[LeetCode] 310. Minimum Height Trees 解题思路

For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels.

Format
The graph contains

n

nodes which are labeled from

0

to

n - 1

. You will be given the number

n

and a list of undirected

edges

(each edge is a pair of labels).

You can assume that no duplicate edges will appear in

edges

. Since all edges are undirected,

[0, 1]

is the same as

[1, 0]

and thus will not appear together in

edges

.

Example 1:

Given

n = 4

,

edges = [[1, 0], [1, 2], [1, 3]]

0
|
1
/ \
2   3

return

[1]

Example 2:

Given

n = 6

,

edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]

0  1  2
\ | /
3
|
4
|
5

return

[3, 4]

Hint:

How many MHTs can a graph have at most?

Note:

(1) According to the definition of tree on Wikipedia: “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.”

(2) The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.

问题：给定一个拥有树性质的无向图，图的每一个节点都可以视为一棵树的根节点。在所有可能的树中，找出高度最小的树，并返回他们的树根。

思路一 ：

求最小高度的树，实际上是一个求最优解题目。求最优解，我首先想到的是动态规划(DP)思路。这个题目也确实满足 DP 的两个主要性质：overlapping substructure & optimal substructure 。思路比较直观：

视某个节点为树根节点，求这棵树的树高。

对所有节点进行上面的求解，得到 n 个树高，其中高度最小的树的树根即为原题目的解。

对于1，如何求解节点 A 为树根的树高？

假设已知 A 的所有相邻节点分别为树根的各个子树的树高，那么 A根的树高等于 已知的各个子树树高中的最大值 加一。方程式表达如下，即状态转换方程：

height(A) = max(height(A.next0), height(A.next2),... height(A.nextk)) + 1

对于2， 存在大量重复计算。可以借助表格，将已计算的树分支高度保存下来后续使用，避免重复计算。将当前节点以及其中一个相邻节点组合分支方向，求得该分支高度后存入 map<string, int> direc_height 中，其中 direc 有这两个节点组成作为 key ，height 表示高度。

若不使用表格优化，时间复杂度为 O(V*E)。使用表格，相当于进行了剪枝，会快不少。跑 LeetCode 的大集合 V = 1000, E =2000 ，测试耗时是 820ms，可惜没能过 submit 要求。

class TNode{
public:
int val;
unordered_set<TNode*> neighbours;
TNode(int val){
this->val = val;
}
};

vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
map<int, TNode*> val_node;

for (int i = 0 ; i < n ; i++){
TNode* tmp = new TNode(i);
val_node[i] = tmp;
}

for (int i = 0 ; i < edges.size(); i++){
pair<int, int> pp = edges[i];
val_node[pp.first]->neighbours.insert(val_node[pp.second]);
val_node[pp.second]->neighbours.insert(val_node[pp.first]);
}

map<int, TNode*>::iterator m_iter;

while(val_node.size() > 2){

// obtain all leaves in current graph;
list<TNode*> listg;
for ( m_iter = val_node.begin(); m_iter != val_node.end(); m_iter++){
if(m_iter->second->neighbours.size() == 1){
listg.push_back(m_iter->second);
}
}

// remove all leaves
list<TNode*>::iterator l_iter;
for(l_iter = listg.begin(); l_iter != listg.end(); l_iter++){
TNode* p = (*(*l_iter)->neighbours.begin());
p->neighbours.erase(*l_iter);
(*l_iter)->neighbours.erase(p);

val_node.erase((*l_iter)->val);
}
}

vector<int> res;
for ( m_iter = val_node.begin(); m_iter != val_node.end(); m_iter++){
res.push_back(m_iter->first);

}

return res;
}

View Code

参考资料：

C++ Solution. O(n)-Time, O(n)-Space, LeetCode OJ