Print all Palindromic Levels Of a Binary Tree

Given a Binary Tree, the task is to print all palindromic levels of this tree. 

Palindrome Level Any level of a Binary tree is said to be a palindromic level if on traversing it from left to right, the result is same as traversing that level from right to left.

Examples:

Input: 
                 1
                /  \ 
              12    13 
             /     /   \ 
            11    6     11 
                   \    / 
                   2   2  
Output:
1
11 6 11
2 2
Explanation:
First, third and fourth levels are palindrome.

Input: 
                 7
                /  \ 
              22      22 
             /  \      \
            3     6     3 
           / \     \    / 
          1   5     8   1    
                   /
                  23 
Output:
7
22 22
3 6 3
23

Explanation:
First, second, third and fifth levels are palindrome.

Approach: In order to check if each level is a Palindromic level, we need to first do the Level Order Traversal of the Binary tree to get the values at each level. Then for each level, check if it a palindrome or not.
Here a Queue data structure is used to store the levels of the Tree while doing the Level Order Traversal.

Below is the implementation of the above approach: 

C++

// C++ program for printing a
// Palindromic Levels of Binary Tree

#include <bits/stdc++.h>
using namespace std;

// A Tree node
struct Node {
    int key;
    struct Node *left, *right;
};

// Utility function to create a new node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return (temp);
}

// Function to print a Palindromic level
void printLevel(struct Node* queue[],
                int index, int size)
{
    for (int i = index; i < size; i++) {
        cout << queue[i]->key << " ";
    }

    cout << endl;
}

// Function to check whether given level
// is Palindromic level or not
bool isPalindrome(struct Node* queue[],
                  int index, int size)
{

    while (index < size) {
        // check value of two nodes are
        // equal or not
        if (queue[index++]->key
            != queue[size--]->key)
            return false;
    }

    return true;
}

// Utility function to get palindromic
// level of a given Binary tree
void printPalLevel(struct Node* node,
                   struct Node* queue[],
                   int index, int size)
{

    // Print root node value
    // as a single value in a
    // level is also a Palindrome
    cout << queue[index]->key << endl;

    // Level order traversal of Tree
    while (index < size) {
        int curr_size = size;
        while (index < curr_size) {
            struct Node* temp = queue[index];

            if (temp->left != NULL) {
                queue[size++] = temp->left;
            }

            if (temp->right != NULL) {
                queue[size++] = temp->right;
            }

            index++;
        }

        // Check if level is Palindrome or not
        if (isPalindrome(queue, index, size - 1)) {
            printLevel(queue, index, size);
        }
    }
}

// Function to find total no of nodes
int findSize(struct Node* node)
{

    if (node == NULL)
        return 0;

    return 1
           + findSize(node->left)
           + findSize(node->right);
}

// Function to find palindromic level
// In a given binary tree
void printPalindromicLevel(struct Node* node)
{
    int t_size = findSize(node);
    struct Node* queue[t_size];
    queue[0] = node;
    printPalLevel(node, queue, 0, 1);
}

// Driver Code
int main()
{
    /*     10
        /    \
       13     13
        /       \
        14        15
        / \        / \
       21 22      22 21
                     /
                    8 */

    // Create Binary Tree as shown
    Node* root = newNode(10);
    root->left = newNode(13);
    root->right = newNode(13);

    root->right->left = newNode(14);
    root->right->right = newNode(15);

    root->right->left->left = newNode(21);
    root->right->left->right = newNode(22);
    root->right->right->left = newNode(22);
    root->right->right->right = newNode(21);
    root->right->right->right->left = newNode(8);

    // Print Palindromic Levels
    printPalindromicLevel(root);

    return 0;
}

Java

// Java program for printing a
// Palindromic Levels of Binary Tree
class GFG{

// A Tree node
static class Node {
    int key;
    Node left, right;
};

// Utility function to create a new node
static Node newNode(int key)
{
    Node temp = new Node();
    temp.key = key;
    temp.left = temp.right = null;
    return (temp);
}

// Function to print a Palindromic level
static void printLevel(Node queue[],
                int index, int size)
{
    for (int i = index; i < size; i++) {
        System.out.print(queue[i].key+ " ");
    }

    System.out.println();
}

// Function to check whether given level
// is Palindromic level or not
static boolean isPalindrome(Node queue[],
                  int index, int size)
{

    while (index < size) {
        // check value of two nodes are
        // equal or not
        if (queue[index++].key
            != queue[size--].key)
            return false;
    }

    return true;
}

// Utility function to get palindromic
// level of a given Binary tree
static void printPalLevel(Node node,
                   Node queue[],
                   int index, int size)
{

    // Print root node value
    // as a single value in a
    // level is also a Palindrome
    System.out.print(queue[index].key +"\n");

