[LeetCode] Unique Binary Search Trees II dfs 深度搜索

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.

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Tree Dynamic Programming

　　这个嘛，对于1 to n ，如果要用某个值做节点，那么这个值左部分的全部可能的树，递归调用获得，右部分同理，这样便可以获取结果。

#include <iostream>
#include <vector>
using namespace std;

/**
* 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 n) {
return help_f(1,n);
}
vector<TreeNode *> help_f(int l,int r)
{
vector<TreeNode *> ret;
if(l>r){
ret.push_back(NULL);
return ret;
}
for(int i=l;i<=r;i++){
vector<TreeNode *> lPart = help_f(l,i-1);
vector<TreeNode *> rPart = help_f(i+1,r);
for(int lidx=0;lidx<lPart.size();lidx++){
for(int ridx=0;ridx<rPart.size();ridx++){
TreeNode * pNode = new TreeNode(i);
pNode->left = lPart[lidx];
pNode->right = rPart[ridx];
ret.push_back(pNode);
}
}
}
return ret;
}
};

int main()
{
return 0;
}