LeetCode 95. Unique Binary Search Trees II

本博客转载自：[1]https://segmentfault.com/a/1190000007443961
[2]http://www.cnblogs.com/grandyang/p/4301096.html
Solution1:
转载自链接[1]。思路：递归的方式处理，对于N各节点的二叉树，分别对应的就是左边i个，右边N-i-1个，根节点一个。i的取值范围是0到N-1. 初始为0个的场景即可。这里为了降低一次递归，把一个的场景也放在初始值里。

/*** 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) {if (n == 0) return vector<TreeNode* >();return generateTrees(1, n);}vector<TreeNode*> generateTrees(int start, int end) {vector<TreeNode*> subTree;if (start > end) { subTree.push_back(NULL);return subTree;}for (int k = start; k <= end; k++) {vector<TreeNode*> left = generateTrees(start, k - 1);vector<TreeNode*> right = generateTrees(k + 1, end);for (auto i : left) {for (auto j : right) {TreeNode* root = new TreeNode(k);root->left = i;root->right = j;subTree.push_back(root);}}}return subTree;}
};

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【重点递归构造二叉树】LeetCode 95. Unique Binary Search Trees II

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