# Reverse Level Order Traversal

• Difficulty Level : Easy
• Last Updated : 23 Aug, 2021

We have discussed the level order traversal of a post in the previous post. The idea is to print the last level first, then the second last level, and so on. Like Level order traversal, every level is printed from left to right. Become a success story instead of just reading about them. Prepare for coding interviews at Amazon and other top product-based companies with our Amazon Test Series. Includes topic-wise practice questions on all important DSA topics along with 10 practice contests of 2 hours each. Designed by industry experts that will surely help you practice and sharpen your programming skills. Wait no more, start your preparation today!

Reverse Level order traversal of the above tree is “4 5 2 3 1”.
Both methods for normal level order traversal can be easily modified to do reverse level order traversal.

METHOD 1 (Recursive function to print a given level)
We can easily modify the method 1 of the normal level order traversal. In method 1, we have a method printGivenLevel() which prints a given level number. The only thing we need to change is, instead of calling printGivenLevel() from the first level to the last level, we call it from the last level to the first level.

## C++

 `// A recursive C++ program to print``// REVERSE level order traversal``#include ``using` `namespace` `std;` `/* A binary tree node has data,``pointer to left and right child */``class` `node``{``    ``public``:``    ``int` `data;``    ``node* left;``    ``node* right;``};` `/*Function prototypes*/``void` `printGivenLevel(node* root, ``int` `level);``int` `height(node* node);``node* newNode(``int` `data);` `/* Function to print REVERSE``level order traversal a tree*/``void` `reverseLevelOrder(node* root)``{``    ``int` `h = height(root);``    ``int` `i;``    ``for` `(i=h; i>=1; i--) ``//THE ONLY LINE DIFFERENT FROM NORMAL LEVEL ORDER``        ``printGivenLevel(root, i);``}` `/* Print nodes at a given level */``void` `printGivenLevel(node* root, ``int` `level)``{``    ``if` `(root == NULL)``        ``return``;``    ``if` `(level == 1)``        ``cout << root->data << ``" "``;``    ``else` `if` `(level > 1)``    ``{``        ``printGivenLevel(root->left, level - 1);``        ``printGivenLevel(root->right, level - 1);``    ``}``}` `/* Compute the "height" of a tree -- the number of``    ``nodes along the longest path from the root node``    ``down to the farthest leaf node.*/``int` `height(node* node)``{``    ``if` `(node == NULL)``        ``return` `0;``    ``else``    ``{``        ``/* compute the height of each subtree */``        ``int` `lheight = height(node->left);``        ``int` `rheight = height(node->right);` `        ``/* use the larger one */``        ``if` `(lheight > rheight)``            ``return``(lheight + 1);``        ``else` `return``(rheight + 1);``    ``}``}` `/* Helper function that allocates a new node with the``given data and NULL left and right pointers. */``node* newNode(``int` `data)``{``    ``node* Node = ``new` `node();``    ``Node->data = data;``    ``Node->left = NULL;``    ``Node->right = NULL;` `    ``return``(Node);``}` `/* Driver code*/``int` `main()``{``    ``node *root = newNode(1);``    ``root->left = newNode(2);``    ``root->right = newNode(3);``    ``root->left->left = newNode(4);``    ``root->left->right = newNode(5);` `    ``cout << ``"Level Order traversal of binary tree is \n"``;``    ``reverseLevelOrder(root);` `    ``return` `0;` `}` `// This code is contributed by rathbhupendra`

