Reverse alternate levels of a perfect binary tree using Stack

Given a Perfect Binary Tree, the task is to reverse the alternate level nodes of the binary tree.

Examples:

Input:
               a
            /     \
           b       c
         /  \     /  \
        d    e    f    g
       / \  / \  / \  / \
       h  i j  k l  m  n  o  
Output:
Inorder Traversal of given tree
h d i b j e k a l f m c n g o 
Inorder Traversal of modified tree
o d n c m e l a k f j b i g h

Input:
               a
              / \
             b   c
Output:
Inorder Traversal of given tree
b a c 
Inorder Traversal of modified tree
c a b

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

Approach: Another approach of the problem is discussed here. In this article, we discuss an approach involving stack.
Traverse the tree in a depth first fashion and for each level,

If level is odd, push the left and right child(if exists) in a stack.
If level is even, replace the value of the current node with the top of stack.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// A tree node 
struct Node { 
    char key; 
    struct Node *left, *right; 
}; 
  
// Utility function to create new Node 
Node* newNode(int key) 
{ 
    Node* temp = new Node; 
    temp->key = key; 
    temp->left = temp->right = NULL; 
    return (temp); 
} 
  
// Utility function to perform 
// inorder traversal of the tree 
void inorder(Node* root) 
{ 
    if (root != NULL) { 
        inorder(root->left); 
        cout << root->key << " "; 
        inorder(root->right); 
    } 
} 
  
// Function to reverse alternate nodes 
void reverseAlternate(Node* root) 
{ 
  
    // Queue for depth first traversal 
    queue<Node*> q; 
    q.push(root); 
    Node* temp; 
  
    // Level of root considered to be 1 
    int n, level = 1; 
  
    // Stack to store nodes of a level 
    stack<int> s; 
  
    while (!q.empty()) { 
        n = q.size(); 
        while (n--) { 
            temp = q.front(); 
            q.pop(); 
  
            // If level is odd 
            if (level % 2) { 
  
                // Store the left and right child 
                // in the stack 
                if (temp->left) { 
                    q.push(temp->left); 
                    s.push(temp->left->key); 
                } 
  
                if (temp->right) { 
                    q.push(temp->right); 
                    s.push(temp->right->key); 
                } 
            } 
  
            // If level is even 
            else { 
  
                // Replace the value of node 
                // with top of the stack 
                temp->key = s.top(); 
                s.pop(); 
  
                if (temp->left) 
                    q.push(temp->left); 
                if (temp->right) 
                    q.push(temp->right); 
            } 
        } 
  
        // Increment the level 
        level++; 
    } 
} 
  
// Driver code 
int main() 
{ 
    struct Node* root = newNode('a'); 
    root->left = newNode('b'); 
    root->right = newNode('c'); 
    root->left->left = newNode('d'); 
    root->left->right = newNode('e'); 
    root->right->left = newNode('f'); 
    root->right->right = newNode('g'); 
    root->left->left->left = newNode('h'); 
    root->left->left->right = newNode('i'); 
    root->left->right->left = newNode('j'); 
    root->left->right->right = newNode('k'); 
    root->right->left->left = newNode('l'); 
    root->right->left->right = newNode('m'); 
    root->right->right->left = newNode('n'); 
    root->right->right->right = newNode('o'); 
  
    cout << "Inorder Traversal of given tree\n"; 
    inorder(root); 
  
    reverseAlternate(root); 
  
    cout << "\nInorder Traversal of modified tree\n"; 
    inorder(root); 
  
    return 0; 
} 

Java

// Java implementation of the approach  
import java.util.*; 
  
class GfG 
{  
  
// A tree node  
static class Node 
{  
    char key;  
    Node left, right;  
} 
  
// Utility function to create new Node  
static Node newNode(char key)  
{  
    Node temp = new Node();  
    temp.key = key;  
    temp.left = temp.right = null;  
    return (temp);  
}  
  
// Utility function to perform  
// inorder traversal of the tree  
static void inorder(Node root)  
{  
    if (root != null) 
    {  
        inorder(root.left);  
        System.out.print(root.key + " ");  
        inorder(root.right);  
    }  
}  
  
// Function to reverse alternate nodes  
static void reverseAlternate(Node root)  
{  
  
    // Queue for depth first traversal  
    Queue<Node> q = new LinkedList<Node> ();  
    q.add(root);  
    Node temp;  
  
    // Level of root considered to be 1  
    int n, level = 1;  
  
    // Stack to store nodes of a level  
    Stack<Character> s = new Stack<Character> ();  
  
    while (!q.isEmpty()) 
    {  
        n = q.size();  
        while (n != 0) 
        {  
            if(!q.isEmpty())  
            { 
                temp = q.peek();  
                q.remove();  
            } 
            else
            temp = null; 
              
