# Print middle level of perfect binary tree without finding height

Given a perfect binary tree, print nodes of middle level without computing its height. A perfect binary tree is a binary tree in which all interior nodes have two children and all leaves have the same depth or same level. Output : 4 5 6 7

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

The idea is similar to method 2 of finding middle of singly linked list.

Use fast and slow (or tortoise) pointers in each route of a the tree.
1. Advance fast pointer towards leaf by 2.
3. If fast pointer reaches the leaf print value at slow pointer
4. Call recursively the next route.

## C++

 `#include ` `using` `namespace` `std; ` `  `  `/* A binary tree node has key, pointer to left ` `   ``child and a pointer to right child */` `struct` `Node ` `{ ` `    ``int` `key; ` `    ``struct` `Node* left, *right; ` `}; ` `  `  `/* To create a newNode of tree and return pointer */` `struct` `Node* newNode(``int` `key) ` `{ ` `    ``Node* temp = ``new` `Node; ` `    ``temp->key = key; ` `    ``temp->left = temp->right = NULL; ` `    ``return` `(temp); ` `} ` ` `  `// Takes two parameters - same initially and ` `// calls recursively ` `void` `printMiddleLevelUtil(Node* a, Node* b) ` `{ ` `    ``// Base case e ` `    ``if` `(a == NULL || b == NULL) ` `            ``return``; ` `  `  `    ``// Fast pointer has reached the leaf so print ` `    ``// value at slow pointer ` `    ``if` `((b->left == NULL) && (b->right == NULL)) ` `    ``{ ` `        ``cout << a->key << ``" "``; ` `        ``return``; ` `    ``} ` `  `  `    ``// Recursive call ` `    ``// root.left.left and root.left.right will ` `    ``// print same value ` `    ``// root.right.left and root.right.right ` `    ``// will print same value ` `    ``// So we use any one of the condition ` `    ``printMiddleLevelUtil(a->left, b->left->left); ` `    ``printMiddleLevelUtil(a->right, b->left->left);     ` `} ` ` `  `// Main printing method that take a Tree as input ` `void` `printMiddleLevel(Node* node) ` `{ ` `    ``printMiddleLevelUtil(node, node); ` `}  ` ` `  ` `  `// Driver program to test above functions ` `int` `main() ` `{ ` `     `  `    ``Node* n1 = newNode(1); ` `    ``Node* n2 = newNode(2); ` `    ``Node* n3 = newNode(3); ` `    ``Node* n4 = newNode(4); ` `    ``Node* n5 = newNode(5); ` `    ``Node* n6 = newNode(6); ` `    ``Node* n7 = newNode(7); ` ` `  `    ``n2->left = n4; ` `    ``n2->right = n5; ` `    ``n3->left = n6; ` `    ``n3->right = n7; ` `    ``n1->left = n2; ` `    ``n1->right = n3; ` ` `  `    ``printMiddleLevel(n1);    ` `} ` ` `  `// This code is contributed by Prasad Kshirsagar        `

## Java

 `// Tree node definition ` `class` `Node ` `{ ` `    ``public` `int` `key; ` `    ``public` `Node left; ` `    ``public` `Node right; ` `    ``public` `Node(``int` `val) ` `    ``{ ` `        ``this``.left = ``null``; ` `        ``this``.right = ``null``; ` `        ``this``.key = val; ` `    ``} ` `} ` ` `  `public` `class` `PrintMiddle ` `{ ` `    ``// Takes two parameters - same initially and ` `    ``// calls recursively ` `    ``private` `static` `void` `printMiddleLevelUtil(Node a, ` `                                             ``Node b) ` `    ``{ ` `        ``// Base case e ` `        ``if` `(a == ``null` `|| b == ``null``) ` `            ``return``; ` ` `  `        ``// Fast pointer has reached the leaf so print ` `        ``// value at slow pointer ` `        ``if` `((b.left == ``null``) && (b.right == ``null``)) ` `        ``{ ` `            ``System.out.print(a.key + ``" "``); ` `            ``return``; ` `        ``} ` ` `  `        ``// Recursive call ` `        ``// root.left.left and root.left.right will ` `        ``// print same value ` `        ``// root.right.left and root.right.right ` `        ``// will print same value ` `        ``// So we use any one of the condition ` `        ``printMiddleLevelUtil(a.left, b.left.left); ` `        ``printMiddleLevelUtil(a.right, b.left.left); ` `    ``} ` ` `  `    ``// Main printing method that take a Tree as input ` `    ``public` `static` `void` `printMiddleLevel(Node node) ` `    ``{ ` `        ``printMiddleLevelUtil(node, node); ` `    ``} ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``Node n1 = ``new` `Node(``1``); ` `        ``Node n2 = ``new` `Node(``2``); ` `        ``Node n3 = ``new` `Node(``3``); ` `        ``Node n4 = ``new` `Node(``4``); ` `        ``Node n5 = ``new` `Node(``5``); ` `        ``Node n6 = ``new` `Node(``6``); ` `        ``Node n7 = ``new` `Node(``7``); ` ` `  `        ``n2.left = n4; ` `        ``n2.right = n5; ` `        ``n3.left = n6; ` `        ``n3.right = n7; ` `        ``n1.left = n2; ` `        ``n1.right = n3; ` ` `  `        ``printMiddleLevel(n1); ` `    ``} ` `} `

Output:

```2 3
```

This article is contributed by Balkishan. You could hit me an email – kishan020696@gmail.com If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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