计算二叉树中节点个数,叶节点个数,满节点个数的函数

下面这三个是我写的.

static int node_count (const Tree tree){int the_number_of_left_subtree = 0, the_number_of_right_subtree = 0, the_number_of_node = 0 ;if (tree != NULL){the_number_of_left_subtree = node_count (tree -> left) ;the_number_of_right_subtree = node_count (tree -> right) ;the_number_of_node = the_number_of_left_subtree + the_number_of_right_subtree + 1 ;}return the_number_of_node ;}static int leaf_count (const Tree tree){int the_number_of_left_subtree = 0, the_number_of_right_subtree = 0, the_number_of_leaf = 0 ;if (tree != NULL){the_number_of_left_subtree = leaf_count (tree -> left) ;the_number_of_right_subtree = leaf_count (tree -> right) ;the_number_of_leaf = the_number_of_left_subtree + the_number_of_right_subtree ;if (0 == the_number_of_left_subtree && 0 == the_number_of_right_subtree)the_number_of_leaf++ ;}return the_number_of_leaf ;}static int full_count (const Tree tree){int the_number_of_left_subtree = 0, the_number_of_right_subtree = 0, the_number_of_full_node = 0 ;if (tree != NULL){the_number_of_left_subtree = full_count (tree -> left) ;the_number_of_right_subtree = full_count (tree -> right) ;if (the_number_of_left_subtree != 0 && the_number_of_right_subtree != 0)the_number_of_full_node += the_number_of_left_subtree + the_number_of_right_subtree + 1 ;else if (tree -> left != NULL && tree -> right != NULL)the_number_of_full_node++ ;}return the_number_of_full_node ;}

下面这三个是答案给出的,看了让人惭愧啊.

static int new_node_count (const Tree tree){if (NULL == tree)return 0;return 1 + new_node_count (tree -> left) + new_node_count (tree -> right) ;}static int new_leaf_count (const Tree tree){if (NULL == tree)return 0;else if (NULL == tree -> left && NULL == tree -> right)return 1;return new_leaf_count (tree -> left) + new_leaf_count (tree -> right) ;}static int new_full_count (const Tree tree){if (NULL == tree)return 0;return (tree -> left != NULL && tree -> right != NULL) + new_full_count (tree -> left) + new_full_count (tree -> right) ;}

虽然看懂了,但是叫我自己写还是写不出.总结些经验的话,就是这种递归函数多利用return...