    // Level order traversal of Tree
    while (index < size) {
        int curr_size = size;
        while (index < curr_size) {
            Node temp = queue[index];

            if (temp.left != null) {
                queue[size++] = temp.left;
            }

            if (temp.right != null) {
                queue[size++] = temp.right;
            }

            index++;
        }

        // Check if level is Palindrome or not
        if (isPalindrome(queue, index, size - 1)) {
            printLevel(queue, index, size);
        }
    }
}

// Function to find total no of nodes
static int findSize(Node node)
{

    if (node == null)
        return 0;

    return 1
           + findSize(node.left)
           + findSize(node.right);
}

// Function to find palindromic level
// In a given binary tree
static void printPalindromicLevel(Node node)
{
    int t_size = findSize(node);
    Node []queue = new Node[t_size];
    queue[0] = node;
    printPalLevel(node, queue, 0, 1);
}

// Driver Code
public static void main(String[] args)
{
    /*     10
        /    \
       13     13
        /       \
        14        15
        / \        / \
       21 22      22 21
                     /
                    8 */

    // Create Binary Tree as shown
    Node root = newNode(10);
    root.left = newNode(13);
    root.right = newNode(13);

    root.right.left = newNode(14);
    root.right.right = newNode(15);

    root.right.left.left = newNode(21);
    root.right.left.right = newNode(22);
    root.right.right.left = newNode(22);
    root.right.right.right = newNode(21);
    root.right.right.right.left = newNode(8);

    // Print Palindromic Levels
    printPalindromicLevel(root);

}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program for printing a
# Palindromic Levels of Binary Tree

# A BST node
class Node:

    def __init__(self, x):

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

# Function to print a Palindromic level
def printLevel(queue, index, size):

    for i in range(index, size):
        print(queue[i].key, end = " ")

    print()

# Function to check whether given level
# is Palindromic level or not
def isPalindrome(queue, index, size):

    #print(index,size)

    while (index < size):

        # Check value of two nodes are
        # equal or not
        if (queue[index].key != queue[size].key):
            return False

        index += 1
        size -= 1

    return True

# Utility function to get palindromic
# level of a given Binary tree
def printPalLevel(node, queue, index, size):

    # Print root node value
    # as a single value in a
    # level is also a Palindrome
    print(queue[index].key)

    # Level order traversal of Tree
    while (index < size):
        curr_size = size

        while (index < curr_size):
            temp = queue[index]

            if (temp.left != None):
                queue[size] = temp.left
                size += 1

            if (temp.right != None):
                queue[size] = temp.right
                size += 1

            index += 1

        # Check if level is Palindrome or not
        if (isPalindrome(queue, index, size - 1) == True):
            printLevel(queue, index, size)

# Function to find total no of nodes
def findSize(node):

    if (node == None):
        return 0

    return (1 + findSize(node.left) +
                findSize(node.right))

# Function to find palindromic level
# In a given binary tree
def printPalindromicLevel(node):

    t_size = findSize(node)
    queue = [None for i in range(t_size)]
    queue[0] = node
    printPalLevel(node, queue, 0, 1)

# Driver Code
if __name__ == '__main__':

    #       10
    #     /    \
    #    13     13
    #     /       \
    #     14        15
    #     / \        / \
    #    21 22      22 21
    #                  /
    #                 8

    # Create Binary Tree as shown
    root = Node(10)
    root.left = Node(13)
    root.right = Node(13)

    root.right.left = Node(14)
    root.right.right = Node(15)

    root.right.left.left = Node(21)
    root.right.left.right = Node(22)
    root.right.right.left = Node(22)
    root.right.right.right = Node(21)
    root.right.right.right.left = Node(8)

    # Print Palindromic Levels
    printPalindromicLevel(root)

# This code is contributed by mohit kumar 29

C#

// C# program for printing a
// Palindromic Levels of Binary Tree
using System;

class GFG{

// A Tree node
class Node {
    public int key;
    public Node left, right;
};

// Utility function to create a new node
static Node newNode(int key)
{
    Node temp = new Node();
    temp.key = key;
    temp.left = temp.right = null;
    return (temp);
}

// Function to print a Palindromic level
static void printLevel(Node []queue,
                int index, int size)
{
    for (int i = index; i < size; i++) {
        Console.Write(queue[i].key+ " ");
    }

    Console.WriteLine();
}

// Function to check whether given level
// is Palindromic level or not
static bool isPalindrome(Node []queue,
                  int index, int size)
{

    while (index < size) {
        // check value of two nodes are
        // equal or not
        if (queue[index++].key
            != queue[size--].key)
            return false;
    }

    return true;
}

// Utility function to get palindromic
// level of a given Binary tree
static void printPalLevel(Node node,
                   Node []queue,
                   int index, int size)
{

