# Reverse Level Order traversal in spiral form

• Difficulty Level : Hard
• Last Updated : 12 Jan, 2023

Given a binary tree, the task is to print the reverse level order in spiral form.

Examples:

```Input:
1
/  \
2    3
/ \  / \
4   5 6  7
Output: 4 5 6 7 3 2 1

Input:
5
/   \
6     4
/ \   /
7   1 8
\     \
3     10
Output: 3 10 8 1 7 6 4 5```

Approach: The idea is to traverse the tree in a Reverse Level Order manner but with a slight modification. We will use a variable flag and initially set it’s value to one. As we complete the reverse level order traversal of the tree, from left to right we will set the value of flag to zero, so that next time we traverse the Tree from right to left and as we complete the traversal we set it’s value back to one. We will repeat this whole step until we have traversed the Binary Tree completely.

Below is the implementation of the above approach:

## C++

 `// C++ program to print reverse level order``// traversal of a binary tree in spiral form``#include ``using` `namespace` `std;` `// Binary tree Node``struct` `Node {``    ``struct` `Node* left;``    ``struct` `Node* right;``    ``int` `data;` `    ``Node(``int` `data)``    ``{``        ``this``->data = data;``        ``this``->left = NULL;``        ``this``->right = NULL;``    ``}``};` `// Recursive Function to find height``// of binary tree``int` `height(``struct` `Node* root)``{``    ``if` `(root == NULL)``        ``return` `0;` `    ``// Compute the height of each subtree``    ``int` `lheight = height(root->left);``    ``int` `rheight = height(root->right);` `    ``// Return the maximum of two``    ``return` `max(lheight + 1, rheight + 1);``}` `// Function to Print Nodes from right to left``void` `rightToLeft(``struct` `Node* root, ``int` `level)``{``    ``if` `(root == NULL)``        ``return``;` `    ``if` `(level == 1)``        ``cout << root->data << ``" "``;` `    ``else` `if` `(level > 1) {``        ``rightToLeft(root->right, level - 1);``        ``rightToLeft(root->left, level - 1);``    ``}``}` `// Function to Print Nodes from left to right``void` `leftToRight(``struct` `Node* root, ``int` `level)``{``    ``if` `(root == NULL)``        ``return``;` `    ``if` `(level == 1)``        ``cout << root->data << ``" "``;` `    ``else` `if` `(level > 1) {``        ``leftToRight(root->left, level - 1);``        ``leftToRight(root->right, level - 1);``    ``}``}` `// Function to print Reverse level order traversal``// of a Binary tree in spiral form``void` `reverseSpiral(``struct` `Node* root)``{``    ``// Flag is used to mark the change``    ``// in level``    ``int` `flag = 1;` `    ``// Height of tree``    ``int` `h = height(root);` `    ``for` `(``int` `i = h; i >= 1; i--) {` `        ``// If flag value is one print Nodes``        ``// from left to right``        ``if` `(flag == 1) {``            ``leftToRight(root, i);` `            ``// Mark flag as zero so that next time``            ``// tree is traversed from right to left``            ``flag = 0;``        ``}` `        ``// If flag is zero print Nodes``        ``// from right to left``        ``else` `if` `(flag == 0) {``            ``rightToLeft(root, i);` `            ``// Mark flag as one so that next time``            ``// Nodes are printed from left to right``            ``flag = 1;``        ``}``    ``}``}` `// Driver code``int` `main()``{``    ``struct` `Node* root = ``new` `Node(5);``    ``root->left = ``new` `Node(9);``    ``root->right = ``new` `Node(3);``    ``root->left->left = ``new` `Node(6);``    ``root->right->right = ``new` `Node(4);``    ``root->left->left->left = ``new` `Node(8);``    ``root->left->left->right = ``new` `Node(7);` `    ``reverseSpiral(root);` `    ``return` `0;``}`

## Java

 `// Java program to print reverse level order``// traversal of a binary tree in spiral form``class` `Solution {` `    ``// Binary tree Node``    ``static` `class` `Node {``        ``Node left;``        ``Node right;``        ``int` `data;` `        ``Node(``int` `data)``        ``{``            ``this``.data = data;``            ``this``.left = ``null``;``            ``this``.right = ``null``;``        ``}``    ``};` `    ``// Recursive Function to find height``    ``// of binary tree``    ``static` `int` `height(Node root)``    ``{``        ``if` `(root == ``null``)``            ``return` `0``;` `        ``// Compute the height of each subtree``        ``int` `lheight = height(root.left);``        ``int` `rheight = height(root.right);` `        ``// Return the maximum of two``        ``return` `Math.max(lheight + ``1``, rheight + ``1``);``    ``}` `    ``// Function to Print Nodes from right to left``    ``static` `void` `rightToLeft(Node root, ``int` `level)``    ``{``        ``if` `(root == ``null``)``            ``return``;` `        ``if` `(level == ``1``)``            ``System.out.print(root.data + ``" "``);` `        ``else` `if` `(level > ``1``) {``            ``rightToLeft(root.right, level - ``1``);``            ``rightToLeft(root.left, level - ``1``);``        ``}``    ``}` `    ``// Function to Print Nodes from left to right``    ``static` `void` `leftToRight(Node root, ``int` `level)``    ``{``        ``if` `(root == ``null``)``            ``return``;` `        ``if` `(level == ``1``)``            ``System.out.print(root.data + ``" "``);` `        ``else` `if` `(level > ``1``) {``            ``leftToRight(root.left, level - ``1``);``            ``leftToRight(root.right, level - ``1``);``        ``}``    ``}` `    ``// Function to print Reverse level order traversal``    ``// of a Binary tree in spiral form``    ``static` `void` `reverseSpiral(Node root)``    ``{``        ``// Flag is used to mark the change``        ``// in level``        ``int` `flag = ``1``;` `        ``// Height of tree``        ``int` `h = height(root);` `        ``for` `(``int` `i = h; i >= ``1``; i--) {` `            ``// If flag value is one print Nodes``            ``// from left to right``            ``if` `(flag == ``1``) {``                ``leftToRight(root, i);` `                ``// Mark flag as zero so that next time``                ``// tree is traversed from right to left``                ``flag = ``0``;``            ``}` `            ``// If flag is zero print Nodes``            ``// from right to left``            ``else` `if` `(flag == ``0``) {``                ``rightToLeft(root, i);` `                ``// Mark flag as one so that next time``                ``// Nodes are printed from left to right``                ``flag = ``1``;``            ``}``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``Node root = ``new` `Node(``5``);``        ``root.left = ``new` `Node(``9``);``        ``root.right = ``new` `Node(``3``);``        ``root.left.left = ``new` `Node(``6``);``        ``root.right.right = ``new` `Node(``4``);``        ``root.left.left.left = ``new` `Node(``8``);``        ``root.left.left.right = ``new` `Node(``7``);` `        ``reverseSpiral(root);``    ``}``}` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 program to print reverse level order``# traversal of a binary tree in spiral form` `class` `newNode:` `    ``# Construct to create a newNode``    ``def` `__init__(``self``, key):``        ``self``.data ``=` `key``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# Recursive Function to find height``# of binary tree``def` `height(root):` `    ``if` `(root ``=``=` `None``):``        ``return` `0` `    ``# Compute the height of each subtree``    ``lheight ``=` `height(root.left)``    ``rheight ``=` `height(root.right)` `    ``# Return the maximum of two``    ``return` `max``(lheight ``+` `1``, rheight ``+` `1``)` `# Function to Print Nodes from right to left``def` `rightToLeft(root, level):``    ``if` `(root ``=``=` `None``):``        ``return` `    ``if` `(level ``=``=` `1``):``        ``print``(root.data, end ``=``" "``)` `    ``elif` `(level > ``1``):``        ``rightToLeft(root.right, level ``-` `1``)``        ``rightToLeft(root.left, level ``-` `1``)``    ` `# Function to Print Nodes from left to right``def` `leftToRight(root, level):``    ``if` `(root ``=``=` `None``):``        ``return` `    ``if` `(level ``=``=` `1``) :``        ``print``(root.data, end ``=` `" "``)` `    ``elif` `(level > ``1``):``        ``leftToRight(root.left, level ``-` `1``)``        ``leftToRight(root.right, level ``-` `1``)``    ` `# Function to print Reverse level order traversal``# of a Binary tree in spiral form``def` `reverseSpiral(root):` `    ``# Flag is used to mark the``    ``# change in level``    ``flag ``=` `1` `    ``# Height of tree``    ``h ``=` `height(root)` `    ``for` `i ``in` `range``(h, ``0``, ``-``1``):` `        ``# If flag value is one print Nodes``        ``# from left to right``        ``if` `(flag ``=``=` `1``):``            ``leftToRight(root, i)` `            ``# Mark flag as