Given a Binary Tree, the task is to perform a Specific Level Order Traversal of the tree such that at each level print 1st element then the last element, then 2nd element and 2nd last element, until all elements of that level is printed and so on.
Examples:
Input: Below is the given tree:
Output: 1 8 3 4 7 5 6
Explanation:
1st level: 1(root)
2nd level: 8(left), 3(right)
3rd level: 4(left), 7(right), 5(left), 6(right)Input: Below is the given tree:
Output: 20 8 22 4 12 10 14
Explanation:
1st level: 20(root)
2nd level: 8(left), 22(right)
3rd level: 4(left), 12(right)
4th level: 10(left), 14(right)
Approach: The idea is to the Level Order Traversal of the given Binary Tree and converts this tree into a Perfect Binary Tree. While traversing if any node whose child Node doesn’t exist then append NULL node as the child node in the queue. Below are the steps:
- Perform the level order traversal of the given tree.
- During traversal, if any node whose child Node doesn’t exist then append NULL node as the child node in the queue
- During traversal for each level do the following:
- Maintain two queues to store the nodes for the two half of each level.
- For the first half of the level store nodes in a left to right manner.
- For the second half of the level store nodes in a right to left manner.
- Iterate the above two queues formed and print in an alternate manner for specific level order traversal.
Below is the implementation of the above approach:
// C++ Program for the above approach #include <bits/stdc++.h> using namespace std;
struct Node {
int data;
Node* left;
Node* right;
}; // Creates and initialize a new node Node* newNode( int ch)
{ // Allocating memory to a new node
Node* n = new Node();
n->data = ch;
n->left = NULL;
n->right = NULL;
return n;
} // Function to find the height of tree int Height(Node* root)
{ if (root == NULL)
return 0;
return 1 + max(Height(root->left), Height(root->right));
} // Given a perfect binary tree // print its node in specific order void printSpecificLevelOrder(queue<Node*> A, queue<Node*> B,
int height)
{ while (height != 0) {
// Get each one front
// element of both queue
Node* X = A.front();
A.pop();
Node* Y = B.front();
B.pop();
// check if X exist or not
if (X == NULL) {
// Assume is has and put
// their both child as none
A.push(NULL);
A.push(NULL);
}
else {
// print the data and store
// their child first left
// then right
cout << X->data << " " ;
A.push(X->left);
A.push(X->right);
}
// checkY exist or not
if (Y == NULL) {
// Assume is has and put
// their both child as none
B.push(NULL);
B.push(NULL);
}
else {
// print the data and store
// their child first left
// then right
cout << Y->data << " " ;
B.push(Y->right);
B.push(Y->left);
}
// Decrease by 1 unit
height -= 1;
}
} int main()
{ // Given Tree
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(7);
root->left->left->left = newNode(8);
root->left->left->right = newNode(9);
root->left->right->left = newNode(10);
root->left->right->right = newNode(11);
root->right->right->left = newNode(14);
root->right->right->right = newNode(15);
root->left->left->left->left = newNode(16);
root->left->left->left->right = newNode(17);
root->left->left->right->left = newNode(18);
root->left->left->right->right = newNode(19);
root->left->right->left->left = newNode(20);
root->left->right->left->right = newNode(21);
root->left->right->right->left = newNode(22);
root->left->right->right->right = newNode(23);
root->right->right->left->left = newNode(28);
root->right->right->left->right = newNode(29);
root->right->right->right->left = newNode(30);
root->right->right->right->right = newNode(31);
// Initialise Queue
queue<Node*> A;
queue<Node*> B;
int height = 0;
// check top root manually
if (root != NULL) {
cout << root->data << " " ;
A.push(root->left);
B.push(root->right);
height = Height(root);
height = pow (2, (height - 1)) - 1;
}
// Function call
printSpecificLevelOrder(A, B, height);
} // This code is contributed by Yash Agarwal(yashagarwal2852002) |
// Java program for the above approach import java.util.*;
class GFG{
// Node structure static class node
{ int data;
node left = null ;
node right = null ;
} // Creates and initialize a new node static node newNode( int ch)
{ // Allocating memory to a new node
node n = new node();
n.data = ch;
n.left = null ;
n.right = null ;
return n;
} // Function to find the height of tree static int Height(node root)
{ if (root == null )
return 0 ;
return 1 + Math.max(Height(root.left),
Height(root.right));
} // Given a perfect binary tree // print its node in Specific order static void printSpecificLevelOrder(Queue<node> A,
Queue<node> B,
int height)
{ while (height != 0 )
