给出 n，问由 1…n 为节点组成的不同的二叉查找树有多少种？
例如，
给出 n = 3，则有 5 种不同形态的二叉查找树：
   1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3
详见：https://leetcode.com/problems/unique-binary-search-trees-ii/description/

Java实现：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<TreeNode> generateTrees(int n) {
        List<TreeNode> res=new ArrayList<TreeNode>();
        if(n<1){
            return res;
        }
        return generateTrees(1,n);
    }
    private ArrayList<TreeNode> generateTrees(int left, int right){
        ArrayList<TreeNode> res = new ArrayList<TreeNode>();
        if (left > right){
            res.add(null);
            return res;
        }
        for (int i = left; i <= right; i++){
            ArrayList<TreeNode> lefts = generateTrees(left, i-1);//以i作为根节点，左子树由[1,i-1]构成
            ArrayList<TreeNode> rights = generateTrees(i+1, right);//右子树由[i+1, n]构成
            for (int j = 0; j < lefts.size(); j++){
                for (int k = 0; k < rights.size(); k++){
                    TreeNode root = new TreeNode(i);
                    root.left = lefts.get(j);
                    root.right = rights.get(k);
                    res.add(root);//存储所有可能行
                }
            }
        }
        return res;
    }
}

 C++实现：

/**
 * 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 buildTree(1,n);
    }
    vector<TreeNode*> buildTree(int start,int end)
    {
        vector<TreeNode*> trees;
        if(start>end)
        {
            trees.push_back(nullptr);
            return trees;
        }
        if(start==end)
        {
            trees.push_back(new TreeNode(start));
            return trees;
        }
        for(int i=start;i<=end;++i)
        {
            vector<TreeNode*> leftTree=buildTree(start,i-1);
            vector<TreeNode*> rightTree=buildTree(i+1,end);
            for(int j=0;j<leftTree.size();++j)
            {
                for(int k=0;k<rightTree.size();++k)
                {
                    TreeNode *root=new TreeNode(i);
                    root->left=leftTree[j];
                    root->right=rightTree[k];
                    trees.push_back(root);
                }
            }
        }
        return trees;
    }
};