# Unique Binary Search Trees
### Source
- leetcode: [Unique Binary Search Trees | LeetCode OJ](https://leetcode.com/problems/unique-binary-search-trees/)
- lintcode: [(163) Unique Binary Search Trees](http://www.lintcode.com/en/problem/unique-binary-search-trees/)
~~~
Given n, how many structurally unique BSTs (binary search trees)
that store values 1...n?
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
~~~
### 题解1 - 两重循环
挺有意思的一道题，与数据结构和动态规划都有点关系。这两天在骑车路上和睡前都一直在想，始终未能找到非常明朗的突破口，直到看到这么一句话——『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。』这不就是 BST 的定义嘛！灵活运用下就能找到递推关系了。
容易想到这道题的动态规划状态为 count[n], count[n] 表示到正整数 i 为止的二叉搜索树个数。容易得到 count[1] = 1, 根节点为1，count[2] = 2, 根节点可为1或者2。那么 count[3] 的根节点自然可为1，2，3. 如果以1为根节点，那么根据 BST 的定义，2和3只可能位于根节点1的右边；如果以2为根节点，则1位于左子树，3位于右子树；如果以3为根节点，则1和2必位于3的左子树。
抽象一下，如果以 i 作为根节点，由基本的排列组合知识可知，其唯一 BST 个数为左子树的 BST 个数乘上右子树的 BST 个数。故对于 i 来说，其左子树由[0, i - 1]构成，唯一的 BST 个数为 count[i - 1], 右子树由[i + 1, n] 构成，其唯一的 BST 个数没有左子树直观，但是也有迹可循。对于两组有序数列「1, 2, 3] 和 [4, 5, 6]来说，**这两个有序数列分别组成的 BST 个数必然是一样的，因为 BST 的个数只与有序序列的大小有关，而与具体值没有关系。**所以右子树的 BST 个数为 count[n - i]，于是乎就得到了如下递推关系：count[i]=∑j=0i−1(count[j]⋅count[i−j−1])count[i] = \sum _{j = 0} ^{i - 1} (count[j] \cdot count[i - j - 1])count[i]=∑j=0i−1(count[j]⋅count[i−j−1])
网上有很多用 count[3] 的例子来得到递推关系，恕本人愚笨，在没有从 BST 的定义和有序序列个数与 BST 关系分析的基础上，我是不敢轻易说就能得到如上状态转移关系的。
### Python
~~~
class Solution:
# @paramn n: An integer
# @return: An integer
def numTrees(self, n):
if n < 0:
return -1
count = [0] * (n + 1)
count[0] = 1
for i in xrange(1, n + 1):
for j in xrange(i):
count[i] += count[j] * count[i - j - 1]
return count[n]
~~~
### C++
~~~
class Solution {
public:
/**
* @paramn n: An integer
* @return: An integer
*/
int numTrees(int n) {
if (n < 0) {
return -1;
}
vector<int> count(n + 1);
count[0] = 1;
for (int i = 1; i != n + 1; ++i) {
for (int j = 0; j != i; ++j) {
count[i] += count[j] * count[i - j - 1];
}
}
return count[n];
}
};
~~~
### Java
~~~
public class Solution {
/**
* @paramn n: An integer
* @return: An integer
*/
public int numTrees(int n) {
if (n < 0) {
return -1;
}
int[] count = new int[n + 1];
count[0] = 1;
for (int i = 1; i < n + 1; ++i) {
for (int j = 0; j < i; ++j) {
count[i] += count[j] * count[i - j - 1];
}
}
return count[n];
}
}
~~~
### 源码分析
1. 对 n 小于0特殊处理。
1. 初始化大小为 n + 1 的数组，初始值为0，但对 count[0] 赋值为1.
1. 两重 for 循环递推求得 count[i] 的值。
1. 返回 count[n] 的值。
由于需要处理空节点的子树，故初始化 count[0] 为1便于乘法处理。其他值必须初始化为0，因为涉及到累加操作。
### 复杂度分析
一维数组大小为 n + 1, 空间复杂度为 O(n+1)O(n + 1)O(n+1). 两重 for 循环等差数列求和累计约 n2/2n^2 / 2n2/2, 故时间复杂度为 O(n2)O(n^2)O(n2). 此题为 Catalan number 的一种，除了平方时间复杂度的解法外还存在 O(n)O(n)O(n) 的解法，欲练此功，先戳 [Wikipedia](http://en.wikipedia.org/wiki/Catalan_number) 的链接。
### Reference
- fisherlei
> .
[水中的鱼: [LeetCode] Unique Binary Search Trees, Solution](http://fisherlei.blogspot.com/2013/03/leetcode-unique-binary-search-trees.html)[ ↩](# "Jump back to footnote [fisherlei] in the text.")
