"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

Since 1~n is the ascending sequence, the tree that is built naturally is BST.

Recursive thinking, select the root node, and then separate the left and right sub-sequences.

Since the results of the left and right sub-sequences may be more than one, consideration should be given to all collocation situations.

/** Definition for binary tree * struct TreeNode {* int val; * TreeNode *left; * TreeNode *right; * T Reenode (int x): Val (x), left (null), right (NULL) {} *}; */classSolution { Public: Vector<treenode *> Generatetrees (intN) {vector<treenode *>Res; if(n = =0) {res.push_back (NULL); returnRes; }        Else         {            returnHelper (1, N); }} vector<TreeNode*> Helper (intBeginintend) {Vector<TreeNode*>Res; if(Begin >end)            {Res.push_back (NULL); returnRes; }        Else if(Begin = =end) {TreeNode* Root =NewTreeNode (begin);            Res.push_back (root); returnRes; }        Else        {             for(inti = begin; I <= end; i + +)            {                //TagVector<treenode*> left = Helper (begin, I-1); Vector<TreeNode*> right = Helper (i+1, end);  for(intL =0; L < Left.size (); L + +)                {//All BST in left                     for(intR =0; R < Right.size (); R + +)                    {//All BST on right//New root here! not the tag position.treenode* root =NewTreeNode (i); Root->left =Left[l]; Root->right =Right[r];                    Res.push_back (root); }                }                            }            returnRes; }    }};

"Leetcode" Unique Binary Search Trees II