zero so that next time``            ``# tree is traversed from right to left``            ``flag ``=` `0``        ` `        ``# If flag is zero print Nodes``        ``# from right to left``        ``elif` `(flag ``=``=` `0``):``            ``rightToLeft(root, i)` `            ``# Mark flag as one so that next time``            ``# Nodes are printed from left to right``            ``flag ``=` `1``    ` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``root ``=` `newNode(``5``)``    ``root.left ``=` `newNode(``9``)``    ``root.right ``=` `newNode(``3``)``    ``root.left.left ``=` `newNode(``6``)``    ``root.right.right ``=` `newNode(``4``)``    ``root.left.left.left ``=` `newNode(``8``)``    ``root.left.left.right ``=` `newNode(``7``)` `    ``reverseSpiral(root)``    ` `# This code is contributed``# by SHUBHAMSINGH10`

## C#

 `// C# program to print reverse level order``// traversal of a binary tree in spiral form``using` `System;` `class` `GFG {` `    ``// Binary tree Node``    ``public` `class` `Node {``        ``public` `Node left;``        ``public` `Node right;``        ``public` `int` `data;` `        ``public` `Node(``int` `data)``        ``{``            ``this``.data = data;``            ``this``.left = ``null``;``            ``this``.right = ``null``;``        ``}``    ``};` `    ``// Recursive Function to find height``    ``// of binary tree``    ``static` `int` `height(Node root)``    ``{``        ``if` `(root == ``null``)``            ``return` `0;` `        ``// Compute the height of each subtree``        ``int` `lheight = height(root.left);``        ``int` `rheight = height(root.right);` `        ``// Return the maximum of two``        ``return` `Math.Max(lheight + 1, rheight + 1);``    ``}` `    ``// Function to Print Nodes from right to left``    ``static` `void` `rightToLeft(Node root, ``int` `level)``    ``{``        ``if` `(root == ``null``)``            ``return``;` `        ``if` `(level == 1)``            ``Console.Write(root.data + ``" "``);` `        ``else` `if` `(level > 1) {``            ``rightToLeft(root.right, level - 1);``            ``rightToLeft(root.left, level - 1);``        ``}``    ``}` `    ``// Function to Print Nodes from left to right``    ``static` `void` `leftToRight(Node root, ``int` `level)``    ``{``        ``if` `(root == ``null``)``            ``return``;` `        ``if` `(level == 1)``            ``Console.Write(root.data + ``" "``);` `        ``else` `if` `(level > 1) {``            ``leftToRight(root.left, level - 1);``            ``leftToRight(root.right, level - 1);``        ``}``    ``}` `    ``// Function to print Reverse level order traversal``    ``// of a Binary tree in spiral form``    ``static` `void` `reverseSpiral(Node root)``    ``{``        ``// Flag is used to mark the change``        ``// in level``        ``int` `flag = 1;` `        ``// Height of tree``        ``int` `h = height(root);` `        ``for` `(``int` `i = h; i >= 1; i--) {` `            ``// If flag value is one print Nodes``            ``// from left to right``            ``if` `(flag == 1) {``                ``leftToRight(root, i);` `                ``// Mark flag as zero so that next time``                ``// tree is traversed from right to left``                ``flag = 0;``            ``}` `            ``// If flag is zero print Nodes``            ``// from right to left``            ``else` `if` `(flag == 0) {``                ``rightToLeft(root, i);` `                ``// Mark flag as one so that next time``                ``// Nodes are printed from left to right``                ``flag = 1;``            ``}``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``Node root = ``new` `Node(5);``        ``root.left = ``new` `Node(9);``        ``root.right = ``new` `Node(3);``        ``root.left.left = ``new` `Node(6);``        ``root.right.right = ``new` `Node(4);``        ``root.left.left.left = ``new` `Node(8);``        ``root.left.left.right = ``new` `Node(7);` `        ``reverseSpiral(root);``    ``}``}` `/* This code is contributed by PrinciRaj1992 */`

## Javascript

 ``

Output:

`8 7 4 6 9 3 5`

Time Complexity Analysis : The best case time complexity of reverseSpiral() is O(n+n/2+n/4…) = O(n) in case of a perfect binary tree. And the worst case time complexity is O(n2) [O((n)+(n-1)+(n-2)+…] in case of a skewed tree taken as input.

Auxiliary Space: O(n) for call stack since using recursion.

