Leetcode Unique Binary Search Trees

1. Topics

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

2. Solution

Class Solution {public      :          int numtrees (int n) {              vector<int> num;              Num.push_back (1);              for (int i=1; i<=n; i++) {                  num.push_back (0);                  if (i<3)                      num[i]=i;                  else{for                      (int j=1; j<=i; j + +)                          num[i]+=num[j-1]*num[i-j];}              }              return num[n];          }      ;  

Ideas：

We'll start with the 0 analysis.

0 when there is no one binary search tree without any node is also a case, that is tree[0] = 1

1 when a binary search tree has only one node is also a case, i.e. tree[1] = 1

2 when a binary search tree has 2 nodes, it must have a node as the root node, so the total quantity is, tree[2] = tree[0] * tree[1] + tree[1] * [0]

3 When a binary search tree has 3 nodes, there must be a node as the root node, so the total quantity is, tree[3] = tree[0] * tree[2] + tree[1] * [1] + TREE[2] * [0]

The idea of dynamic planning is used here to save the previously counted in an array.

http://www.waitingfy.com/archives/1617

Leetcode Unique Binary Search Trees