            // If level is odd  
            if (level % 2 != 0) 
            {  
  
                // Store the left and right child  
                // in the stack  
                if (temp != null && temp.left != null)  
                {  
                    q.add(temp.left);  
                    s.push(temp.left.key);  
                }  
  
                if (temp != null && temp.right != null) 
                {  
                    q.add(temp.right);  
                    s.push(temp.right.key);  
                }  
            }  
  
            // If level is even  
            else
            {  
  
                // Replace the value of node  
                // with top of the stack  
                temp.key = s.peek();  
                s.pop();  
  
                if (temp.left != null)  
                    q.add(temp.left);  
                if (temp.right != null)  
                    q.add(temp.right);  
            } 
            n--; 
        }  
  
        // Increment the level  
        level++;  
    }  
}  
  
// Driver code  
public static void main(String[] args)  
{  
    Node root = newNode('a');  
    root.left = newNode('b');  
    root.right = newNode('c');  
    root.left.left = newNode('d');  
    root.left.right = newNode('e');  
    root.right.left = newNode('f');  
    root.right.right = newNode('g');  
    root.left.left.left = newNode('h');  
    root.left.left.right = newNode('i');  
    root.left.right.left = newNode('j');  
    root.left.right.right = newNode('k');  
    root.right.left.left = newNode('l');  
    root.right.left.right = newNode('m');  
    root.right.right.left = newNode('n');  
    root.right.right.right = newNode('o');  
  
    System.out.println("Inorder Traversal of given tree");  
    inorder(root);  
  
    reverseAlternate(root);  
  
    System.out.println("\nInorder Traversal of modified tree");  
    inorder(root);  
} 
}  
  
// This code is contributed by Prerna Saini 

C#

// C# implementation of the approach  
using System; 
using System.Collections; 
  
class GfG  
{  
  
// A tree node  
public class Node  
{  
    public char key;  
    public Node left, right;  
}  
  
// Utility function to create new Node  
static Node newNode(char key)  
{  
    Node temp = new Node();  
    temp.key = key;  
    temp.left = temp.right = null;  
    return (temp);  
}  
  
// Utility function to perform  
// inorder traversal of the tree  
static void inorder(Node root)  
{  
    if (root != null)  
    {  
        inorder(root.left);  
        Console.Write(root.key + " ");  
        inorder(root.right);  
    }  
}  
  
// Function to reverse alternate nodes  
static void reverseAlternate(Node root)  
{  
  
    // Queue for depth first traversal  
    Queue q = new Queue ();  
    q.Enqueue(root);  
    Node temp;  
  
    // Level of root considered to be 1  
    int n, level = 1;  
  
    // Stack to store nodes of a level  
    Stack s = new Stack ();  
  
    while (q.Count > 0)  
    {  
        n = q.Count;  
        while (n != 0)  
        {  
            if(q.Count > 0)  
            {  
                temp = (Node)q.Peek();  
                q.Dequeue();  
            }  
            else
            temp = null;  
              
            // If level is odd  
            if (level % 2 != 0)  
            {  
  
                // Store the left and right child  
                // in the stack  
                if (temp != null && temp.left != null)  
                {  
                    q.Enqueue(temp.left);  
                    s.Push(temp.left.key);  
                }  
  
                if (temp != null && temp.right != null)  
                {  
                    q.Enqueue(temp.right);  
                    s.Push(temp.right.key);  
                }  
            }  
  
            // If level is even  
            else
            {  
  
                // Replace the value of node  
                // with top of the stack  
                temp.key =(char)s.Peek();  
                s.Pop();  
  
                if (temp.left != null)  
                    q.Enqueue(temp.left);  
                if (temp.right != null)  
                    q.Enqueue(temp.right);  
            }  
            n--;  
        }  
  
        // Increment the level  
        level++;  
    }  
}  
  
// Driver code  
public static void Main(String []args)  
{  
    Node root = newNode('a');  
    root.left = newNode('b');  
    root.right = newNode('c');  
    root.left.left = newNode('d');  
    root.left.right = newNode('e');  
    root.right.left = newNode('f');  
    root.right.right = newNode('g');  
    root.left.left.left = newNode('h');  
    root.left.left.right = newNode('i');  
    root.left.right.left = newNode('j');  
    root.left.right.right = newNode('k');  
    root.right.left.left = newNode('l');  
    root.right.left.right = newNode('m');  
    root.right.right.left = newNode('n');  
    root.right.right.right = newNode('o');  
  
    Console.WriteLine("Inorder Traversal of given tree");  
    inorder(root);  
  
    reverseAlternate(root);  
  
    Console.WriteLine("\nInorder Traversal of modified tree");  
    inorder(root);  
}  
}  
  
// This code is contributed by Arnab Kundu 

Output:

Inorder Traversal of given tree
h d i b j e k a l f m c n g o 
Inorder Traversal of modified tree
o d n c m e l a k f j b i g h

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My Personal Notes

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

C++

Java

C#

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