Iterative Implementation –
The above problem can be solved with the help of queue and stack. Just perform a normal spiral traversal, use stack to use reverse the elements alternatively. Below is the implementation of the same –

## C++14

 `#include ``using` `namespace` `std;` `// Node structure``struct` `Node {``    ``int` `data;``    ``Node* left;``    ``Node* right;``    ``Node(``int` `val)``    ``{``        ``data = val;``        ``left = right = NULL;``    ``}``};` `// function to perform``// reversal spiral traversal``void` `reverseSpiral(Node* root)``{``    ``if` `(root == NULL) {``        ``return``;``    ``}``    ``queue q;``    ``stack<``int``> s;``    ``bool` `reverse = ``true``;``    ``q.push(root);``    ``// iterate until queue is empty``    ``while` `(!q.empty()) {``        ``// calculate size of level``        ``int` `count = q.size();` `        ``// iterate over level nodes``        ``for` `(``int` `i = 0; i < count; i++) {``            ``Node* curr = q.front();``            ``q.pop();``            ``s.push(curr->data);``            ``if` `(reverse) {``                ``if` `(curr->right) {``                    ``q.push(curr->right);``                ``}``                ``if` `(curr->left) {``                    ``q.push(curr->left);``                ``}``            ``}``            ``else` `{``                ``if` `(curr->left) {``                    ``q.push(curr->left);``                ``}``                ``if` `(curr->right) {``                    ``q.push(curr->right);``                ``}``            ``}``        ``}``        ``reverse = !reverse;``    ``}``    ``// pop and print reversed nodes``    ``while` `(!s.empty()) {``        ``int` `curr = s.top();``        ``cout << curr << ``" "``;``        ``s.pop();``    ``}``}` `// Driver code``int` `main()``{``    ``Node* root = ``new` `Node(5);``    ``root->left = ``new` `Node(9);``    ``root->right = ``new` `Node(3);``    ``root->left->left = ``new` `Node(6);``    ``root->left->right = ``new` `Node(4);``    ``root->left->left->left = ``new` `Node(8);``    ``root->left->left->right = ``new` `Node(7);``    ``reverseSpiral(root);``    ``return` `0;``}`

## Java

 `// Java code for the above approach` `import` `java.io.*;``import` `java.util.*;` `class` `GFG {` `    ``static` `class` `Node {``        ``int` `data;``        ``Node left, right;``        ``Node(``int` `val)``        ``{``            ``this``.data = val;``            ``left = right = ``null``;``        ``}``    ``}` `    ``// Function to perform reversal spiral traversal``    ``static` `void` `reverseSpiral(Node root)``    ``{``        ``if` `(root == ``null``) {``            ``return``;``        ``}``        ``Queue q = ``new` `LinkedList<>();``        ``Stack s = ``new` `Stack<>();``        ``boolean` `reverse = ``true``;``        ``q.add(root);` `        ``// Iterate until queue is empty``        ``while` `(!q.isEmpty()) {``            ``// Calculate size of level``            ``int` `count = q.size();` `            ``// Iterate over level nodes``            ``for` `(``int` `i = ``0``; i < count; i++) {``                ``Node curr = q.peek();``                ``q.remove();``                ``s.push(curr.data);``                ``if` `(reverse) {``                    ``if` `(curr.right != ``null``) {``                        ``q.add(curr.right);``                    ``}``                    ``if` `(curr.left != ``null``) {``                        ``q.add(curr.left);``                    ``}``                ``}``                ``else` `{``                    ``if` `(curr.left != ``null``) {``                        ``q.add(curr.left);``                    ``}``                    ``if` `(curr.right != ``null``) {``                        ``q.add(curr.right);``                    ``}``                ``}``            ``}``            ``reverse = !reverse;``        ``}``        ``// pop and print reversed nodes``        ``while` `(!s.isEmpty()) {``            ``int` `curr = s.peek();``            ``System.out.print(curr + ``" "``);``            ``s.pop();``        ``}``    ``}` `    ``public` `static` `void` `main(String[] args)``    ``{``        ``Node root = ``new` `Node(``5``);``        ``root.left = ``new` `Node(``9``);``        ``root.right = ``new` `Node(``3``);``        ``root.left.left = ``new` `Node(``6``);``        ``root.left.right = ``new` `Node(``4``);``        ``root.left.left.left = ``new` `Node(``8``);``        ``root.left.left.right = ``new` `Node(``7``);` `        ``reverseSpiral(root);``    ``}``}` `// This code is contributed by lokeshmvs21.`

