Level Order Tree Traversal....

// Recursive C program for level order traversal of Binary Tree

#include <stdio.h>

#include <stdlib.h>

/* A binary tree node has data, pointer to left child

   and a pointer to right child */

struct node

{

    int data;

    struct node* left, *right;

};

/* Function protoypes */

void printGivenLevel(struct node* root, int level);

int height(struct node* node);

struct node* newNode(int data);

/* Function to print level order traversal a tree*/

void printLevelOrder(struct node* root)

{

    int h = height(root);

    int i;

    for (i=1; i<=h; i++)

        printGivenLevel(root, i);

}

/* Print nodes at a given level */

void printGivenLevel(struct node* root, int level)

{

    if (root == NULL)

        return;

    if (level == 1)

        printf("%d ", root->data);

    else if (level > 1)

    {

        printGivenLevel(root->left, level-1);

        printGivenLevel(root->right, level-1);

    }

}

/* Compute the "height" of a tree -- the number of

    nodes along the longest path from the root node

    down to the farthest leaf node.*/

int height(struct node* node)

{

    if (node==NULL)

        return 0;

    else

    {

        /* compute the height of each subtree */

        int lheight = height(node->left);

        int rheight = height(node->right);

        /* use the larger one */

        if (lheight > rheight)

            return(lheight+1);

        else return(rheight+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->right = newNode(5);

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

    printLevelOrder(root);

    return 0;

}

Output:

Level order traversal of binary tree is - 
1 2 3 4 5 

Time Complexity: O(n^2) in worst case. For a skewed tree, printGivenLevel() takes O(n) time where n is the number of nodes in the skewed tree. So time complexity of printLevelOrder() is O(n) + O(n-1) + O(n-2) + .. + O(1) which is O(n^2).