- [Unique Binary Search Trees | 九章算法](http://www.jiuzhang.com/solutions/unique-binary-search-trees/)
- [Catalan number - Wikipedia, the free encyclopedia](http://en.wikipedia.org/wiki/Catalan_number)
- Preface
- Part I - Basics
- Basics Data Structure
- String
- Linked List
- Binary Tree
- Huffman Compression
- Queue
- Heap
- Stack
- Set
- Map
- Graph
- Basics Sorting
- Bubble Sort
- Selection Sort
- Insertion Sort
- Merge Sort
- Quick Sort
- Heap Sort
- Bucket Sort
- Counting Sort
- Radix Sort
- Basics Algorithm
- Divide and Conquer
- Binary Search
- Math
- Greatest Common Divisor
- Prime
- Knapsack
- Probability
- Shuffle
- Basics Misc
- Bit Manipulation
- Part II - Coding
- String
- strStr
- Two Strings Are Anagrams
- Compare Strings
- Anagrams
- Longest Common Substring
- Rotate String
- Reverse Words in a String
- Valid Palindrome
- Longest Palindromic Substring
- Space Replacement
- Wildcard Matching
- Length of Last Word
- Count and Say
- Integer Array
- Remove Element
- Zero Sum Subarray
- Subarray Sum K
- Subarray Sum Closest
- Recover Rotated Sorted Array
- Product of Array Exclude Itself
- Partition Array
- First Missing Positive
- 2 Sum
- 3 Sum
- 3 Sum Closest
- Remove Duplicates from Sorted Array
- Remove Duplicates from Sorted Array II
- Merge Sorted Array
- Merge Sorted Array II
- Median
- Partition Array by Odd and Even
- Kth Largest Element
- Binary Search
- Binary Search
- Search Insert Position
- Search for a Range
- First Bad Version
- Search a 2D Matrix
- Search a 2D Matrix II
- Find Peak Element
- Search in Rotated Sorted Array
- Search in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array
- Find Minimum in Rotated Sorted Array II
- Median of two Sorted Arrays
- Sqrt x
- Wood Cut
- Math and Bit Manipulation
- Single Number
- Single Number II
- Single Number III
- O1 Check Power of 2
- Convert Integer A to Integer B
- Factorial Trailing Zeroes
- Unique Binary Search Trees
- Update Bits
- Fast Power
- Hash Function
- Count 1 in Binary
- Fibonacci
- A plus B Problem
- Print Numbers by Recursion
- Majority Number
- Majority Number II
- Majority Number III
- Digit Counts
- Ugly Number
- Plus One
- Linked List
- Remove Duplicates from Sorted List
- Remove Duplicates from Sorted List II
- Remove Duplicates from Unsorted List
- Partition List
- Two Lists Sum
- Two Lists Sum Advanced
- Remove Nth Node From End of List
- Linked List Cycle
- Linked List Cycle II
- Reverse Linked List
- Reverse Linked List II
- Merge Two Sorted Lists
- Merge k Sorted Lists
- Reorder List
- Copy List with Random Pointer
- Sort List
- Insertion Sort List
- Check if a singly linked list is palindrome
- Delete Node in the Middle of Singly Linked List
- Rotate List
- Swap Nodes in Pairs
- Remove Linked List Elements
- Binary Tree
- Binary Tree Preorder Traversal
- Binary Tree Inorder Traversal
- Binary Tree Postorder Traversal
- Binary Tree Level Order Traversal
- Binary Tree Level Order Traversal II
- Maximum Depth of Binary Tree
- Balanced Binary Tree
- Binary Tree Maximum Path Sum
- Lowest Common Ancestor
- Invert Binary Tree
- Diameter of a Binary Tree
- Construct Binary Tree from Preorder and Inorder Traversal
- Construct Binary Tree from Inorder and Postorder Traversal
- Subtree
- Binary Tree Zigzag Level Order Traversal
- Binary Tree Serialization
- Binary Search Tree
- Insert Node in a Binary Search Tree
- Validate Binary Search Tree
- Search Range in Binary Search Tree
- Convert Sorted Array to Binary Search Tree
- Convert Sorted List to Binary Search Tree
- Binary Search Tree Iterator
- Exhaustive Search
- Subsets
- Unique Subsets
- Permutations
- Unique Permutations
- Next Permutation
- Previous Permuation
- Unique Binary Search Trees II
- Permutation Index
- Permutation Index II
- Permutation Sequence
- Palindrome Partitioning
- Combinations
- Combination Sum
- Combination Sum II
- Minimum Depth of Binary Tree
- Word Search
- Dynamic Programming
- Triangle
- Backpack
- Backpack II
- Minimum Path Sum
- Unique Paths
- Unique Paths II
- Climbing Stairs
- Jump Game
- Word Break
- Longest Increasing Subsequence
- Palindrome Partitioning II
- Longest Common Subsequence
- Edit Distance
- Jump Game II
- Best Time to Buy and Sell Stock
- Best Time to Buy and Sell Stock II
- Best Time to Buy and Sell Stock III
- Best Time to Buy and Sell Stock IV
- Distinct Subsequences
- Interleaving String
- Maximum Subarray
- Maximum Subarray II
- Longest Increasing Continuous subsequence
- Longest Increasing Continuous subsequence II
- Graph
- Find the Connected Component in the Undirected Graph
- Route Between Two Nodes in Graph
- Topological Sorting
- Word Ladder
- Bipartial Graph Part I
- Data Structure
- Implement Queue by Two Stacks
- Min Stack
- Sliding Window Maximum
- Longest Words
- Heapify
- Problem Misc
- Nuts and Bolts Problem
- String to Integer
- Insert Interval
- Merge Intervals
- Minimum Subarray
- Matrix Zigzag Traversal
- Valid Sudoku
- Add Binary
- Reverse Integer
- Gray Code
- Find the Missing Number
- Minimum Window Substring
- Continuous Subarray Sum
- Continuous Subarray Sum II
- Longest Consecutive Sequence
- Part III - Contest
- Google APAC
- APAC 2015 Round B
- Problem A. Password Attacker
- Microsoft
- Microsoft 2015 April
- Problem A. Magic Box
- Problem B. Professor Q's Software
- Problem C. Islands Travel
- Problem D. Recruitment
- Microsoft 2015 April 2
- Problem A. Lucky Substrings
- Problem B. Numeric Keypad
- Problem C. Spring Outing
- Microsoft 2015 September 2
- Problem A. Farthest Point
- Appendix I Interview and Resume
- Interview
- Resume