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

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;
    }
};