    // Print root node value
    // as a single value in a
    // level is also a Palindrome
    Console.Write(queue[index].key +"\n");

    // Level order traversal of Tree
    while (index < size) {
        int curr_size = size;
        while (index < curr_size) {
            Node temp = queue[index];

            if (temp.left != null) {
                queue[size++] = temp.left;
            }

            if (temp.right != null) {
                queue[size++] = temp.right;
            }

            index++;
        }

        // Check if level is Palindrome or not
        if (isPalindrome(queue, index, size - 1)) {
            printLevel(queue, index, size);
        }
    }
}

// Function to find total no of nodes
static int findSize(Node node)
{

    if (node == null)
        return 0;

    return 1
           + findSize(node.left)
           + findSize(node.right);
}

// Function to find palindromic level
// In a given binary tree
static void printPalindromicLevel(Node node)
{
    int t_size = findSize(node);
    Node []queue = new Node[t_size];
    queue[0] = node;
    printPalLevel(node, queue, 0, 1);
}

// Driver Code
public static void Main(String[] args)
{
    /*     10
        /    \
       13     13
        /       \
        14        15
        / \        / \
       21 22      22 21
                     /
                    8 */

    // Create Binary Tree as shown
    Node root = newNode(10);
    root.left = newNode(13);
    root.right = newNode(13);

    root.right.left = newNode(14);
    root.right.right = newNode(15);

    root.right.left.left = newNode(21);
    root.right.left.right = newNode(22);
    root.right.right.left = newNode(22);
    root.right.right.right = newNode(21);
    root.right.right.right.left = newNode(8);

    // Print Palindromic Levels
    printPalindromicLevel(root);

}
}

// This code is contributed by sapnasingh4991

Javascript

<script>

// Javascript program for printing a
// Palindromic Levels of Binary Tree

// A Tree node
class Node
{

    // Utility function to create
    // a new node
    constructor(key)
    {
        this.key = key;
        this.left = this.right = null;
    }
}

// Function to print a Palindromic level
function printLevel(queue, index, size)
{
    for(let i = index; i < size; i++)
    {
        document.write(queue[i].key + " ");
    }
    document.write("<br>");
}

// Function to check whether given level
// is Palindromic level or not
function isPalindrome(queue, index, size)
{
    while (index < size)
    {

        // Check value of two nodes are
        // equal or not
        if (queue[index++].key !=
            queue[size--].key)
            return false;
    }
    return true;
}

// Utility function to get palindromic
// level of a given Binary tree
function printPalLevel(node, queue,
                       index, size)
{

    // Print root node value
    // as a single value in a
    // level is also a Palindrome
    document.write(queue[index].key + "<br>");

    // Level order traversal of Tree
    while (index < size)
    {
        let curr_size = size;
        while (index < curr_size)
        {
            let temp = queue[index];

            if (temp.left != null)
            {
                queue[size++] = temp.left;
            }

            if (temp.right != null)
            {
                queue[size++] = temp.right;
            }
            index++;
        }

        // Check if level is Palindrome or not
        if (isPalindrome(queue, index, size - 1))
        {
            printLevel(queue, index, size);
        }
    }
}

// Function to find total no of nodes
function findSize(node)
{
    if (node == null)
        return 0;

    return 1 + findSize(node.left) +
               findSize(node.right);
}

// Function to find palindromic level
// In a given binary tree
function printPalindromicLevel(node)
{
    let t_size = findSize(node);
    let queue = new Array(t_size);
    queue[0] = node;

    printPalLevel(node, queue, 0, 1);
}

// Driver Code
/*     10
        /    \
       13     13
        /       \
        14        15
        / \        / \
       21 22      22 21
                     /
                    8 */

// Create Binary Tree as shown
let root = new Node(10);
root.left = new Node(13);
root.right = new Node(13);

root.right.left = new Node(14);
root.right.right = new Node(15);

root.right.left.left = new Node(21);
root.right.left.right = new Node(22);
root.right.right.left = new Node(22);
root.right.right.right = new Node(21);
root.right.right.right.left = new Node(8);

// Print Palindromic Levels
printPalindromicLevel(root);

// This code is contributed by unknown2108

</script>

Output

10
13 13 
21 22 22 21 
8 

Time Complexity: time complexity of this code is O(n^2), as the isPalindrome function takes O(n) time to check whether the current level is a palindrome or not, and this function is called for every level of the tree. The total number of levels in the tree is equal to the height of the tree, which is also O(n) in the worst case. Therefore, the overall time complexity is O(n^2).

Auxiliary Space: O(n), as the size of the queue array, is equal to the total number of nodes in the binary tree. This is because the queue array is used to store the nodes during the level order traversal of the tree.