{
// Get each one front
// element of both queue
node X = A.poll();
node Y = B.poll();
// Check if X exist or not
if (X == null )
{
// Assume is has and put
// their both child as none
A.add( null );
A.add( null );
}
else
{
// print the data and store
// their child first left
// then right
System.out.print(X.data + " " );
A.add(X.left);
A.add(X.right);
}
// Check Y exist or not
if (Y == null )
{
// Assume is has and put
// their both child as none
B.add( null );
B.add( null );
}
else
{
// Print the data and store their
// child first left then right
System.out.print(Y.data + " " );
B.add(Y.right);
B.add(Y.left);
}
// Decrease by 1 unit
height -= 1 ;
}
} // Driver Code public static void main (String[] args)
{ // Given tree
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( 7 );
root.left.left.left = newNode( 8 );
root.left.left.right = newNode( 9 );
root.left.right.left = newNode( 10 );
root.left.right.right = newNode( 11 );
root.right.right.left = newNode( 14 );
root.right.right.right = newNode( 15 );
root.left.left.left.left = newNode( 16 );
root.left.left.left.right = newNode( 17 );
root.left.left.right.left = newNode( 18 );
root.left.left.right.right = newNode( 19 );
root.left.right.left.left = newNode( 20 );
root.left.right.left.right = newNode( 21 );
root.left.right.right.left = newNode( 22 );
root.left.right.right.right = newNode( 23 );
root.right.right.left.left = newNode( 28 );
root.right.right.left.right = newNode( 29 );
root.right.right.right.left = newNode( 30 );
root.right.right.right.right = newNode( 31 );
// Initialise Queue
Queue<node> A = new LinkedList<>();
Queue<node> B = new LinkedList<>();
int height = 0 ;
// Check top root manually
if (root != null )
{
System.out.print(root.data + " " );
A.add(root.left);
B.add(root.right);
height = Height(root);
height = ( int )Math.pow( 2 , (height - 1 )) - 1 ;
}
// Function Call
printSpecificLevelOrder(A, B, height);
} } // This code is contributed by offbeat |
# Python3 program for the above approach from queue import Queue
# A binary tree node class Node:
# A constructor for making
# a new node
def __init__( self , key):
self .data = key
self .left = None
self .right = None
# Function to find the height of tree def Height(root):
if root = = None :
return 0
return 1 + max (Height(root.left), Height(root.right))
# Given a perfect binary tree # print its node in Specific order def printSpecificLevelOrder(A, B, height):
while height ! = 0 :
# Get each one front
# element of both queue
X = A.get()
Y = B.get()
# Check if X exist or not
if X = = None :
# Assume is has and put
# their both child as none
A.put( None )
A.put( None )
else :
# print the data and store
# their child first left
# then right
print (X.data, end = " " )
A.put(X.left)
A.put(X.right)
# Check Y exist or not
if Y = = None :
# Assume is has and put
# their both child as none
B.put( None )
B.put( None )
else :
# print the data and store their
# child first left then right
print (Y.data, end = " " )
B.put(Y.right)
B.put(Y.left)
# Decrease by 1 unit
height - = 1
# Driver Code # Given Tree root = Node( 1 )
root.left = Node( 2 )
root.right = Node( 3 )
root.left.left = Node( 4 )
root.left.right = Node( 5 )
root.right.right = Node( 7 )
root.left.left.left = Node( 8 )
root.left.left.right = Node( 9 )
root.left.right.left = Node( 10 )
root.left.right.right = Node( 11 )
root.right.right.left = Node( 14 )
root.right.right.right = Node( 15 )
root.left.left.left.left = Node( 16 )
root.left.left.left.right = Node( 17 )
root.left.left.right.left = Node( 18 )
root.left.left.right.right = Node( 19 )
root.left.right.left.left = Node( 20 )
root.left.right.left.right = Node( 21 )
root.left.right.right.left = Node( 22 )
root.left.right.right.right = Node( 23 )
root.right.right.left.left = Node( 28 )
root.right.right.left.right = Node( 29 )
root.right.right.right.left = Node( 30 )
root.right.right.right.right = Node( 31 )
# Initialise Queue A = Queue( 100 )
B = Queue( 100 )
# Check top root manually if root ! = None :
print (root.data, end = " " )
A.put(root.left)
B.put(root.right)
height = Height(root)
height = 2 * * (height - 1 ) - 1
# Function Call printSpecificLevelOrder(A, B, height)
|
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG{
// Node structure class node
{ public int data;
public node left = null ;
public node right = null ;
} // Creates and initialize a new node static node newNode( int ch)
{ // Allocating memory to a new node
node n = new node();
n.data = ch;
n.left = null ;
n.right = null ;
return n;
} // Function to find the height of tree static int Height(node root)
{ if (root == null )
return 0;
return 1 + Math.Max(Height(root.left),
