WAP to Print a Binary tree in level order with a new line after every level.

Data Structure Used Queue

Algorithm Shown Below

printLevelorder(tree)
Initialize the global variable size=0
1) Create an empty queue q
2) temp_node = root /*start from root*/ & initialize the local variable cur_size=0
3) Loop while temp_node is not NULL
A) print temp_node->data.
B) Enqueue temp_node’s children (first left then right children) to q
C) After this check cur_size is equals to zero or not
if(cur_size==0)set cur_size=size of queue(no of elements at that time in queue
D) Dequeue a node from q and assign it’s value to temp_node
E) Decrement the cur_size until it become zero so that we can re-initialize it
again with new size of queue.


#include
#include
#define MAX_Q_SIZE 100

/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct node
{
int data;
struct node* left;
struct node* right;
};

/* frunction prototypes */
struct node** createQueue(int *, int *);
void enQueue(struct node **, int *, struct node *);
struct node *deQueue(struct node **, int *);

int size=0;

/* Given a binary tree, print its nodes in level order
using array for implementing queue */
void printLevelOrder(struct node* root)
{
int rear, front;
struct node **queue = createQueue(&front, &rear);
struct node *temp_node = root;
int nsize=0;

while(temp_node)
{
printf("%d ", temp_node->data);

/*Enqueue left child */
if(temp_node->left)
enQueue(queue, &rear, temp_node->left);

/*Enqueue right child */
if(temp_node->right)
enQueue(queue, &rear, temp_node->right);
if(nsize==0)
{
nsize=size;
printf( " \n ");
}

/*Dequeue node and make it temp_node*/
temp_node = deQueue(queue, &front);
nsize--;
}
}

/*UTILITY FUNCTIONS*/
struct node** createQueue(int *front, int *rear)
{
struct node **queue =
(struct node **)malloc(sizeof(struct node*)*MAX_Q_SIZE);

*front = *rear = 0;
return queue;
}

void enQueue(struct node **queue, int *rear, struct node *new_node)
{
queue[*rear] = new_node;
(*rear)++;
size++;
}

struct node *deQueue(struct node **queue, int *front)
{
(*front)++;
size--;
return queue[*front - 1];
}

/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
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);
}

/* Driver program to test above functions*/
int main()
{
struct node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->left->left = newNode(41);
root->left->right = newNode(5);
root->right->left = newNode(6);
root->right->right = newNode(7);
root->right->right->right = newNode(71);


printf("Level Order traversal of binary tree is \n");
printLevelOrder(root);

getchar();
return 0;
}

Time Complexity O(n)
Space Complexity O(1)
Run Here https://ideone.com/FLgSI