## Python3

 `from` `collections ``import` `deque` `# Node structure``class` `Node:``    ``def` `__init__(``self``, x):``        ``self``.data ``=` `x``        ``self``.right ``=` `None``        ``self``.left ``=` `None` `# function to perform``# reversal spiral traversal``def` `reverseSpiral(root):``    ``if` `(root ``=``=` `None``):``        ``return` `    ``q ``=`  `deque()``    ``s ``=` `[]``    ``reverse ``=` `True``    ``q.append(root)``    ` `    ``# iterate until queue is empty``    ``while` `(``len``(q) > ``0``):``      ` `        ``# calculate size of level``        ``count ``=` `len``(q)` `        ``# iterate over level nodes``        ``for` `i ``in` `range``(count):``            ``curr ``=` `q.popleft()``            ``s.append(curr.data)``            ``if` `(reverse):``                ``if` `(curr.right):``                    ``q.append(curr.right)``                ``if` `(curr.left):``                    ``q.append(curr.left)``            ``else``:``                ``if` `(curr.left):``                    ``q.append(curr.left)``                ``if` `(curr.right):``                    ``q.append(curr.right)` `        ``reverse ``=` `not` `reverse``        ` `    ``# pop and print reversed nodes``    ``while` `(``len``(s) > ``0``):``        ``curr ``=` `s[``-``1``]``        ``print``(curr, end ``=` `" "``)``        ``del` `s[``-``1``]` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``root ``=` `Node(``5``)``    ``root.left ``=` `Node(``9``)``    ``root.right ``=` `Node(``3``)``    ``root.left.left ``=` `Node(``6``)``    ``root.left.right ``=` `Node(``4``)``    ``root.left.left.left ``=` `Node(``8``)``    ``root.left.left.right ``=` `Node(``7``)``    ``reverseSpiral(root)` `# This code is contributed by mohit kumar 29.`

## C#

 `//C# code for the above approach``using` `System;``using` `System.Collections.Generic;` `namespace` `ConsoleApp``{``    ``class` `Program``    ``{``        ``// Structure for a binary tree node``        ``class` `Node``        ``{``            ``public` `int` `data;``            ``public` `Node left, right;``            ``public` `Node(``int` `val)``            ``{``                ``this``.data = val;``                ``left = right = ``null``;``            ``}``        ``}` `        ``// Function to perform reversal spiral traversal``        ``static` `void` `ReverseSpiral(Node root)``        ``{``            ``if` `(root == ``null``)``            ``{``                ``return``;``            ``}``            ``Queue q = ``new` `Queue();``            ``Stack<``int``> s = ``new` `Stack<``int``>();``            ``bool` `reverse = ``true``;``            ``q.Enqueue(root);` `            ``// Iterate until queue is empty``            ``while` `(q.Count > 0)``            ``{``                ``// Calculate size of level``                ``int` `count = q.Count;` `                ``// Iterate over level nodes``                ``for` `(``int` `i = 0; i < count; i++)``                ``{``                    ``Node curr = q.Peek();``                    ``q.Dequeue();``                    ``s.Push(curr.data);``                    ``if` `(reverse)``                    ``{``                        ``if` `(curr.right != ``null``)``                        ``{``                            ``q.Enqueue(curr.right);``                        ``}``                        ``if` `(curr.left != ``null``)``                        ``{``                            ``q.Enqueue(curr.left);``                        ``}``                    ``}``                    ``else``                    ``{``                        ``if` `(curr.left != ``null``)``                        ``{``                            ``q.Enqueue(curr.left);``                        ``}``                        ``if` `(curr.right != ``null``)``                        ``{``                            ``q.Enqueue(curr.right);``                        ``}``                    ``}``                ``}``                ``reverse = !reverse;``            ``}``            ``// pop and print reversed nodes``            ``while` `(s.Count > 0)``            ``{``                ``int` `curr = s.Peek();``                ``Console.Write(curr + ``" "``);``                ``s.Pop();``            ``}``        ``}` `        ``static` `void` `Main(``string``[] args)``        ``{``            ``Node root = ``new` `Node(5);``            ``root.left = ``new` `Node(9);``            ``root.right = ``new` `Node(3);``            ``root.left.left = ``new` `Node(6);``            ``root.left.right = ``new` `Node(4);``            ``root.left.left.left = ``new` `Node(8);``            ``root.left.left.right = ``new` `Node(7);` `            ``ReverseSpiral(root);``        ``}``    ``}``}``//This code is contributed by Potta Lokesh`

## Javascript

 ``

Output:

`8 7 4 6 9 3 5`

Time Complexity: O(N)
Auxiliary Space: O(N)

Improved by :HarendraSingh22

My Personal Notes arrow_drop_up