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
/**
 * 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 *> generateTrees(int n) 
    {
        if(n==0) 
        {
            vector<TreeNode *> listTree;
            TreeNode * root=NULL;
            listTree.push_back(root);
            return listTree;
        }
        return generate(1,n);
    }
    
    vector<TreeNode*> generate(int l,int r)
    {
        vector<TreeNode*> listTree;
        if(l>r) return listTree;
        
        for(int i=l;i<=r;i++)
        {            
            vector<TreeNode*> left=generate(l,i-1);
            vector<TreeNode*> right=generate(i+1,r);
            if(left.size()==0 && right.size()==0)
            {
                TreeNode* root=new TreeNode(i);
                listTree.push_back(root);
                continue;
            }
            if(left.size()==0)
            {
                for(int i1=0;i1<right.size();i1++)
                {
                    TreeNode* root=new TreeNode(i);
                    root->right=right[i1];
                    listTree.push_back(root);
                }
                continue;
            }
            if(right.size()==0)
            {
                for(int i2=0;i2<left.size();i2++)
                {
                    TreeNode* root=new TreeNode(i);
                    root->left=left[i2];
                    listTree.push_back(root);
                }
                continue;
            }
            for(int i1=0;i1<right.size();i1++)
                for(int i2=0;i2<left.size();i2++)
                {
                    TreeNode* root=new TreeNode(i);
                    root->right=right[i1];
                    root->left=left[i2];
                    listTree.push_back(root);
                }
        }
        return listTree;
    }
};
原文地址:https://www.cnblogs.com/erictanghu/p/3759576.html