## C

 `// A recursive C program to print REVERSE level order traversal``#include ``#include ` `/* A binary tree node has data, pointer to left and right child */``struct` `node``{``    ``int` `data;``    ``struct` `node* left;``    ``struct` `node* right;``};` `/*Function prototypes*/``void` `printGivenLevel(``struct` `node* root, ``int` `level);``int` `height(``struct` `node* node);``struct` `node* newNode(``int` `data);` `/* Function to print REVERSE level order traversal a tree*/``void` `reverseLevelOrder(``struct` `node* root)``{``    ``int` `h = height(root);``    ``int` `i;``    ``for` `(i=h; i>=1; i--) ``//THE ONLY LINE DIFFERENT FROM NORMAL LEVEL ORDER``        ``printGivenLevel(root, i);``}` `/* Print nodes at a given level */``void` `printGivenLevel(``struct` `node* root, ``int` `level)``{``    ``if` `(root == NULL)``        ``return``;``    ``if` `(level == 1)``        ``printf``(``"%d "``, root->data);``    ``else` `if` `(level > 1)``    ``{``        ``printGivenLevel(root->left, level-1);``        ``printGivenLevel(root->right, level-1);``    ``}``}` `/* Compute the "height" of a tree -- the number of``    ``nodes along the longest path from the root node``    ``down to the farthest leaf node.*/``int` `height(``struct` `node* node)``{``    ``if` `(node==NULL)``        ``return` `0;``    ``else``    ``{``        ``/* compute the height of each subtree */``        ``int` `lheight = height(node->left);``        ``int` `rheight = height(node->right);` `        ``/* use the larger one */``        ``if` `(lheight > rheight)``            ``return``(lheight+1);``        ``else` `return``(rheight+1);``    ``}``}` `/* Helper function that allocates a new node with the``   ``given data and NULL left and right pointers. */``struct` `node* newNode(``int` `data)``{``    ``struct` `node* node = (``struct` `node*)``                        ``malloc``(``sizeof``(``struct` `node));``    ``node->data = data;``    ``node->left = NULL;``    ``node->right = NULL;` `    ``return``(node);``}` `/* Driver program to test above functions*/``int` `main()``{``    ``struct` `node *root = newNode(1);``    ``root->left        = newNode(2);``    ``root->right       = newNode(3);``    ``root->left->left  = newNode(4);``    ``root->left->right = newNode(5);` `    ``printf``(``"Level Order traversal of binary tree is \n"``);``    ``reverseLevelOrder(root);` `    ``return` `0;``}`

## Java

 `// A recursive java program to print reverse level order traversal`` ` `// A binary tree node``class` `Node``{``    ``int` `data;``    ``Node left, right;``     ` `    ``Node(``int` `item)``    ``{``        ``data = item;``        ``left = right;``    ``}``}`` ` `class` `BinaryTree``{``    ``Node root;`` ` `    ``/* Function to print REVERSE level order traversal a tree*/``    ``void` `reverseLevelOrder(Node node)``    ``{``        ``int` `h = height(node);``        ``int` `i;``        ``for` `(i = h; i >= ``1``; i--)``        ``//THE ONLY LINE DIFFERENT FROM NORMAL LEVEL ORDER``        ``{``            ``printGivenLevel(node, i);``        ``}``    ``}`` ` `    ``/* Print nodes at a given level */``    ``void` `printGivenLevel(Node node, ``int` `level)``    ``{``        ``if` `(node == ``null``)``            ``return``;``        ``if` `(level == ``1``)``            ``System.out.print(node.data + ``" "``);``        ``else` `if` `(level > ``1``)``        ``{``            ``printGivenLevel(node.left, level - ``1``);``            ``printGivenLevel(node.right, level - ``1``);``        ``}``    ``}`` ` `    ``/* Compute the "height" of a tree -- the number of``     ``nodes along the longest path from the root node``     ``down to the farthest leaf node.*/``    ``int` `height(Node node)``    ``{``        ``if` `(node == ``null``)``            ``return` `0``;``        ``else``        ``{``            ``/* compute the height of each subtree */``            ``int` `lheight = height(node.left);``            ``int` `rheight = height(node.right);`` ` `            ``/* use the larger one */``            ``if` `(lheight > rheight)``                ``return` `(lheight + ``1``);``            ``else``                ``return` `(rheight + ``1``);``        ``}``    ``}`` ` `    ``// Driver program to test above functions``    ``public` `static` `void` `main(String args[])``    ``{``        ``BinaryTree tree = ``new` `BinaryTree();`` ` `        ``// Let us create trees shown in above diagram``        ``tree.root = ``new` `Node(``1``);``        ``tree.root.left = ``new` `Node(``2``);``        ``tree.root.right = ``new` `Node(``3``);``        ``tree.root.left.left = ``new` `Node(``4``);``        ``tree.root.left.right = ``new` `Node(``5``);``         ` `        ``System.out.println(``"Level Order traversal of binary tree is : "``);``        ``tree.reverseLevelOrder(tree.root);``    ``}``}`` ` `// This code has been contributed by Mayank Jaiswal`

