# Print level order traversal line by line | Set 1

Given a binary tree, print level order traversal in a way that nodes of all levels are printed in separate lines.

For example consider the following tree

```          1
/     \
2       3
/   \       \
4     5       6
/  \     /
7    8   9

Output for above tree should be
1
2 3
4 5 6
7 8 9```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Note that this is different from simple level order traversal where we need to print all nodes together. Here we need to print nodes of different levels in different lines.

A simple solution is to print use the recursive function discussed in the level order traversal post and print a new line after every call to printGivenLevel().

## C++

```/* Function to line by line 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);
printf("\n");
}
}

/* 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);
}
}
```

## Java

```/* Function to line by line print level order traversal a tree*/
static void printLevelOrder(Node root)
{
int h = height(root);
int i;
for (i=1; i<=h; i++)
{
printGivenLevel(root, i);
System.out.println();
}
}
/* Print nodes at a given level */
void printGivenLevel(Node root, int level)
{
if (root == null)
return;
if (level == 1)
System.out.println(root.data);
else if (level > 1)
{
printGivenLevel(root.left, level-1);
printGivenLevel(root.right, level-1);
}
}
```

The time complexity of the above solution is O(n2)

How to modify the iterative level order traversal (Method 2 of this) to levels line by line?
The idea is similar to this post. We count the nodes at current level. And for every node, we enqueue its children to queue.

## C++

```/* Iterative program to print levels line by line */
#include <iostream>
#include <queue>
using namespace std;

// A Binary Tree Node
struct node
{
struct node *left;
int data;
struct node *right;
};

// Iterative method to do level order traversal line by line
void printLevelOrder(node *root)
{
// Base Case
if (root == NULL)  return;

// Create an empty queue for level order tarversal
queue<node *> q;

// Enqueue Root and initialize height
q.push(root);

while (1)
{
// nodeCount (queue size) indicates number of nodes
// at current lelvel.
int nodeCount = q.size();
if (nodeCount == 0)
break;

// Dequeue all nodes of current level and Enqueue all
// nodes of next level
while (nodeCount > 0)
{
node *node = q.front();
cout << node->data << " ";
q.pop();
if (node->left != NULL)
q.push(node->left);
if (node->right != NULL)
q.push(node->right);
nodeCount--;
}
cout << endl;
}
}

// Utility function to create a new tree node
node* newNode(int data)
{
node *temp = new node;
temp->data = data;
temp->left = NULL;
temp->right = NULL;
return temp;
}

// Driver program to test above functions
int main()
{
// Let us create binary tree shown in above diagram
node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->right = newNode(6);

printLevelOrder(root);
return 0;
}```

## Java

```/* An Iterative Java program to print levels line by line */

import java.util.Queue;

public class LevelOrder
{
// A Binary Tree Node
static class Node
{
int data;
Node left;
Node right;

// constructor
Node(int data){
this.data = data;
left = null;
right =null;
}
}

// Iterative method to do level order traversal line by line
static void printLevelOrder(Node root)
{
// Base Case
if(root == null)
return;

// Create an empty queue for level order tarversal

// Enqueue Root and initialize height

while(true)
{

// nodeCount (queue size) indicates number of nodes
// at current level.
int nodeCount = q.size();
if(nodeCount == 0)
break;

// Dequeue all nodes of current level and Enqueue all
// nodes of next level
while(nodeCount > 0)
{
Node node = q.peek();
System.out.print(node.data + " ");
q.remove();
if(node.left != null)
if(node.right != null)
nodeCount--;
}
System.out.println();
}
}

// Driver program to test above functions
public static void main(String[] args)
{
// Let us create binary tree shown in above diagram
/*               1
/     \
2       3
/   \       \
4     5       6
*/

Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.right = new Node(6);

printLevelOrder(root);

}

}
//This code is contributed by Sumit Ghosh
```

Output:
```1
2 3
4 5 6```

### Asked in: Amazon, Microsoft, Morgan-Stanley

Time complexity of this method is O(n) where n is number of nodes in given binary tree.

Level order traversal line by line | Set 2 (Using Two Queues)

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