LeetCode Unique Binary Search Trees
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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
思路分析:这题关键在于把握BST的结构特征,从i=0,1,2...逐渐找出递推规律,用count[i]表示有i个节点是BST的数量,那么
i = 0 时 count[i] = 1// 空树
i = 1 时 count[i] = 1//以1为root
i = 2 时 count[i] = count[0]* count[1]//以1为root
+ count[1]*count[0]//以2为root
i = 3 是 count[i] = count[0]*count[2]//以1为root 都在右子树
+ count[1]*count[1]//以2为root 左边右边各一个
+ count[2]*count[0]//以3为root 都在左子树
....
也就是对于 节点1,2,3...i,i+1,...n而言,如果选择i当作root,左边有1,2...i-1,右边有i+1...n。 于是左右子树又划归成了规模较小的已经解决的问题,很容易找到如下递推公式,用DP求解。
count[i] = ∑ count[k] *count [ i-1-k] 0<=k<=i-1
这个就是卡特兰数的一个定义方式。
DP实现,时间复杂度O(n^2),空间复杂度O(n)
AC Code
public class Solution { public int numTrees(int n) { int [] C = new int[n+1]; C[0] = 1; C[1] = 1; for(int i = 2; i <= n; i++){ C[i] = 0; for(int j = 0; j < i; j++){ C[i] += C[j]*C[i-j-1]; } } return C[n]; }}
其他参考题解
http://www.programcreek.com/2014/05/leetcode-unique-binary-search-trees-java/
http://blog.csdn.net/linhuanmars/article/details/24761459
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