Unique binary search trees,unique binary search Trees2 generate two-fork sort tree

Unique Binary Search Trees: To generate the number of two-fork sort trees.

Given N, how many structurally unique BST's (binary search trees) that store values 1 ... n?

For example,
Given N = 3, there is a total of 5 unique BST ' s.

   1         3     3      2      1    \//      /\           3     2     1      1   3      2    /     /       \                    2     1         2                 3

Algorithm analysis: Similar to the ladder, simple dynamic planning problems. When the root node is I, the node that is smaller than I is i-1, and the node of I is n-i, so I can generate a two-fork sort tree for the root node.

Nums[n] + = Nums[i-1]*nums[n-i],i from 1 to N.

public class Uniquebinarysearchtrees{public int numtrees (int n) {        if (n <= 0)        {        return 0;        }        int[] res = new Int[n+1];        Res[0] = 1;        RES[1] = 1;        for (int i = 2; I <= n; i + +)        {        for (int j = 1; J <= I; j + +)//j as root node        {        Res[i] + = res[j-1]*res[i-j];< c12/>}        }        return res[n];}}    

Unique Binary Search Trees2: A collection of root nodes that generate a two-fork sort tree

Given an integer 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

Algorithm analysis: This is not the number, but to generate root nodes. Use recursion.

public class Uniquebinarysearchtreesii {public list<treenode> generatetrees (int n) {if (n <= 0) {return new Arrayl Ist<treenode> ();} return helper (1, n);    } Public list<treenode> Helper (int m, int n) {list<treenode> res = new arraylist<> (); if (M > N) {res.add ( NULL); return res;} for (int i = m; i <= N; i + +) {//i is the root node list<treenode> ls = helper (m, i-1);//i node's left subtree list<treenode> rs = helper ( I+1, n); Right subtree of//i node for (TreeNode l:ls) {TreeNode r:rs = new TreeNode (i); curr = TreeNode = Curr.left S.add (Curr);}}} return res;}}

Unique binary search trees,unique binary search Trees2 generate two-fork sort tree