A binary tree problem

[zahit]

Hi guys, below there is a c code which caluclates given data repetition in a binary tree.

typedef struct node *tree_pointer;
typedef struct node{
char data;
tree pointer left_child, right_child;
}

int repetition(tree_pointer L, char x){
if(L){
if(L->data == x){
return 1 + repetition(L->left_child, x) + repetition(L->right_child, x);
}
else{
return repetition(L->left_child, x) + repetition(L->right_child, x);
}
else{
return 0;
}
}

The problem is I don't understand how it works. The all recursive and return process confused me. For example; what happens when we track the code on the tree attached(for data 'a'). If anyone can explain me step by step, I would appreciate.

[Sick0Fant]

Hi guys, below there is a c code which caluclates given data repetition in a binary tree.

typedef struct node *tree_pointer;
typedef struct node{
char data;
tree pointer left_child, right_child;
}

int repetition(tree_pointer L, char x){
if(L){
if(L->data == x){
return 1 + repetition(L->left_child, x) + repetition(L->right_child, x);
}
else{
return repetition(L->left_child, x) + repetition(L->right_child, x);
}
else{
return 0;
}
}

The problem is I don't understand how it works. The all recursive and return process confused me. For example; what happens when we track the code on the tree attached(for data 'a'). If anyone can explain me step by step, I would appreciate.

It's recursive. It calls itself with both leaves and adds 1 when the datum is repeated, otherwise, it just calls itself with its leaves

[JosAH]

The problem is I don't understand how it works. The all recursive and return process confused me. For example; what happens when we track the code on the tree attached(for data 'a'). If anyone can explain me step by step, I would appreciate.

Suppose I have a pile of notes; each note contains a letter. Someone asks me
to count the number of 'a's in that pile of notes. I divide that pile in two (not
necessarily equal) parts and give those parts to two friends. I ask them the same
thing to do and I keep one note for myself. In your example I keep an 'a' note.

My friends L (Leo) and R (Rudolph) come up with the two totals: 1 and 1; I had
one 'a' myself so there are 1+1+1 == 3 'a's in the pile.

Now suppose Leo and Rudolph did the same thing I did and divided there piles
as well ... we still can get the end result; that's what recursion is all about.

kind regards,

Jos