https://leetcode-cn.com/problems/unique-binary-search-trees-ii/ https://leetcode-cn.com/problems/unique-binary-search-trees-ii/solution/xiang-xi-tong-su-de-si-lu-fen-xi-duo-jie-fa-by-2-7/
* struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<TreeNode*> generateTrees(int n) { return generateTrees(1, n); } vector<TeeNode*> generateTrees(int begin...
classSolution:defgenerateTrees(self,n:int)->List[TreeNode]:ret=[]# 拷贝二叉树 defcopyTree(node):ifnode is None:returnNone u=TreeNode(node.val)u.left=copyTree(node.left)u.right=copyTree(node.right)returnu defdfs(n):# n=1只有一种情况ifn==1:ret.append(TreeNode(1))returndfs(n-1)# ...
leetcode刷题笔记九十五题 不同的二叉搜索树II 源地址:95. 不同的二叉搜索树 II 问题描述: 给定一个整数 n,生成所有由 1 ... n 为节点所组成的 二叉搜索树 。 示例: 输入:3 输出: [ [1,null,3,2], [3,2,null,1], [3,1,null,null
Solution 思路:回溯法。因为二叉树搜索树的特性,每次选择一个数做根结点,然后获取左子树的所有组合和右子树的所有组合方式,对它们进行组合就能得到当前树的所有组合方式。 Language:java /** * Definition for a binary tree node. * public class TreeNode { ...
* };*/classSolution {public: vector<TreeNode*> Build(intlow,inthight) {if(low > hight)return{}; vector<TreeNode* >tmp;//if(low == hight)//{//TreeNode* now = new TreeNode(low);//tmp.push_back(now);//return tmp;//}for(inti=low;i<=hight;i++) ...
LeetCode 95. Unique Binary Search Trees II 题目 DFS, classSolution{public: int vis[100005]; int m; vector<TreeNode*> generateTrees(int n) { vector<TreeNode*> ans;if(n==0)returnans;returnfun(1,n); } vector<TreeNode*>fun(int l,int r){...
# class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def generateTrees(self, n: int) -> List[TreeNode]: ret = [] # 拷贝二叉树 def copyTree(node): if node is None: return None ...
class Solution { public: vector<TreeNode*> generateTrees(int n) { if (n == 0) return {}; return helper(1, n); } vector<TreeNode*> helper(int start, int end) { if (start > end) return {nullptr}; vector<TreeNode*> res; for (int i = start; i <= end; ++i) { auto left...
* struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<TreeNode*> generateTrees(int n) { vector<TreeNode*> res; if(n==0)//边界情况 return res; return get(1...