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

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 the tree. 

1. Advance fast pointer towards leaf by 2.
2. Advance slow pointer towards lead by 1.
3. If fast pointer reaches the leaf print value at the slow pointer
4. Check if the fast->left->left exists, then recursively move slow pointer by one step and fast pointer by two steps.
5. If the fast->left->left doesn’t exist (in case of even number of levels), the move both the pointers by one step.

Implementation:

C++

#include <bits/stdc++.h>
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
    if (b->left->left) {
        printMiddleLevelUtil(a->left, b->left->left);
        printMiddleLevelUtil(a->right, b->left->left);
    }
    else {
        printMiddleLevelUtil(a->left, b->left);
        printMiddleLevelUtil(a->right, b->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
        if (b.left.left!=null)
        {
            printMiddleLevelUtil(a.left, b.left.left);
            printMiddleLevelUtil(a.right, b.left.left);
        }
        else
        {
            printMiddleLevelUtil(a.left, b.left);
            printMiddleLevelUtil(a.right, b.left);
        }
    }

    // Main printing method that take a Tree as input
    public static void printMiddleLevel(Node node)
    {
        printMiddleLevelUtil(node, node);
    }

    // Driver code
    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);
    }
}

Python3

''' A binary tree node has key, pointer to left
   child and a pointer to right child '''

class Node:

    def __init__(self, key):

        self.key=key
        self.left = None
        self.right = None

# To create a newNode of tree and return pointer
def newNode(key):

    temp = Node(key)
    return temp

# Takes two parameters - same initially and
# calls recursively
def  printMiddleLevelUtil(a, b):

    # Base case e
    if (a == None or b == None):
        return;

    # Fast pointer has reached the leaf so print
    # value at slow pointer
    if ((b.left == None) and (b.right == None)):
        print(a.key, end=' ')

        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
    if (b.left.left):
        printMiddleLevelUtil(a.left, b.left.left);
        printMiddleLevelUtil(a.right, b.left.left);

    else:
        printMiddleLevelUtil(a.left, b.left);
        printMiddleLevelUtil(a.right, b.left);

# Main printing method that take a Tree as input
def printMiddleLevel(node):

    printMiddleLevelUtil(node, node);

# Driver program to test above functions
if __name__=='__main__':


    n1 = newNode(1);
    n2 = newNode(2);
    n3 = newNode(3);
    n4 = newNode(4);
    n5 = newNode(5);
    n6 = newNode(6);
    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 rutvik_56

C#

using System;
// Tree node definition
public 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))
        {
            Console.Write(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
        if (b.left.left!=null)
        {
            printMiddleLevelUtil(a.left, b.left.left);
            printMiddleLevelUtil(a.right, b.left.left);
        }
        else
        {
            printMiddleLevelUtil(a.left, b.left);
            printMiddleLevelUtil(a.right, b.left);
        }
    }

    // Main printing method that take a Tree as input
    public static void printMiddleLevel(Node node)
    {
        printMiddleLevelUtil(node, node);
    }

    // Driver code
    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);
    }
}

// This code is contributed by Amit Katiyar

Javascript

<script>
    // Tree node definition

    class Node
    {
        constructor(val) {
           this.left = null;
           this.right = null;
           this.key = val;
        }
    }

    // Takes two parameters - same initially and
    // calls recursively
    function printMiddleLevelUtil(a, 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))
        {
            document.write(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
        if (b.left.left != null)
        {
            printMiddleLevelUtil(a.left, b.left.left);
            printMiddleLevelUtil(a.right, b.left.left);
        }
        else
        {
            printMiddleLevelUtil(a.left, b.left);
            printMiddleLevelUtil(a.right, b.left);
        }
    }

    // Main printing method that take a Tree as input
    function printMiddleLevel(node)
    {
        printMiddleLevelUtil(node, node);
    }

    let n1 = new Node(1);
    let n2 = new Node(2);
    let n3 = new Node(3);
    let n4 = new Node(4);
    let n5 = new Node(5);
    let n6 = new Node(6);
    let n7 = new Node(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 suresh07.
</script>

Output

2 3

Time Complexity: O(n), As we are doing normal preorder traversal, every node can be visited atmost once.
Auxiliary Space: O(h), Here h is the height of the tree and the extra space is used due to recursive function call stack.

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