标签：style blog color 使用 io strong for ar
Given n, how many structurally unique BST‘s (binary search trees) that store values 1...n?
For example,
Given n = 3, there are a total of 5 unique BST‘s.
1 3 3 2 1
\ / / / \ 3 2 1 1 3 2
/ / \ 2 1 2 3
思路：若取第i个作为顶点，其对应二叉树个数为：numTrees(i-1)*numTrees(n-i)。加和所有不同i即可求解该题。此处，为了方便递归，令numTrees(0)取1。递归解法如下：
1 class Solution { 2 public: 3 int numTrees( int n ) { 4 if( n == 0 ) { return 1; } 5 if( n <= 2 ) { return n; } 6 int num = 0; 7 for( int i = 1; i <= n/2; ++i ) { 8 num += 2 * numTrees( i-1 ) * numTrees( n-i ); 9 } 10 if( n % 2 == 1 ) { num += numTrees( n/2 ) * numTrees( n/2 ); } 11 return num; 12 } 13 };
使用num[i]保存numTrees(i)，动态规划解法如下：
1 class Solution { 2 public: 3 int numTrees(int n) { 4 if( n <= 2 ) { return n; } 5 vector<int> num( n, 0 ); 6 num[0] = 1; num[1] = 2; 7 for( int i = 2; i < n; ++i ) { 8 num[i] += 2*num[i-1]; 9 for( int j = 1; j < i; ++j ) { 10 num[i] += num[j-1] * num[i-j-1]; 11 } 12 } 13 return num.back(); 14 } 15 };
结果满足卡特兰递归式：h(n) = h(0)*h(n-1) + h(1)*h(n-2) + h(2)*h(n-3) + … + h(i)*h(n-i-1) + … + h(n-1)*h(0) ( n >= 2 )，且h(0) = 1，h(1) = 1。另一种递归形式为：h(n) = h(n-1) * (4*n-2) / (n+1)。
1 class Solution { 2 public: 3 int numTrees( int n ) { 4 if( n <= 1 ) { return n; } 5 int catalan = 1; 6 for( int i = 2; i <= n; ++i ) { 7 catalan = catalan * (4*i-2) / (i+1); 8 } 9 return catalan; 10 } 11 };
Unique Binary Search Trees,布布扣,bubuko.com
标签：style blog color 使用 io strong for ar
原文地址：http://www.cnblogs.com/moderate-fish/p/3918461.html
