# Print Binary Tree levels in sorted order

Given a Binary tree, the task is to print its all level in sorted order

Examples:

```Input :     7
/    \
6       5
/ \     / \
4  3    2   1
Output :
7
5 6
1 2 3 4

Input :     7
/    \
16       1
/ \
4   13
Output :
7
1 16
4 13
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Here we can use two Priority queue for print in sorted order. We create an empty queue q and two priority queues, current_level and next_level. We use NULL as a separator between two levels. Whenever we encounter NULL in normal level order traversal, we swap current_level and next_level.

 `// CPP program to print levels in sorted order. ` `#include ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// A Binary Tree Node ` `struct` `Node { ` `    ``int` `data; ` `    ``struct` `Node *left, *right; ` `}; ` ` `  `// Iterative method to find height of Binary Tree ` `void` `printLevelOrder(Node* root) ` `{ ` `    ``// Base Case ` `    ``if` `(root == NULL) ` `        ``return``; ` ` `  `    ``// Create an empty queue for level order traversal ` `    ``queue q; ` ` `  `    ``// A priority queue (or min heap) of integers for  ` `    ``// to store all elements of current level.  ` `    ``priority_queue<``int``, vector<``int``>, greater<``int``> > current_level; ` ` `  `    ``// A priority queue (or min heap) of integers for  ` `    ``// to store all elements of next level.  ` `    ``priority_queue<``int``, vector<``int``>, greater<``int``> > next_level; ` ` `  `    ``// push the root for traverse all next level nodes ` `    ``q.push(root); ` ` `  `    ``// for go level by level ` `    ``q.push(NULL); ` ` `  `    ``// push the first node data in previous_level queue ` `    ``current_level.push(root->data); ` ` `  `    ``while` `(q.empty() == ``false``) { ` ` `  `        ``// Get top of priority queue  ` `        ``int` `data = current_level.top(); ` ` `  `        ``// Get top of queue ` `        ``Node* node = q.front(); ` ` `  `        ``// if node == NULL (Means this is boundary ` `        ``// between two levels), swap current_level ` `        ``// next_level priority queues. ` `        ``if` `(node == NULL) { ` `            ``q.pop(); ` ` `  `            ``// here queue is empty represent ` `            ``// no element in the actual ` `            ``// queue ` `            ``if` `(q.empty()) ` `                ``break``; ` ` `  `            ``q.push(NULL); ` `            ``cout << ``"\n"``; ` ` `  `            ``// swap next_level to current_level level ` `            ``// for print in sorted order ` `            ``current_level.swap(next_level); ` ` `  `            ``continue``; ` `        ``} ` ` `  `        ``// print the current_level data ` `        ``cout << data << ``" "``; ` ` `  `        ``q.pop(); ` `        ``current_level.pop(); ` ` `  `        ``/* Enqueue left child */` `        ``if` `(node->left != NULL) { ` `            ``q.push(node->left); ` ` `  `            ``// Enqueue left child in next_level queue ` `            ``next_level.push(node->left->data); ` `        ``} ` ` `  `        ``/*Enqueue right child */` `        ``if` `(node->right != NULL) { ` `            ``q.push(node->right); ` ` `  `            ``// Enqueue right child in next_level queue ` `            ``next_level.push(node->right->data); ` `        ``} ` `    ``} ` `} ` ` `  `// Utility function to create a new tree node ` `Node* newNode(``int` `data) ` `{ ` `    ``Node* temp = ``new` `Node; ` `    ``temp->data = data; ` `    ``temp->left = 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(7); ` `    ``root->left = newNode(6); ` `    ``root->right = newNode(5); ` `    ``root->left->left = newNode(4); ` `    ``root->left->right = newNode(3); ` `    ``root->right->left = newNode(2); ` `    ``root->right->right = newNode(1); ` ` `  `    ``/*     7 ` `         ``/    \ ` `        ``6       5 ` `       ``/ \     / \ ` `      ``4  3     2  1          */` ` `  `    ``cout << ``"Level Order traversal of binary tree is \n"``; ` `    ``printLevelOrder(root); ` `    ``return` `0; ` `} `

Output:

```Level Order traversal of binary tree is
7
5 6
1 2 3 4
```

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