c - 二叉树函数

Question

有人可以解释在下面的二叉树上调用 mystery(root) 的输出吗？

struct treenode {
  int data;
  struct treenode* left;
  struct treenode* right;
}

void mystery(struct treenode* root) {
  if (root == NULL) return;
  mystery(root->right);
  printf("%d ", root->data%25);
  mystery(root->left);
}

树:

       35
    /      \
  23       17
 /  \     /
89  135  56
        /   \
       44    89
              \
              74
             /
            287     

Answer 1

见下面的痕迹:

You call mystery(35)
the root is not null which calls mystery(17)
    mystery(17) is not null and calls mystery(17-Right)
        This is null so it Returns
    printf(17%25 == 17)
    call mystery(56)
        mystery(56) is not null and calls mystery(89)
            mystery(89) is not null and calls mystery(74)
                    mystery(74) is not null and calls mystery(74-Right)
                    mystery(74-Right) is null; return;
                printf(74%25 == 24)
                mystery(74) calls mystery(287)
                    printf(287%25 == 12)
                printf(89%25 == 14)
        printf(56%25 == 6)
        mystery(56) calls mystery(44)
                mystery(44) has no right or left child
                printf(44%25 == 19)
printf(35%25 == 10)
root calls mystery(23)
    mystery(23) calls mystery(135)
        printf(135%25 == 10)
    printf(23%25 == 23)
    mystery(23) calls mystery(89)
        printf(89%25 == 14)