Return value of a recursive function - recursion

Question

In the following code, I'm trying to get a better hand at understanding how recursion actually work. I've always been a bit confused about it's actual working. I want to know what value does the inorder() function actually return in every step. From where does it get these values of 0,0,11,0,0,11,12,0,0,11 respectively. Could someone tell me the logic? It's a basic inorder tree traversal program.The reason why I'm trying to understand these outputs is because the same logic is somehow used to find the depth of the tree( I think) where with every recursion the value of depth increases without initialization.
#include<stdio.h>
#include<stdlib.h>
struct node
{
int data;
struct node *left;
struct node *right;
};
struct node* newNode(int data)
{
struct node* node=(struct node*)malloc(sizeof(struct node));
node->data=data;
node->left=NULL;
node->right=NULL;
return node;
}
int inorder(struct node *temp) {
if (temp != NULL) {
printf("\nleft %d\n",inorder(temp->left));
printf("\n%d\n", temp->data);
printf("\nright %d\n",inorder(temp->right));
}
}
int main()
{
struct node *root=newNode(1);
root->left=newNode(2);
root->right=newNode(3);
root->left->left=newNode(4);
root->left->right=newNode(5);
inorder(root);
getchar();
return 0;
}

Answer 1

This function should be changed to the following (the first and last print in the original code will only get you more confused!):
int inorder(struct node *temp) {
if (temp != NULL) {
inorder(temp->left);
printf("%d\n", temp->data);
inorder(temp->right);
}
}
The recursion starts with the left branch of a specific node (usually the "root") - printing recursively (in-order) all the nodes on that left-branch, then printing the current node, moving on to printing recursively (in-order) all the nodes in the right branch.
By the way, if you want to keep that tree "ordered" (meaning, all the nodes on the left branch are smaller than the node, and all the nodes on the right branch are bigger or equal to the node) you should change:
root->left->left=newNode(4);
root->left->right=newNode(5);
to:
root->right->right=newNode(4);
root->right->right->right=newNode(5);