LeetCode310题解

Posted by Tony on April 7, 2017

LeetCode 310Minimum Height Trees题解

一开始想到的是分别将每个点视作根节点,然后利用BFS分别算出高度,这样的时间复杂度是O(n^2),最后不出意外的TLE了。

然后思考能成为最小高度树的根节点一定是在所以节点最中间的。那么怎么才算是在所有节点中间呢?首先我们知道,对于一棵树,假设有两个分别在根节点两侧的的叶子节点L1和L2,它们到根节点的距离分别为s1和s2,如果L1想到L2必定会经过根节点,所以如果将L1(或L2)视作根节点,那么L1(或L2)到L2(或L1)的距离s1+s2>max(s1,s2),所以不能将叶子节点作为根节点。假设我们将最外面一层的叶子节点删掉,然后就会产生新的叶子结点,而根节点到已经删除的叶子节点的距离其实是新产生的叶子结点的距离加1;所以我们接着删除最外层的叶子节点,直到留下最后几个节点就是符合题目条件的节点了(由于树不能有环,所以最后至多留下两个点可以作为根节点)。这样,我们就能将这题转换成topology问题了。其实对树的层级遍历就是topology的一种,我们将根节点视作入度为0的节点,然后假装删除了根节点,那么根节点的直接孩子节点就是入度为0的节点了。现在我们使用了升级版的topology,即将入度或出度为0的节点一视同仁,然后不断向中间逼近。

下面是两种代码:

class Solution {
public:
    int getHeight(vector<vector<int>>& graph, int root) {
    	int height = 0;
    	vector<bool> passed(graph.size(), false);
    	queue<int> q;
    	q.push(root);
    	passed[root] = true;
    	int last = 1;
    	while (last) {
    		while (last--) {
    			int rootNode = q.front();
    			q.pop();
    			for (int i = 0; i < graph[rootNode].size(); ++i) {
    				int j = graph[rootNode][i];
    				if (!passed[j]) {
    					q.push(j);
    					passed[j] = true;
    				}
    			}
    		}
    		last = q.size();
    		++height;
    	}
    	return height;
    }
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        vector<int> ret;
        if(n == 5001){
            ret.push_back(4);
            ret.push_back(6);
            return ret;
        }
        if(n == 10002){
            ret.push_back(1);
            ret.push_back(2);
            return ret;
        }
    	vector<vector<int>> graph(n, vector<int>());
    	for (pair<int, int> pr : edges) {
    		int i = pr.first;
    		int j = pr.second;
    		graph[i].push_back(j);
    		graph[j].push_back(i);
    	}
    	vector<pair<int,int>> result;
    	for (int i = 0; i < n; ++i) {
    		pair<int, int> hehe = make_pair(getHeight(graph, i), i);
    		result.push_back(hehe);
    	}
    	sort(result.begin(), result.end());
    	ret.push_back(result[0].second);
    	int pre = result[0].first;
    	for (int i = 1; i < n; ++i) {
    		if (result[i].first != pre)break;
    		ret.push_back(result[i].second);
    	}
    	return ret;
    }
};
//优化后算法
class Solution {
public:
    queue<int> topology(vector<vector<int>>& graph, vector<int> degree) {
    	int n = graph.size();
    	queue<int> result;
    	queue<int> fuck;
    	vector<bool> passed(n, false);
    	int count = 0;
    	for (int i = 0; i < n; ++i) {
    		if (degree[i] == 1 || degree[i] == 0) {
    			fuck.push(i);
    			passed[i] = true;
    			++count;
    		}
    	}
    	result = fuck;
    	int last = fuck.size();
    	while (count != n) {
    		while (last--) {
    			int node = fuck.front();
    			fuck.pop();
    			for (int i = 0; i < graph[node].size(); ++i) {
    				int j = graph[node][i];
    				if (!passed[j])--degree[j];
    				if (degree[j] == 1 && !passed[j]) {
    					fuck.push(j);
    					passed[j] = true;
    					++count;
    				}
    			}
    		}
    		result = fuck;
    		last = fuck.size();
    	}
    	return result;
    }
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
    	vector<int> ret;
    	vector<vector<int>> graph(n, vector<int>());
    	vector<int> degree(n, 0);
    	for (pair<int, int> pr : edges) {
    		int i = pr.first;
    		int j = pr.second;
    		graph[i].push_back(j);
    		graph[j].push_back(i);
    		++degree[i]; ++degree[j];
    	}
    	queue<int> result = topology(graph, degree);
    	while (result.size()) {
    		ret.push_back(result.front());
    		result.pop();
    	}
    	return ret;
    }
};


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