使用邻接矩阵在树中查找叶节点

Question

嘿伙计们我正在编写一个代码，它使用邻接矩阵检测有向图和未加权图是否为树，还打印叶节点。但我对解决问题的方式不太满意。那么，我怎样才能改进下面的代码片段。提前谢谢。

int check_tree(int graph[][MAX], int numberOfNodes, int leafNodes[]){
    int noOfEdges = 0, i, j;
    int visited[numberOfNodes];

    for(i = 0; i < numberOfNodes; i++)
        for(j = 0; j < numberOfNodes; j++)
            if(graph[i][j]) noOfEdges++;


    if(noOfEdges != numberOfNodes - 1) return 0; //in a tree, number of edges are always equal to no. of nodes - 1.

    for(i = 0; i < numberOfNodes; i++)
        visited[i] = 0; //setting all edges to zero meaning they are not visited.
    visited[0] = 1; //parent node is always 1.

    for(i = 0; i < numberOfNodes; i++)
        for(j = 0; j < numberOfNodes; j++)
            if(graph[i][j]) visited[j]++; //if [i][j] is 1 this means there is an edge an it is visited.

    for(i = 0; i < numberOfNodes; i++)
        if(visited[i] > 1) return 0; //this means the node is visited more than once which is not the case in a tree.

    //if the code made it this far, we understand that it's a tree. so operations below are for tree.

    int numberOfZeroIndex;

    for(i = 0; i < numberOfNodes; i++){ // this loop detects the leaf nodes from adjacency matrix.
        numberOfZeroIndex = 0; //make it zero in this step so that it will count correctly.
        for(j = 0; j < numberOfNodes; j++){
            if(!(graph[i][j])) numberOfZeroIndex++;
        }
        if(numberOfZeroIndex == numberOfNodes){
            leafNodes[noOfLeafNodes] = i;
            noOfLeafNodes++; //this one is a global variable to keep track of leaf nodes.
        }
    }

    return 1;
}