思路:递归,由于是二叉查找树,先选择任一结点根结点,假设为结点i,则[1,i-1]范围的结点为结点i的左子树结点,[i+1,n]范围的结点为结点i的右子树结点,则以结点i为根结点的BST个数为左,右子树可构成BST个数的乘积,基于这个思路,可以写出以下递归程序。 第一种方法使用动态规划 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
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 ...
比如,以1为根的树有几个,完全取决于有二个元素的子树有几种。同理,2为根的子树取决于一个元素的子树有几个。以3为根的情况,则与1相同。 定义Count[i] 为以[0,i]能产生的Unique Binary Tree的数目, 如果数组为空,毫无疑问,只有一种BST,即空树,Count[0] =1 如果数组仅有一个元素{1},只有一种BST,...
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...
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 ...
Unique Binary Search Trees II: https://leetcode.com/problems/unique-binary-search-trees-ii/description/ 解题思路: 思想如上题,但是在每一步需要create TreeeNode class Solution { public List<TreeNode> generateTrees(int n) { List<TreeNode>[]dp=newList[n+1];dp[0]=newArrayList<TreeNode>();if...
* Definition for a binary tree node. * 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) { ...
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...
LeetCode 938. Range Sum of BST 2019-12-12 08:00 − 原题链接在这里:https://leetcode.com/problems/range-sum-of-bst/ 题目: Given the root node of a binary search tree, return the sum of values of all node... Dylan_Java_NYC 0 490 【leetcode】1289. Minimum Falling Path Sum ...
(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...