Re: K&R2, Binary-Tree, section 6.5

arnuld <sunrise@[EMAIL PROTECTED]
> writes:

>                    now
>                 /       \
>               is         the
>             /   \       /   \
>          for    men    of    time
>         /   \           \    /    \
>      all   good      party  their  to
>     /   \
>   aid   come
>
>
>
> This is an unbalanced-tree, where balanced factor is:
>
>   (1 - 3)  = -2
>
> balance factor = (height of right-subtree) - (height of left-subtree)

I don't see how you get a balance factor of -2 from that tree.  I
see a balance factor of -1:

                             now
                          /       \
        ^               is         the           ^
        |             /   \       /   \          |
height  |          for    men    of    time      |  height
  4     |         /   \           \    /    \    |    3
        |      all   good      party  their  to  V
        |     /   \
        V   aid   come

The right child of "now" has height 3.  The left child of "now"
has height 4.  The difference is -1.

> right-subtree starts with word "the" which has only 1 child-node.

"the" has two children: "of" and "time".

> left-subtree starts with "is" and it has 3 nodes

"is" also has two children: "for" and "men".
--
char a[]="\n .CJacehknorstu";int putchar(int);int main(void){unsigned long
b[]
={0x67dffdff,0x9aa9aa6a,0xa77ffda9,0x7da6aa6a,0xa67f6aaa,0xaa9aa9f6,0x11f6},*p
=b,i=24;for(;p+=!*p;*p/=4)switch(0[p]&3)case 0:{return
0;for(p--;i--;i--)case+
2:{i++;if(i)break;else default:continue;if(0)case
1:putchar(a[i&15]);break;}}}