小编srb*_*rbh的帖子

用于计算树中节点数的函数.

int count(node *t)
{    
    int i;  
    if (t == NULL)         
        return(0);  
    i = 1 + count(t->left) + count(t->right); // recursion occurs address of left node is passed and 
    return(i);                                // continue to pass until whole left node
}                                             // is traversed and the value of t is
                                              // NULL and 0 is returned same for right node counting
                                              // i = 1 + 0 + 0 = 1

如何计算节点数？