95. Unique Binary Search Trees II
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Given an integer 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
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分析:
由于1~n是升序列,因此建起来的树天然就是BST。
递归思想,依次选择根节点,对左右子序列再分别建树。
由于左右子序列建树的结果也可能不止一种,需要考虑所有搭配情况。
vector<TreeNode *> left代表所有valid左子树。
vector<TreeNode *> right代表所有valid右子树。
ac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) {
vector<TreeNode*>rets;
if(n==0)
return rets;
return helper(1,n);
//return rets;
}
vector<TreeNode*> helper(int begin,int end)
{
vector<TreeNode*>rets;
if(begin>end)
{
rets.push_back(nullptr);
}
else if(begin==end)
{
TreeNode *node=new TreeNode(begin);
rets.push_back(node);
}
else
{
for(int i=begin;i<=end;i++)
{
vector<TreeNode*>left=helper(begin,i-1);
vector<TreeNode*>right=helper(i+1,end);
for(int l=0;l<left.size();l++)
{
for(int r=0;r<right.size();r++)
{
TreeNode *root=new TreeNode(i);
root->left=left[l];
root->right=right[r];
rets.push_back(root);
}
}
}
}
return rets;
}
};
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