您的位置:首页 > 产品设计 > UI/UE

Unique Binary Search Trees II

2015-01-22 10:49 337 查看
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.

生成不相同的二叉树

/**
* 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 *> generateUtil(int start, int end){
vector<TreeNode *> ans;
if(start > end){
ans.push_back(NULL);
return ans;
}
for(int i = start; i <= end; ++i){
vector<TreeNode*> leftTrees = generateUtil(start,i-1);
vector<TreeNode*> rightTrees = generateUtil(i+1,end);
for(int  l = 0; l < leftTrees.size(); ++l)
for(int r = 0; r < rightTrees.size(); ++r){
TreeNode* root = new TreeNode(i);
root->left = leftTrees[l];
root->right = rightTrees[r];
ans.push_back(root);
}
}
return ans;
}
vector<TreeNode*>generateTrees(int n){
return generateUtil(1,n);
}
};
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: