[LeetCode] 310. Minimum Height Trees 解題思路

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:

1. 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.

問題：給定一個擁有樹性質的無向圖，圖的每一個節點都可以視為一棵樹的根節點。在所有可能的樹中，找出高度最小的樹，並返回他們的樹根。

思路一 ： 

求最小高度的樹，實際上是一個求最優解題目。求最優解，我首先想到的是動態規划(DP)思路。這個題目也確實滿足 DP 的兩個主要性質：overlapping substructure & optimal substructure 。思路比較直觀：

1. 視某個節點為樹根節點，求這棵樹的樹高。
2. 對所有節點進行上面的求解，得到 n 個樹高，其中高度最小的樹的樹根即為原題目的解。

對於1，如何求解節點 A 為樹根的樹高？

假設已知 A 的所有相鄰節點分別為樹根的各個子樹的樹高，那么 A根的樹高等於  已知的各個子樹樹高中的最大值 加一。方程式表達如下，即狀態轉換方程：

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

對於2， 存在大量重復計算。可以借助表格，將已計算的樹分支高度保存下來后續使用，避免重復計算。將當前節點以及其中一個相鄰節點組合分支方向，求得該分支高度后存入 map<string, int> direc_height 中，其中 direc 有這兩個節點組成作為 key ，height 表示高度。

若不使用表格優化，時間復雜度為 O(V*E)。使用表格，相當於進行了剪枝，會快不少。跑 LeetCode 的大集合 V = 1000, E =2000 ，測試耗時是 820ms，可惜沒能過 submit 要求。

private:
    
    // dirct consists of two node cur_next, height means the height of the subtree which souce node is the current node and walk forward on the cur_next direction
    map<string, int> direc_height;
    
public:

    /**
     * calulate the height of a tree which parent node is p and current node is node
     * 
     */ 
    int getHeight(gnode* node, gnode* p){
        int h = 0;
        for (int i = 0; i < node->neighbours.size(); i++){
            gnode* tmpn = node->neighbours[i];
            if (tmpn == p){
                continue;
            }
            
            int tmph;
            string str = to_string(node->val) + "_" + to_string(tmpn->val);
            if(direc_height.count(str) != 0){
                tmph = direc_height[str];
            }else{
                tmph = getHeight(tmpn, node);
                direc_height[str] = tmph;
            }
            
            h = max(h, tmph);
        }
        
        return h + 1;
    }

    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        
        map<int, gnode*> val_node;
        for(int i = 0; i < n ; i++){
            gnode* node = new gnode(i);
            val_node[i] = node;
        }
        
        for (int i = 0; i < edges.size(); i++){
            pair<int, int> pp = edges[i];
            val_node[pp.first]->neighbours.push_back(val_node[pp.second]);
            val_node[pp.second]->neighbours.push_back(val_node[pp.first]);
        }
        
        int minh = INT_MAX;
        
        map<int, int> node_height;
        
        map<int, gnode*>::iterator m_iter;
        for(m_iter = val_node.begin(); m_iter != val_node.end(); m_iter++){
            int h = getHeight(m_iter->second, NULL);
            node_height[m_iter->first] = h;
            minh = min(minh, h);
        }
        
        vector<int> res;
        map<int, int>::iterator mii_iter;
        for(mii_iter = node_height.begin(); mii_iter != node_height.end(); mii_iter++){
            if(mii_iter->second == minh){
                res.push_back(mii_iter->first);
            }
        }
  
        return res;
    }

 

思路二 ： 

除了 DP 方案，沒有想到其他思路，在網上借鑒了其他了的想法，理解后實現通過。

這個思路實際上是一個 BFS 思路。和常見的從根節點進行 BFS 不同，這里從葉子節點開始進行 BFS。

所有入度（即相連邊數）為 1 的節點即是葉子節點。找高度最小的節點，即找離所有葉子節點最遠的節點，也即找最中心的節點。

找最中心的節點的思路很簡單：

• 每次去掉當前圖的所有葉子節點，重復此操作直到只剩下最后的根。

根據這個思路可以回答題目中的 [ hint : How many MHTs can a graph have at most? ]，只能有一個或者兩個最小高度樹樹根。證明省略。

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;
    }

 

參考資料：

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