* Unique Binary Search Trees && 95. Unique Binary Search Trees II && 241. Different Ways to Add parentheses

. Unique Binary Search Treesgiven NHow 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

Tree Dynamic Programming

 Public classSolution { Public intNumtrees (intN) {//numtrees (n) = Sum (Numtrees (n-1-i) * Numtrees (n-1-i)), where I is [0,n-1]        if(n==0)            return1; int[] Nums =New int[N+1]; nums[0] = 1; NUMS[1] = 1;  for(inti = 2; i<n+1; ++i) {intTotalchildren = i-1;  for(intleft = 0; left<=totalchildren;++Left ) {Nums[i]+=nums[left]*nums[totalchildren-Left ]; }        }        returnNums[n]; /*public int numtrees (int n) {if (n==0) return 1;                        if (n <= 2) return n;            int total = 0;                for (int i = 1; i<=n;++i) {int left = Numtrees (i-1);                int right = Numtrees (n-i);            Total + = left * RIGHT;        } return total; }        */    }}

Unique Binary Search Trees II

Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n.

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

Tree Dynamic Programming

/*** Definition for a binary tree node. * public class TreeNode {* int val; * TreeNode left; * TreeNode RI Ght * TreeNode (int x) {val = x;} }*/ Public classSolution { PublicList<treenode> Generatetrees (intN) {returnGeneratetrees (1, N); }        PrivateList<treenode> Generatetrees (intFromintTo ) {List<TreeNode> results =NewArraylist<treenode>(); if(From >to )returnresults; if(From = =To ) {Results.add (NewTreeNode (from)); returnresults; }                 for(inti = from; i<=to; ++i) {List<TreeNode> left = Generatetrees (from, i-1); List<TreeNode> right = Generatetrees (i+1, to); if(left.size () = = 0) {                 for(TreeNode r:right) {TreeNode root=NewTreeNode (i); Root.right=R;                Results.add (root); }            }            Else if(right.size () = = 0) {                 for(TreeNode l:left) {TreeNode root=NewTreeNode (i); Root.left=l;                Results.add (root); }            }            Else{                 for(TreeNode l:left) { for(TreeNode r:right) {TreeNode root=NewTreeNode (i); Root.left=l; Root.right=R;                    Results.add (root); }                }            }        }        returnresults; }}

241. Different Ways to ADD parentheses

Given A string of numbers and operators, return all possible results from computing all the different possible ways to Gro Up numbers and operators. The valid operators + are, - and * .

Example 1

Input: "2-1-1" .

((2-1)-1) = 0 ((1-1)) = 2

Output:[0, 2]

Example 2

Input:"2*3-4*5"

((4*5)) =-34 ((2*3)-(4*5)) = 14 ((3-4)) = 10 (((3-4))) =-10 (((2*3) 4) = 10

Output:[-34, -14, -10, -10, 10]

Divide and Conquer

* Unique Binary Search Trees && 95. Unique Binary Search Trees II && 241. Different Ways to Add parentheses