Binary tree search algorithm - algorithm

Question

I think there is something I am not understanding with an algorithm I'm working with. I have a binary tree where "*" is the root, "C" is on the left, "+" is on the right and + has "a" to the left and "b" to the right.
This is the algorithm:
Module Find_Value ( List, Root, Value ) {
If ( Root is NULL ){
Return NULL;
}Else{
If ( List [ Root ] .data equals Value )
Return Root;
Temp = Find_Value ( List, List[ Root ].left, Value )
If ( Temp is NULL )
Return Find_Value ( List, List[ Root ].right, Value );
Else
Return Temp;
}
}
And I'm supposed to carry out Find_Value(List, 0, b)
I could be just reading it wrong, but I do not see any instruction to go back up the tree. I get find_value(List,1,C) the first run-through and then I see temp as looking to the left of C (which is null) and the the right of C (also null). Can anyone tell me I am just interpreting the instructions incorrectly?

Answer 1

This is what is happening here:
(I have replaced * with 0,C with 2 for better understanding and also used them as identifiers for each temp at each step)
1) Temp(*) = Find_Value ( List, *.left, b );
2) Temp(C) = Find_value ( List,C.left,b); //returns null ,
therefore Temp(C) = NULL;
3) If ( Temp(C) is NULL )
Return Find_Value ( List, C.right, Value ); //returns null;
4) therefore the 1st step returns null;
Temp(*) = null;
5) If ( Temp(*) is NULL )
Return Find_Value ( List, *.right, Value );
6) Temp(+) = Find_Value ( List, +.left, Value );
7) Temp(a) = Find_Value ( List, a.left, Value ); // returns null
8) If ( Temp(a) is NULL )
Return Find_Value ( List, a.right, Value ); //returns null
therefore Temp(+) = NULL;
9) If ( Temp(+) is NULL )
Return Find_Value ( List, +.right, Value );
10) If ( List [ Root ].data equals Value )
Return Root;
where List[Root].data = b;
Step 9 returns b,followed by Step 5 returning b;
and overall the function return b;