解法一:递归 classSolution {public:intnumTrees(intn) {if(n ==0)return1;elseif(n ==1)return1;else{intcount =0;for(inti =0; i <= (n-1)/2; i ++) {if(i < n-1-i) count+=2*numTrees(i)*numTrees(n-1-i);elsecount+= numTrees(i)*numTrees(n-1-i); }returncount; } } ...
3. Binary Tree Zigzag Level Order Traversal Given a binary tree, return thezigzag level ordertraversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example: Given binary tree[3,9,20,null,null,15,7], 3 / \ 9 20 ...
题目地址:https://leetcode.com/problems/unique-binary-search-trees/description/ 题目描述 Givenn, how many structurally unique BST’s (binary search trees) that store values1...n? For example, Given n = 3, there are a total of 5 unique BST's. 1 3 3 2 1 \ / / / \ \ 3 2 1 1...
由于比較直接,这里就不列举代码了。 假设是求解全部满足要求的二叉树(而不不过数量)那么时间复杂度是就取决于结果的数量了,不再是一个多项式的解法了,有兴趣的朋友能够看看Unique Binary Search Trees II。
初看推导公式,很容易的想到的是递归解法。我也是采用的这种方式。 为了避免重复计算,使用了map来缓存已计算过的值。 js代码如下: varnumTrees=function(n){// 预设值letmap=newMap()map.set(0,1)map.set(1,1)map.set(2,2)map.set(3,5)const result=recursive(n,map)returnresult};// 89.50%varrecursi...
Unique Binary Search Trees I: https://leetcode.com/problems/unique-binary-search-trees/description/ 解题思路: 对于n:把从1到n分别作为根节点,然后左子树*右子树,最后对其全部相加即可 class Solution { public int numTrees(int n) { int[]dp=newint[n+1];dp[0]=1;for(inti=1;i<=n;i++){for...
leetcode:Unique Binary Search Trees II 问题描述: Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1...n. For example,Given n = 3, your program should return all 5 unique BST's shown below. 1 3 3 2 1 \ / / / \ \ 3 2 1 1 ...
Given an integern, returnthe number of structurally uniqueBST's (binary search trees) which has exactlynnodes of unique values from1ton. Example 1: Input:n = 3Output:5 Example 2: Input:n = 1Output:1 Constraints: 1 <= n <= 19 ...
class Solution(object): def generateTrees(self, n): if n == 0: return [] return self.helper(1, n) def helper(self, start, end): result = [] if start > end: result.append(None) return result for i in range(start, end + 1): # generate left and right sub tree leftTree = ...
* Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */classSolution { public: vector<TreeNode*> generateTrees(intn){if(n==0) { ...