(); ++j )
{
for ( int k=; k<r.size(); ++k )
{
TreeNode* root = new TreeNode(i);
root->left = l[j];
root->right = r[k];
ret.push_back(root);
}
}
}
return ret;
}
};
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