Check if a binary tree is a sum tree - binary-tree

Question

Method:
1) If the node is a leaf node then sum of subtree rooted with this node is equal to value of this node.
2) If the node is not a leaf node then sum of subtree rooted with this node is twice the value of this node (Assuming that the tree rooted with this node is SumTree).
[a link]http://www.geeksforgeeks.org/check-if-a-given-binary-tree-is-sumtree/
Qstn: why do we need to initialize ls=0 or rs=0 when it's a leaf node. Considering the tree as given in the link, if we reach node 4
if(node == NULL || isLeaf(node))
return 1;
the above code returns 1 (true) back to the function from where it is called i.e., node 10 similarly the right side returns true back to node 10 so we can now enter the loop below as both conditions are true
if( isSumTree(node->left) && isSumTree(node->right))
the node being 10 we calculate the data on the left and right as given here in the else if conditions, so why is the condition if(node->left==NULL) then ls=0 necessary(isn't it already taken care of as it is a leaf node)?
because return (4=0+0) will be false and the whole loop will become false?
int isLeaf(struct node *node)
{
if(node == NULL)
return 0;
if(node->left == NULL && node->right == NULL)
return 1;
return 0;
}
int isSumTree(struct node* node)
{
int ls; // for sum of nodes in left subtree
int rs; // for sum of nodes in right subtree
/* If node is NULL or it's a leaf node then
return true */
if(node == NULL || isLeaf(node))
return 1;
if( isSumTree(node->left) && isSumTree(node->right))
{
// Get the sum of nodes in left subtree
if(node->left == NULL)
ls = 0;
else if(isLeaf(node->left))
ls = node->left->data;
else
ls = 2*(node->left->data);
// Get the sum of nodes in right subtree
if(node->right == NULL)
rs = 0;
else if(isLeaf(node->right))
rs = node->right->data;
else
rs = 2*(node->right->data);
/* If root's data is equal to sum of nodes in left
and right subtrees then return 1 else return 0*/
return(node->data == ls + rs);
}
return 0;
}

Answer 1

I assume you are referring to lines
if(node->left == NULL)
ls = 0;
and
if(node->right == NULL)
rs = 0;
The author of that solution initialized rs and ls to 0 when the left and right child doesn't exist. This initialization is needed because there was no previous assignment to ls and rs. If you don't have this step, your ls and rs could be garbage value.
You can save the aforementioned checks by initializing your ls and rs values when you declare them.
int ls = 0,rs = 0;
Hope this helps!