我们知道二叉查找树能够随意取根。从处理子问题的角度来看,选取一个结点为根。可把结点分成左右子树,以这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,所以总数量是将以全部结点为根的可行结果累加起来。 看到有人说到卡特兰数,这正是卡特兰数的一种定义方式。是一个典型的动态规划的定义方式。 卡特兰数的一半公式为Cn
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 注意:二分查找树的...
LeetCode: 96. Unique Binary Search Trees 题目描述 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...
0 当没有一个二分搜索树没有任何节点也是一种情况,即Tree[0] = 1 1 当一个二分搜索树只有一个节点也是一种情况,即Tree[1] = 1 2 当一个二分搜索树有2个节点时,肯定要有一个节点作为根节点,那样总的数量就是,Tree[2] = Tree[0] * Tree[1] + Tree[1] * [0] 3 当一个二分搜索树有3个节...
@(LeetCode) 问题描述 给定一个整数n,能构造出多少种BST,使其节点值包括1~n? 栗子: 输入:3 输出:5 解释: 可构造出 5 种 BST,如下所示: 1 3 3 2 1 \ / / / \ \ 3 2 1 1 3 2 / / \ \ 2 1 2 3 想看英文原文的戳这里。
95. Unique Binary Search Trees II Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n. Example: Input:3Output:[ [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 ...
* TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<TreeNode*> generateTrees(int n) { if (n == 0) { vector<TreeNode*> ret; return ret; ...
Leetcode-Medium 98. Validate Binary Search Tree 编程算法python 判定一棵树是否满足二叉搜索树的性质。二叉查找树(Binary Search Tree),(又:二叉搜索树,二叉排序树)它或者是一棵空树,或者是具有下列性质的二叉树: 若它的左子树不空,则左子树上所有结点的值均小于它的根结点的值; 若它的右子树不空,则右子树...
(n == 0) { return ans; } TreeNode root = new TreeNode(0); //作为一个哨兵节点 getAns(n, ans, root, 0); return ans; } private void getAns(int n, List<TreeNode> ans, TreeNode root, int count) { if (count == n) { //复制当前树并且加到结果中 TreeNode newRoot = treeCopy...
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