Height of a Binary Tree: Recursion Unrolled & Explained


tags: >-
development, C++, tree, height, programming, recursion, internals, complete
flow
categories: development

Height of a Binary Tree is the longest path in the tree.

Code
int getHeight(node *root)
{
    if (root == NULL) {
        return -1;
    }
    int leftHeight = getHeight(root->left);
    int rightHeight = getHeight(root->right);

    return (max(leftHeight, rightHeight) + 1);     
}

Suppose the tree is as follows:

         [10]
          |
       [8]  [12]
              |
           [4] [5]

The recursion flow is as following:

1. getHeight(10)
2. getHeight(left of 10) ==> getHeight(8)
3. getHeight(left of 8) ==> getHeight(NULL)
4. getHeight(NULL) returns -1
5. leftHeight = -1 and flow goes back to line # 3
6. Now, it calls right of 8, getHeight(right of 8)
7. getHeight(NULL) returns -1 to rHeight
8. Both subtree of 8 are traversed and leftHeight = -1, rightHeight = -1
9. It compares both values and return the max + 1
10. max( -1, -1) + 1 = 0
11. The node 8 is left subtree of 10, and returns to line # 2.
12. At node 10, leftHeight = 0
13. Now, rightHeight is calculated by moving to right of 10.
14. getHeight(right of 10) ==> getHeight(12)
15. Similar to above, height at node 12 is calculated as 1.
16. At the end, node 12 returns its height at line #14.
17. Again we compare max (0, 1) + 1
18. The answer is 2   
Summary

Usually, all recursion steps are not visualized and just assumed to be working. This post tries to exhaust a recursion flow and help understand it better.

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File system in userspace (FUSE) on Linux : C implementation

  • BBFS is a good starting point to develop a file system in C
  • Application should keep in mind file offset before issuing a request. A write should be followed with a seek to offset zero, before issuing read.
  • FUSE 2.7 reads 8K data by default, in two 4K chunks
  • Read happens in last 4K and the first 4K data block

Example:

I had set the file size as 4MB in getattr () implementation.

int bb_getattr(const char *path, struct stat *statbuf)
{
    int retstat = 0;
    char fpath[PATH_MAX];
    bb_fullpath(fpath, path);

    memset(statbuf, 0, sizeof(struct stat));
    if (strcmp(path, "/") == 0) {
        statbuf->st_mode = S_IFDIR | 0755;
        statbuf->st_nlink = 2;
    } else {
        statbuf->st_mode = S_IFREG | 0444;
        statbuf->st_nlink = 1;
        statbuf->st_size = 4 * 1024* 1024;
    }   
    return retstat;
}

The sequence of calls and their arguments is as follows:

bb_getattr(path="/abcd.txt", statbuf=0xc5387960)
    bb_fullpath:  rootdir = "/tmp", path = "/abcd.txt", fpath = "/tmp/abcd.txt"
bb_open(path"/abcd.txt", fi=0xc5daaa50)
    bb_fullpath:  rootdir = "/tmp", path = "/abcd.txt", fpath = "/tmp/abcd.txt"
    fi:
    flags = 0x00008002
    fh_old = 0x00000000
    writepage = 0
    direct_io = 0
    keep_cache = 0
    fh = 0x0000000000000001
    lock_owner = 0x0000000000000000
bb_write(path="/abcd.txt", buf=0xc4966050, size=10, offset=0, fi=0xc5387a50)
    fi:
    flags = 0x00000000
    fh_old = 0x00000001
    writepage = 0
    direct_io = 0
    keep_cache = 0
    fh = 0x0000000000000001
    lock_owner = 0x0000000000000000
bb_read(path="/abcd.txt", buf=0x06ccbd90, size=12288, offset=4096, fi=0xc5daaa50)  <- Here
    fi:
    flags = 0x00000000
    fh_old = 0x00000001
    writepage = 0
    direct_io = 0
    keep_cache = 0
    fh = 0x0000000000000001
    lock_owner = 0x0000000000000000

bb_read(path="/abcd.txt", buf=0x06ccbd90, size=4096, offset=0, fi=0xc5daaa50)
    fi:
    flags = 0x00000000
    fh_old = 0x00000001
    writepage = 0
    direct_io = 0
    keep_cache = 0
    fh = 0x0000000000000001
    lock_owner = 0x0000000000000000

Forward declaration of a structure in C

What do you think of following code?

