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
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; } |
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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 |
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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 |
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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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