看這篇題解
首先要回憶一下握手定理，所以奇數度點一定有偶數個；其次回憶一下樹的必要條件，每個點的度數都大於一的話肯定不是樹；然後再嘗試構造，從特殊圖，樣例等等出發
Circular Spanning Tree
相關文章
- Spanning-tree的portfast介面會傳送bpdu嗎AST
- [Javascript] Circular dependencyJavaScript
- 2.7 Lab: Circular Shift
- 妙用ConstraintLayout的Circular positioningAI
- [LeetCode] 641. Design Circular DequeLeetCode
- BZOJ3497 : Pa2009 Circular GameGAM
- Error:Cannot build artifact xxx:war exploded‘ because it is included into a circular dependencyErrorUI
- tree
- Decision Tree
- Tree Compass
- A - Distance in Tree
- DSU on Tree
- Rebuild TreeRebuild
- 01 Tree
- 【MySQL（1）| B-tree和B+tree】MySql
- 多路查詢樹:B-tree/b+tree
- LeetCode#110.Balanced Binary Tree(Tree/Height/DFS/Recursion)LeetCode
- Root of AVL Tree
- Tree – Information TheoryORM
- mvn dependency:tree
- Traversals of binary tree
- B-tree
- B+tree
- segment tree beats
- tree-shaking
- Walking the File Tree
- Causal Inference理論學習篇-Tree Based-Causal Tree
- LeetCode C++ 968. Binary Tree Cameras【Tree/DFS】困難LeetCodeC++
- Trie tree實踐
- 100-Same Tree
- 101-Symmetric Tree
- B-tree索引索引
- el-tree-select
- 100. Same Tree
- [leetcode]same-treeLeetCode
- Leetcode Binary Tree PathsLeetCode
- F - Perfect Matching on a Tree
- tmp dbg parse tree