Height(root.right));
} // Given a perfect binary tree // print its node in Specific order static void printSpecificLevelOrder(Queue<node> A,
Queue<node> B,
int height)
{ while (height != 0)
{
// Get each one front
// element of both queue
node X = A.Peek();
A.Dequeue();
node Y = B.Peek();
B.Dequeue();
// Check if X exist or not
if (X == null )
{
// Assume is has and put
// their both child as none
A.Enqueue( null );
A.Enqueue( null );
}
else
{
// print the data and store
// their child first left
// then right
Console.Write(X.data + " " );
A.Enqueue(X.left);
A.Enqueue(X.right);
}
// Check Y exist or not
if (Y == null )
{
// Assume is has and put
// their both child as none
B.Enqueue( null );
B.Enqueue( null );
}
else
{
// Print the data and store their
// child first left then right
Console.Write(Y.data + " " );
B.Enqueue(Y.right);
B.Enqueue(Y.left);
}
// Decrease by 1 unit
height -= 1;
}
} // Driver Code public static void Main()
{ // Given tree
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(7);
root.left.left.left = newNode(8);
root.left.left.right = newNode(9);
root.left.right.left = newNode(10);
root.left.right.right = newNode(11);
root.right.right.left = newNode(14);
root.right.right.right = newNode(15);
root.left.left.left.left = newNode(16);
root.left.left.left.right = newNode(17);
root.left.left.right.left = newNode(18);
root.left.left.right.right = newNode(19);
root.left.right.left.left = newNode(20);
root.left.right.left.right = newNode(21);
root.left.right.right.left = newNode(22);
root.left.right.right.right = newNode(23);
root.right.right.left.left = newNode(28);
root.right.right.left.right = newNode(29);
root.right.right.right.left = newNode(30);
root.right.right.right.right = newNode(31);
// Initialise Queue
Queue<node> A = new Queue<node>();
Queue<node> B = new Queue<node>();
int height = 0;
// Check top root manually
if (root != null )
{
Console.Write(root.data + " " );
A.Enqueue(root.left);
B.Enqueue(root.right);
height = Height(root);
height = ( int )Math.Pow(2, (height - 1)) - 1;
}
// Function Call
printSpecificLevelOrder(A, B, height);
} } // This code is contributed by ipg2016107 |
<script> // Javascript program for the above approach
// Node structure
class node
{
constructor(ch) {
this .data = ch;
this .left = this .right = null ;
}
}
// Creates and initialize a new node
function newNode(ch)
{
// Allocating memory to a new node
let n = new node(ch);
return n;
}
// Function to find the height of tree
function Height(root)
{
if (root == null )
return 0;
return 1 + Math.max(Height(root.left),
Height(root.right));
}
// Given a perfect binary tree
// print its node in Specific order
function printSpecificLevelOrder(A, B, height)
{
while (height != 0)
{
// Get each one front
// element of both queue
let X = A[0];
A.shift();
let Y = B[0];
B.shift();
// Check if X exist or not
if (X == null )
{
// Assume is has and put
// their both child as none
A.push( null );
A.push( null );
}
else
{
// print the data and store
// their child first left
// then right
document.write(X.data + " " );
A.push(X.left);
A.push(X.right);
}
// Check Y exist or not
if (Y == null )
{
// Assume is has and put
// their both child as none
B.push( null );
B.push( null );
}
else
{
// Print the data and store their
// child first left then right
document.write(Y.data + " " );
B.push(Y.right);
B.push(Y.left);
}
// Decrease by 1 unit
height -= 1;
}
}
// Given tree
let 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(7);
root.left.left.left = newNode(8);
root.left.left.right = newNode(9);
root.left.right.left = newNode(10);
root.left.right.right = newNode(11);
root.right.right.left = newNode(14);
root.right.right.right = newNode(15);
root.left.left.left.left = newNode(16);
root.left.left.left.right = newNode(17);
root.left.left.right.left = newNode(18);
root.left.left.right.right = newNode(19);
root.left.right.left.left = newNode(20);
root.left.right.left.right = newNode(21);
root.left.right.right.left = newNode(22);
root.left.right.right.right = newNode(23);
root.right.right.left.left = newNode(28);
root.right.right.left.right = newNode(29);
root.right.right.right.left = newNode(30);
root.right.right.right.right = newNode(31);
// Initialise Queue
let A = [];
let B = [];
let height = 0;
// Check top root manually
if (root != null )
{
document.write(root.data + " " );
A.push(root.left);
B.push(root.right);
height = Height(root);
height = Math.pow(2, (height - 1)) - 1;
}
// Function Call
printSpecificLevelOrder(A, B, height);
</script> |
1 2 3 4 7 5 8 15 9 14 10 11 16 31 17 30 18 29 19 28 20 21 22 23
Time Complexity: O(N), N is the number of nodes
Auxiliary Space: O(N), N is the number of nodes