06
08/2015
[LeetCode] Unique Binary Search Trees II
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
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
解题思路：
分别以i(1<=i<=n)为根节点构造BST。采用递归方法。左子树为1...i-1构成的BST，右子树为i+1...n构成的BST，如此递归下去组装即可。
/**
* 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 generateSubTrees(1, n);
}
//产生一棵子树
vector<TreeNode*> generateSubTrees(int start, int end){
vector<TreeNode*> result;
if (start>end){
result.push_back(NULL);
return result;
}
for(int i=start; i<=end; i++){
vector<TreeNode*> leftTree = generateSubTrees(start, i-1);
vector<TreeNode*> rightTree = generateSubTrees(i+1, end);
for(vector<TreeNode*>::iterator it1 = leftTree.begin(); it1!=leftTree.end(); it1++){
for(vector<TreeNode*>::iterator it2 = rightTree.begin(); it2!=rightTree.end(); it2++){
TreeNode* root = new TreeNode(i);
root->left = *it1;
root->right = *it2;
result.push_back(root);
}
}
}
return result;
}
};
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