/*
 * decl.h
 */
struct junk {
    int a;
};

----------------------------------------------
/*
 * fwd.c
 * We have not included the header file decl.h.
 */
#include 
struct junk;

int main()
{
    struct junk *ptr;
    printf("%d", ptr->a);
}

You will get compilation error that structure object ptr is incomplete. This is misleading if you do not go through all included header files. cscope will make you believe that definition exists. You just wonder till you find out that structure was never defined in fwd.c!

GNU gdb for C++ applications

For C programmers, using GDB for C++ applications could be daunting task. I have learned following lessons:

    • Always pass fully qualified name of a function such as namespace_1::class_1::function1
    • You can’t put a breakpoint on a template function.
jango::IndexInterface<jango::treedata_v1>::toIndex
      • The easiest way is to put the breakpoint on the line number of the template function.
    • C++ application’s core is legible if you interpret it ignoring name mangling of functions.
_ZN5ns_111class_a6functionB2EPKcPKvibbbPb+0x15ca 
 ==> ns::class_a::functionB()

Headless link list implementation in C

Singly link list

A link list is a non-linear data structure with a head pointer. The rest of the nodes in the list could be accessed by using “next” of head.

The code to create a list of N node involves creating head separately. Rest of the nodes are appended to the head.

That looked ugly to me and so I wrote it in a generic manner. Using a temporary pointer, I wrote a single loop to create N link list nodes, keeping head intact.

Here is the code:

#define NULL (char*)0

typedef struct node_ {
        int n;
        struct node_ *next;
} node;

int main()
{
        int i = 5;
        node *head = NULL;
        node *temp = NULL;

        while (i-- > 0) {
                head = malloc(sizeof(node));
                head->n = i;
                head->next = temp;
                temp = head;
        }

        temp = head;

Pitfall of C array

CODE

int main()
{
int my_array[4] = {11,12,13,14};
int next = 15;
printf("\nSurprise = %d", my_array[4]);
}

What should it print?

Well, although illegal, the above code does legal things. The array “my_array” has four integers allocated on the stack. The next integer “next” is allocated adjacent to the stack memory. So the address of “next” would be (my_array + 5).


-----------------------
11| 12| 13| 14| 15|
-----------------------
^
|
The memory layout on stack is linear and hence our array and the integer are adjacent to each other.

It is a subtle error and you should know it.

Synchronize two threads to print ordered even and odd numbers in C

Problem:You have two threads, that independently print even and odd values. The goal is to synchronize these threads so as to get an non-decreasing ordered set of numbers and order should preserve natural ordering of numbers. So the output should be 1,2,3,4,5,6,7,8…..

It is an interesting problem and could be solved with a conditional variable.

Following is C implementation of the solution:

#include "stdio.h"
#include "stdlib.h"
#include "pthread.h"

pthread_mutex_t count_mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t condition_var = PTHREAD_COND_INITIALIZER;
void *functionCount1();
void *functionCount2();
int count = 0;
#define COUNT_DONE 200

void main()
{
 pthread_t thread1, thread2;
 pthread_create( &thread1, NULL, &functionCount1, NULL);
 pthread_create( &thread2, NULL, &functionCount2, NULL);
 pthread_join( thread1, NULL);
 pthread_join( thread2, NULL);
 exit(0);
}

// Print odd numbers
void *functionCount1()
{
  for(;;) {
   // Lock mutex and then wait for signal to relase mutex
   pthread_mutex_lock( &count_mutex );
   if ( count % 2 != 0 ) {
     pthread_cond_wait( &condition_var, &count_mutex );
   }
   count++;
   printf("Counter value functionCount1: %d\n",count);
   pthread_cond_signal( &condition_var );
   if ( count >= COUNT_DONE ) {
     pthread_mutex_unlock( &count_mutex );
     return(NULL);
   }
   pthread_mutex_unlock( &count_mutex );
 }
}

// print even numbers
void *functionCount2()
{
  for(;;) {
  // Lock mutex and then wait for signal to relase mutex
  pthread_mutex_lock( &count_mutex );
  if ( count % 2 == 0 ) {
    pthread_cond_wait( &condition_var, &count_mutex );
  }
  count++;
  printf("Counter value functionCount2: %d\n",count);
  pthread_cond_signal( &condition_var );
  if( count >= COUNT_DONE ) {
    pthread_mutex_unlock( &count_mutex );
    return(NULL);
  }
  pthread_mutex_unlock( &count_mutex );
 }
}