leetcode -- Unique Binary Search Trees todo
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
[解题思路]
该题一直没有思路,到网上搜索了之后,得到如下结果
1 1 2 3 3
\ \ / \ / /
3 2 1 3 2 1
/ \ / \
2 3 1 2
比如,以1为根的树有几个,完全取决于有二个元素的子树有几种。同理,2为根的子树取决于一个元素的子树有几个。以3为根的情况,则与1相同。
定义Count[i] 为以[0,i]能产生的Unique Binary Tree的数目,
如果数组为空,毫无疑问,只有一种BST,即空树,
Count[0] =1
如果数组仅有一个元素{1},只有一种BST,单个节点
Count[1] = 1
如果数组有两个元素{1,2}, 那么有如下两种可能
1 2
\ /
2 1
Count[2] = Count[0] * Count[1] (1为根的情况)
+ Count[1] * Count[0] (2为根的情况。
再看一遍三个元素的数组,可以发现BST的取值方式如下:
Count[3] = Count[0]*Count[2] (1为根的情况)
+ Count[1]*Count[1] (2为根的情况)
+ Count[2]*Count[0] (3为根的情况)
所以,由此观察,可以得出Count的递推公式为
Count[i] = ∑ Count[k] * [i-k-1] 0<=k<i-1
问题至此划归为一维动态规划。
public int numTrees(int n) {
// Start typing your Java solution below
// DO NOT write main() function
int[] count = new int[n+1];
count[0] = 1;
count[1] = 1;
for(int i = 2; i <= n; i++){
for(int j = 1; j <=i; j++){
count[i] += count[j - 1] * count[i - j];
}
}
return count[n];
}
[Note]
这是很有意思的一个题。刚拿到这题的时候,完全不知道从那下手,因为对于BST是否Unique,很难判断。最后引入了一个条件以后,立即就清晰了,即
当数组为 1,2,3,4,.. i,.. n时,基于以下原则的BST建树具有唯一性:
以i为根节点的树,其左子树由[0, i-1]构成, 其右子树由[i+1, n]构成。
ref:http://fisherlei.blogspot.com/2013/03/leetcode-unique-binary-search-trees.html
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