Leetcode 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

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解题思路：

dynamic programming. 最重要的就是找规律！！！

这类找combination/permutation的题都需要找找规律。

 

n = 0

 

n = 1

1

 

n = 2

   1                  2

     \                /

      2            1

 

n = 3

 1           3    3      2     1

    \        /     /       / \       \

     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2：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)

 

Let count[i] be the number of unique binary search trees for i. The number of trees are determined by the number of subtrees which have different root node. For example,

i=0, count[0]=1 //empty tree

i=1, count[1]=1 //one tree

i=2, count[2]=count[0]*count[1] // 0 is root
            + count[1]*count[0] // 1 is root

i=3, count[3]=count[0]*count[2] // 1 is root
            + count[1]*count[1] // 2 is root
            + count[2]*count[0] // 3 is root

i=4, count[4]=count[0]*count[3] // 1 is root
            + count[1]*count[2] // 2 is root
            + count[2]*count[1] // 3 is root
            + count[3]*count[0] // 4 is root
..
..
..

Recurrence formula: i=n, count[n] = sum(count[0..k]*count[k+1...n]) 0 <= k < n-1
base case: count[0] = 1 count[1] = 1

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Java code:

public 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; j++){
                count[i] += count[j] * count[i-1-j];
            }
        }
        return count[n];
    }
}

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Reference:

1. http://www.programcreek.com/2014/05/leetcode-unique-binary-search-trees-java/

2. http://bangbingsyb.blogspot.com/2014/11/leetcode-unique-binary-search-trees-i-ii.html