## Python

 `# A recursive Python program to print REVERSE level order traversal` `# A binary tree node``class` `Node:` `    ``# Constructor to create a new node``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# Function to print reverse level order traversal``def` `reverseLevelOrder(root):``    ``h ``=` `height(root)``    ``for` `i ``in` `reversed``(``range``(``1``, h``+``1``)):``        ``printGivenLevel(root,i)` `# Print nodes at a given level``def` `printGivenLevel(root, level):` `    ``if` `root ``is` `None``:``        ``return``    ``if` `level ``=``=``1` `:``        ``print` `root.data,` `    ``elif` `level>``1``:``        ``printGivenLevel(root.left, level``-``1``)``        ``printGivenLevel(root.right, level``-``1``)` `# Compute the height of a tree-- the number of``# nodes along the longest path from the root node``# down to the farthest leaf node``def` `height(node):``    ``if` `node ``is` `None``:``        ``return` `0``    ``else``:` `        ``# Compute the height of each subtree``        ``lheight ``=` `height(node.left)``        ``rheight ``=` `height(node.right)` `        ``# Use the larger one``        ``if` `lheight > rheight :``            ``return` `lheight ``+` `1``        ``else``:``            ``return` `rheight ``+` `1` `# Driver program to test above function``root ``=` `Node(``1``)``root.left ``=` `Node(``2``)``root.right ``=` `Node(``3``)``root.left.left ``=` `Node(``4``)``root.left.right ``=` `Node(``5``)` `print` `"Level Order traversal of binary tree is"``reverseLevelOrder(root)` `# This code is contributed by Nikhil Kumar Singh(nickzuck_007)`

## C#

 `// A recursive C# program to print``// reverse level order traversal``using` `System;` `// A binary tree node``class` `Node``{``    ``public` `int` `data;``    ``public` `Node left, right;``        ` `    ``public` `Node(``int` `item)``    ``{``        ``data = item;``        ``left = right;``    ``}``}``    ` `class` `BinaryTree``{``Node root;` `/* Function to print REVERSE``level order traversal a tree*/``void` `reverseLevelOrder(Node node)``{``    ``int` `h = height(node);``    ``int` `i;``    ``for` `(i = h; i >= 1; i--)``    ` `    ``// THE ONLY LINE DIFFERENT``    ``// FROM NORMAL LEVEL ORDER``    ``{``        ``printGivenLevel(node, i);``    ``}``}` `/* Print nodes at a given level */``void` `printGivenLevel(Node node, ``int` `level)``{``    ``if` `(node == ``null``)``        ``return``;``    ``if` `(level == 1)``        ``Console.Write(node.data + ``" "``);``    ``else` `if` `(level > 1)``    ``{``        ``printGivenLevel(node.left, level - 1);``        ``printGivenLevel(node.right, level - 1);``    ``}``}` `/* Compute the "height" of a tree --``the number of nodes along the longest``path from the root node down to the``farthest leaf node.*/``int` `height(Node node)``{``    ``if` `(node == ``null``)``        ``return` `0;``    ``else``    ``{``        ``/* compute the height of each subtree */``        ``int` `lheight = height(node.left);``        ``int` `rheight = height(node.right);` `        ``/* use the larger one */``        ``if` `(lheight > rheight)``            ``return` `(lheight + 1);``        ``else``            ``return` `(rheight + 1);``    ``}``}` `// Driver Code``static` `public` `void` `Main(String []args)``{``    ``BinaryTree tree = ``new` `BinaryTree();` `    ``// Let us create trees shown``    ``// in above diagram``    ``tree.root = ``new` `Node(1);``    ``tree.root.left = ``new` `Node(2);``    ``tree.root.right = ``new` `Node(3);``    ``tree.root.left.left = ``new` `Node(4);``    ``tree.root.left.right = ``new` `Node(5);``        ` `    ``Console.WriteLine(``"Level Order traversal "` `+``                        ``"of binary tree is : "``);``    ``tree.reverseLevelOrder(tree.root);``}``}``    ` `// This code is contributed``// by Arnab Kundu`

