Thursday, 23 November 2017

C++ Program to Print only Odd Numbered Levels of a Tree


Code:

#include    iostream
#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;
        struct node* 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+=2)
        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 4 5 
------------------
(program exited with code: 0)
Press return to continue



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