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
此题是卡塔兰数的一个应用。注意是BST而不是普通的Binary Tree，所以要满足左比根小，右比根大。
1 n = 1
2 1 n = 2
/ \
1 2
1 3 3 2 1 n = 3
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
定义f(n)为unique BST的数量，以n = 3为例：
构造的BST的根节点可以取{1, 2, 3}中的任一数字。
如以1为节点，则left subtree只能有0个节点，而right subtree有2, 3两个节点。所以left/right subtree一共的combination数量为：f(0) * f(2) = 2
以2为节点，则left subtree只能为1，right subtree只能为3：f(1) * f(1) = 1
以3为节点，则left subtree有1, 2两个节点，right subtree有0个节点：f(2)*f(0) = 2
总结规律：
f(0) = 1
f(n) = f(0)*f(n-1) + f(1)*f(n-2) + ... + f(n-2)*f(1) + f(n-1)*f(0)
Java: DP
class Solution {
public int numTrees(int n) {
int[] count = new int[n + 1];
count[0] = 1;
count[1] = 1;
for (int i = 2; i <= n; i++) {
for (int j = 0; j <= i - 1; j++) {
count[i] = count[i] + count[j] * count[i - j - 1];
}
}
return count[n];
}
}
Python: Math
class Solution(object):
def numTrees(self, n):
if n == 0:
return 1 def combination(n, k):
count = 1
# C(n, k) = (n) / 1 * (n - 1) / 2 ... * (n - k + 1) / k
for i in xrange(1, k + 1):
count = count * (n - i + 1) / i;
return count return combination(2 * n, n) - combination(2 * n, n - 1)
Python: DP
class Solution2:
def numTrees(self, n):
counts = [1, 1]
for i in xrange(2, n + 1):
count = 0
for j in xrange(i):
count += counts[j] * counts[i - j - 1]
counts.append(count)
return counts[-1]
C++:
class Solution {
public:
int numTrees(int n) {
vector<int> dp(n + 1, 0);
dp[0] = 1;
dp[1] = 1;
for (int i = 2; i <= n; ++i) {
for (int j = 0; j < i; ++j) {
dp[i] += dp[j] * dp[i - j - 1];
}
}
return dp[n];
}
};
类似题目：
[LeetCode] 96. Unique Binary Search Trees II 唯一二叉搜索树 II