Minimum Height Trees (DFS)中,如何确定根节点? LeetCode 310. Minimum Height Trees (DFS)的DFS遍历过程是怎样的? 题目 给你一个无向无环图,这个图的任何一个节点都可以当成一个树的根节点。让你找到形成的树的高度最小的那几个根节点。 首先,这个根节点要么只有1个,要么只有2个。 而且就在图中最长
Can you solve this real interview question? Minimum Height Trees - A tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree. Given a tree of n nodes la
https://leetcode.com/problems/minimum-height-trees/ 有trick 对graph不熟悉。再看几遍。 参考http://bookshadow.com/weblog/2015/11/26/leetcode-minimum-height-trees/ 基本思路是“逐层删去叶子节点,直到剩下根节点为止” 有点类似于拓扑排序 最终剩下的节点个数可能为1或者2 时间复杂度:O(n),其中n为...
* TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* dfs(const vector<int> &nums, int L, int R) { if(L > R) { return nullptr; ...