Question
Formatted question description: https://leetcode.ca/all/96.html
96 Unique Binary Search Trees
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
@tag-tree
Algorithm
Let’s first look at the situation when n = 1, only a single binary search tree can be formed. The situations where n is 1, 2, and 3 are as follows:
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
We assign 1 when n = 0, because the empty tree is also a binary search tree, then the situation when n = 1 can be regarded as the number of left subtrees multiplied by the number of right subtrees, and the left and right subtrees Trees are all empty trees, so 1 by 1 is still 1. Then when n = 2, since both 1 and 2 can be roots, you can calculate them separately and add them together. The case of n = 2 can be calculated by the following formula (where dp[i]
represents the number of BST that can be composed of i numbers):
dp[2] =
dp[0] * dp[1]
(If 1 is the root, the left subtree must not exist, and the right subtree can have a number)
+
dp[1] * dp[0]
(If 2 is the root, the left subtree can have a number, and the right subtree must not exist)
In the same way, the calculation method for n = 3 can be written:
dp[3] =
dp[0] * dp[2]
(If 1 is the root, the left subtree must not exist, and the right subtree can have two numbers)
+
dp[1] * dp[1]
(If 2 is the root, the left and right subtrees can each have a number)
+
dp[2] * dp[0]
(If 3 is the root, the left subtree can have two numbers, and the right subtree must not exist)
Code
Java
public class Unique_Binary_Search_Trees {
public static void main(String[] args) {
Unique_Binary_Search_Trees out = new Unique_Binary_Search_Trees();
Solution s = out.new Solution();
System.out.println(s.numTrees(3));
}
public class Solution {
public int numTrees(int n) {
if (n <= 0) {
return 0;
}
// dp[i] represents the number of BST that can be composed of i numbers
int[] dp = new int[n + 1];
dp[0] = 1; // null is counted as one unique tree
for (int i = 1; i <= n; i++) { // 1...n
for (int j = 1; j <= i; j++) {
// for each fixed n, calculate its sum
dp[i] += dp[j - 1] * dp[i - j];
}
}
return dp[n];
}
}
public class Solution_recursion {
public int numTrees(int n) {
if (n < 0) {
return 0;
}
if (n == 0 || n == 1) {
return 1;
}
int sum = 0;
for (int i = 1; i <= n; i++) {
// 2 conditions: unique && BST. => inorder-visit will generate ordered sequence
// so, if decide root, then left and right can be calculated
// @note: root is "i", left has "i-1" nodes, right has "n - (i - 1) - 1"=="n - i" nodes
sum += numTrees(i - 1) * numTrees(n - i);
}
return sum;
}
}
}