[LeetCode] 96. Unique Binary Search Trees 唯一二叉搜索树

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

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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

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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

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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++:

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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];     } };

　

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[LeetCode] 96. Unique Binary Search Trees II 唯一二叉搜索树 II

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