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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 a 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 slow pointer
4. Call recursively the next route.

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
    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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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