In this HackerRank Day 23 BST Level order traversal 30 days of code problem set, we need to complete a levelorder function that can accept a root node as an input and then print the level order traversal of the binary search tree.


HackerRank Day 23 BST Level order traversal 30 days of code solution


Problem solution in Python 2 programming.    

    def levelOrder(self,root):
  	     #Write your code here
        s = ''
        if (root != None):
            q = [root]
            while (len(q) != 0):
                s += str(q[0].data) + ' '
                if (q[0].left != None):
                    q.append(q[0].left)
                if (q[0].right != None):
                    q.append(q[0].right)
                q.pop(0)
        print s


Problem solution in Python 3 programming.

    def levelOrder(self,root):
        if root is None:
            return 
        
        qu = []
        qu.append(root)

        while len(qu) !=0:
            p = qu.pop(0)
            print(p.data, end=' ')
            if p.left is not None:
                qu.append(p.left)
            if p.right is not None:
                qu.append(p.right)
        


Problem solution in java programming.

static LinkedList<Integer> queue = new LinkedList();
static void levelOrder(Node root){
    LinkedList<Node> treeQueue = new LinkedList();
    treeQueue.add(root);
    while(treeQueue.peek() != null) {
        Node toprint = treeQueue.remove();
        System.out.print(toprint.data);
        if(toprint.left != null) {
            treeQueue.add(toprint.left);
        }
        if(toprint.right != null) {
            treeQueue.add(toprint.right);
        }
        if(treeQueue.peek() != null) {
            System.out.print(" ");
        }
    }
    
    }


Problem solution in c++ programming.

	void levelOrder(Node * root){
        std::queue<Node*> q;
        Node* c;
  
        if (root != NULL) {
            q.push(root);
        }
        
        while (!q.empty()) {
            c = q.front();
            q.pop();
            cout << c->data << " ";
            if (c->left!=NULL) q.push(c->left);
            if (c->right!=NULL) q.push(c->right);
        }
	}


Problem solution in c programming.

#define max(a, b) (a > b ? a : b)

int getHeight(Node *root) {
    if (root == NULL)
        return 0;
    else
        return 1 + max(getHeight(root->left), getHeight(root->right));
}

void printGivenLevel(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);
    } 
}

void levelOrder(Node* root){
  //Write your code here
    int height = getHeight(root);
    int i;
    for (i = 1; i <= height; i++) {
        printGivenLevel(root, i);
    }
}


Problem solution in Javascript programming.

      var queue = [root];
      while (queue.length > 0) {
        var node = queue.shift();
        write(node.data + " ");
        if(node.left) {
         queue.push(node.left);
        }
        if (node.right) {
         queue.push(node.right);
        }
      }
      function write(str){
        process.stdout.write(str);
      }