Given n, how many structurally unique BST's (binary search trees) that store values 1 ... n? Input: 3 Output: 5 Explanation: 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 题意: 给定n个节点,可形成多少种不同的BST 思…

[Leetcode] Unique binary search trees 唯一二叉搜索树 Unique Binary Search Trees leetcode java 描述 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…

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,…

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     \       /     /      / \  …

96. Unique Binary Search Trees https://www.cnblogs.com/grandyang/p/4299608.html 3由dp[1]*dp[1].dp[0]*dp[2].dp[2]*dp[0]相加而成 从2开始 class Solution { public: int numTrees(int n) { vector<); dp[] = ; dp[] = ; ;i <= n;i++){ ;j < i;j++){ dp[i] += dp[j] *…

Given n, how many structurally unique BST's (binary search trees) that store values 1 ... n? Example: Input: 3 Output: 5 Explanation: 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 这道题实际上是 卡塔兰数 Cat…

题目描述: 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 解题思路: 动态规划法. 用G(n)表示长度为n组成的二叉搜索树的数目: G(0)…

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  给一个正整数n,然后将这n个数进行二叉排序树的排列,求有多少种组合. public clas…

原题 字母题 思路: 一开始妹有一点思路,去查了二叉查找树,发现有个叫做卡特兰数的东西. 1.求可行的二叉查找树的数量,只要满足中序遍历有序. 2.以一个结点为根的可行二叉树数量就是左右子树可行二叉树数量的乘积. 3.总的数量是将以所有结点为根的可行结果累加起来. n = 0 时,因为空树也算一种二叉搜索树,则dp[0]=1; n = 1时,dp[1]=1; n = 2时 dp[2] = dp[0] * dp[1] (1为根的情况) + dp[1] * dp[0] (2为根的情况) n = 3时…

Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n. Input: 3 Output: [ [1,null,3,2], [3,2,null,1], [3,1,null,null,2], [2,1,3], [1,null,2,null,3] ] Explanation: The above output corresponds to th…