题目链接:https://leetcode.com/problems/sum-of-distances-in-tree/ 题意:给出一棵树,对于每个节点,求出其他所有节点与该节点的距离之和。节点数不超过10000。 思路:如图所示,暴力复杂度太高,考虑进行优化,如图所示,考虑一对父子节点u与v,son[u]表示以u为根节点的子树的节点个数。若已知父节点u到其他所有...
这里是从上而下的更新,可以使用最常用的先序遍历,可以参见这道题Binary Tree Preorder Traversal,这样更新下来,所有的 res[i] 就都是题目中要求的值了,参见代码如下: classSolution{public:vector<int>sumOfDistancesInTree(intN, vector<vector<int>>& edges){vector<int>res(N),count(N); vector<vector<in...
Can you solve this real interview question? Binary Search Tree to Greater Sum Tree - Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus the sum of all key
Can you solve this real interview question? Maximum Sum BST in Binary Tree - Given a binary tree root, return the maximum sum of all keys of any sub-tree which is also a Binary Search Tree (BST). Assume a BST is defined as follows: * The left subtree
目来自leetcode3318。 解决步骤 1. 滑动窗口:我们需要遍历所有长度为 k 的子数组。滑动窗口的起始位置从 0 到n - k。 2. 频率统计:对于每个子数组,统计每个元素的出现频率。 3. 选择前 x 个元素: • 将频率和元素值组合成 (频率, 元素值) 的二元组。 • 对这些二元组排序:先按频率降序,频率相同则...
LeetCode_437. Path Sum III 437. Path Sum III Easy You are given a binary tree in which each node contains an integer value. Find the number of paths that sum to a given value. The path does not need to start or end at the root or a leaf, but it must go downwards (traveling ...
Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays. Note:If n is the length of array, assume the following constraints are satisfied:...
0789-minimum-distance-between-two-bst-nodes.go 0846-hand-of-straights.go 0852-peak-index-in-a-mountain-array.go 0853-car-fleet.go 0875-koko-eating-bananas.go 0876-middle-of-the-linked-list.go 0904-fruit-into-baskets.go 0912-sort-an-array.go 0918-maximum-sum-circular-subarray.go 0926-fli...
LeetCode 404 sum of left leaves Find the sum of all left leaves in a given binary tree think about this, when we reached leaf, how are we gonna to know if it is left leaf or not? of course we can modify the iterate version of preorder traverse, it’s very simple....
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