[LeetCode]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

Top-Down 备忘录的Dp解法

/**
* 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*> ret;
if(n==0){
return ret;
}
vector<vector<vector<TreeNode*>>> Dp(n,vector<vector<TreeNode*>>(n));//n*n 初始化Dp
helper(1,n,ret,Dp);//开始Top－Down Dp，备忘录法递归
return ret;
}
void helper(int start,int end,vector<TreeNode*> &ret,vector<vector<vector<TreeNode*>>> &Dp){
if(start>end){
ret.push_back(NULL);//start>end 没有解，放入NULL节点，递归终点，Dp初始值
return;
}
if(!Dp[start-1][end-1].empty()){
ret = Dp[start-1][end-1]; //把当前存储的Dp放在ret中，最后返回的ret就是Dp[0][n-1];
return;
}
for(int i=start; i<=end; ++i){
vector<TreeNode*> left; //左侧解集
vector<TreeNode*> right;//右侧解集
helper(start,i-1,left,Dp);
helper(i+1,end,right,Dp);
for(int j=0; j<left.size(); ++j){
for(int k=0; k<right.size(); ++k){
TreeNode* node = new TreeNode(i);//Dp[0]
就是遍历i从0到n,把i左指针指向左侧解集的每一个节点，右指针指向右侧解集的每一个节点
node->left = left[j];
node->right = right[k];
ret.push_back(node); //当前解放回ret
}
}
}
Dp[start-1][end-1] = ret;//保存Dp值
}
};