leetcode Unique Binary Search Trees
2014-08-21 20:24
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因为涉及到子问题的子问题会多次被计算,因此开辟空间保存中间结果,动态规划问题
特殊性:
个数为0或1时二叉树的个数为1
个数为2时二叉树的个数为2
个数大于等于3时,二叉树的个数为确定根节点后做右子树的二叉树个数的乘积
代码
特殊性:
个数为0或1时二叉树的个数为1
个数为2时二叉树的个数为2
个数大于等于3时,二叉树的个数为确定根节点后做右子树的二叉树个数的乘积
代码
class Solution { public: int numTrees(int n) { if(n==0) return 1; if(n==1) return 1; int *num = new int[n+1]; num[0]=1; num[1]=1; num[2]=2; for(int i = 3; i <= n; ++i) { int temp = 0; for(int j = 0; j < i; ++j) { temp += num[j]*num[i-j-1]; } num[i] = temp; } return num ; } };
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