<p>There are <code>n</code> cities numbered from <code>0</code> to <code>n - 1</code> and <code>n - 1</code> roads such that there is only one way to travel between two different cities (this network form a tree). Last year, The ministry of transport decided to orient the roads in one direction because they are too narrow.</p>

<p>Roads are represented by <code>connections</code> where <code>connections[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> represents a road from city <code>a<sub>i</sub></code> to city <code>b<sub>i</sub></code>.</p>

<p>This year, there will be a big event in the capital (city <code>0</code>), and many people want to travel to this city.</p>

<p>Your task consists of reorienting some roads such that each city can visit the city <code>0</code>. Return the <strong>minimum</strong> number of edges changed.</p>

<p>It&#39;s <strong>guaranteed</strong> that each city can reach city <code>0</code> after reorder.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/05/13/sample_1_1819.png" style="width: 311px; height: 189px;" />
<pre>
<strong>Input:</strong> n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]
<strong>Output:</strong> 3
<strong>Explanation: </strong>Change the direction of edges show in red such that each node can reach the node 0 (capital).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/05/13/sample_2_1819.png" style="width: 509px; height: 79px;" />
<pre>
<strong>Input:</strong> n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]
<strong>Output:</strong> 2
<strong>Explanation: </strong>Change the direction of edges show in red such that each node can reach the node 0 (capital).
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 3, connections = [[1,0],[2,0]]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>connections.length == n - 1</code></li>
	<li><code>connections[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
</ul>



void doconnections(int n, vector<vector<int>>& connections,int &r,int &count,vector<int> &connected,int i){
    if(r==n || i==connections.size()){
        return;
    }
    if(connected[connections[i][0]]==1 && connected[connections[i][1]]==1){
        if(i==n-1){
            return;
        }else{
            doconnections(n,connections,r,count,connected,i+1);
        }
    }else if(connected[connections[i][0]]==1 || connected[connections[i][1]]==1){
        if(connected[connections[i][0]]==1){
            count++;
            r++;
            connected[connections[i][1]]=1;
        }else if(connected[connections[i][1]]==1){
            r++;
            connected[connections[i][0]]=1;
        }
        doconnections(n,connections,r,count,connected,i+1);

    }else{
        doconnections(n,connections,r,count,connected,i+1);
        if(connected[connections[i][0]]==1){
            count++;
            r++;
            connected[connections[i][1]]=1;
        }else if(connected[connections[i][1]]==1){
            r++;
            connected[connections[i][0]]=1;
        }
        return;

    }

}




class Solution {
public:
    int minReorder(int n, vector<vector<int>>& connections) {
        

        int r = 1;
        int count = 0;
        vector<int> connected(n, 0);
        
        connected[0] = 1;
        for (int i = 0; i < connections.size(); i++) {
            if (connections[i][0] == 0 || connections[i][1] == 0) {
                if (connections[i][0] == 0) {
                    count++;
                    r++;
                    connected[connections[i][1]] = 1;
                    if(r==n){
                        break;
                    }
                } else {
                    r++;
                    connected[connections[i][0]] = 1;
                    if(r==n){
                        break;
                    }
                }
            } else {
                continue;
            }
        }

        doconnections(n,connections,r,count,connected,0);

        // while (r < n) {
        //     for (int i = 0; i < connections.size(); i++) {
        //         if (connections[i][0] == 0 || connections[i][1] == 0) {
        //             continue;
        //         } else {
        //             if (connected[connections[i][0]] == 1 && connected[connections[i][1]] == 1) {
        //                 continue;
        //             }
        //             if (connected[connections[i][0]] == 1) {
        //                 count++;
        //                 r++;
        //                 if(r==n){
        //                     break;
        //                 }
        //                 connected[connections[i][1]] = 1;
        //             } else if (connected[connections[i][1]] == 1) {
        //                 r++;
        //                 if(r==n){
        //                     break;
        //                 }
        //                 connected[connections[i][0]] = 1;
        //             } else {
        //                 continue;
        //             }
        //         }
        //     }
        // }

        return count;
    }
};
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