## Javascript

 ``

Output:

```Level Order traversal of binary tree is
4 5 2 3 1```

Time Complexity: The worst-case time complexity of this method is O(n^2). For a skewed tree, printGivenLevel() takes O(n) time where n is the number of nodes in the skewed tree. So time complexity of printLevelOrder() is O(n) + O(n-1) + O(n-2) + .. + O(1) which is O(n^2).
METHOD 2 (Using Queue and Stack)
The method 2 of normal level order traversal can also be easily modified to print level order traversal in reverse order. The idea is to use a deque(double-ended queue) to get the reverse level order. A deque allows insertion and deletion at both the ends. If we do normal level order traversal and instead of printing a node, push the node to a stack and then print the contents of the deque, we get “5 4 3 2 1” for the above example tree, but the output should be “4 5 2 3 1”. So to get the correct sequence (left to right at every level), we process children of a node in reverse order, we first push the right subtree to the deque, then process the left subtree.

## C++

 `// A C++ program to print REVERSE level order traversal using stack and queue``// This approach is adopted from following link``// http://tech-queries.blogspot.in/2008/12/level-order-tree-traversal-in-reverse.html``#include ``using` `namespace` `std;` `/* A binary tree node has data, pointer to left and right children */``struct` `node``{``    ``int` `data;``    ``struct` `node* left;``    ``struct` `node* right;``};` `/* Given a binary tree, print its nodes in reverse level order */``void` `reverseLevelOrder(node* root)``{``    ``stack S;``    ``queue Q;``    ``Q.push(root);` `    ``// Do something like normal level order traversal order. Following are the``    ``// differences with normal level order traversal``    ``// 1) Instead of printing a node, we push the node to stack``    ``// 2) Right subtree is visited before left subtree``    ``while` `(Q.empty() == ``false``)``    ``{``        ``/* Dequeue node and make it root */``        ``root = Q.front();``        ``Q.pop();``        ``S.push(root);` `        ``/* Enqueue right child */``        ``if` `(root->right)``            ``Q.push(root->right); ``// NOTE: RIGHT CHILD IS ENQUEUED BEFORE LEFT` `        ``/* Enqueue left child */``        ``if` `(root->left)``            ``Q.push(root->left);``    ``}` `    ``// Now pop all items from stack one by one and print them``    ``while` `(S.empty() == ``false``)``    ``{``        ``root = S.top();``        ``cout << root->data << ``" "``;``        ``S.pop();``    ``}``}` `/* Helper function that allocates a new node with the``   ``given data and NULL left and right pointers. */``node* newNode(``int` `data)``{``    ``node* temp = ``new` `node;``    ``temp->data = data;``    ``temp->left = NULL;``    ``temp->right = NULL;` `    ``return` `(temp);``}` `/* Driver program to test above functions*/``int` `main()``{``    ``struct` `node *root = newNode(1);``    ``root->left        = newNode(2);``    ``root->right       = newNode(3);``    ``root->left->left  = newNode(4);``    ``root->left->right = newNode(5);``    ``root->right->left  = newNode(6);``    ``root->right->right = newNode(7);` `    ``cout << ``"Level Order traversal of binary tree is \n"``;``    ``reverseLevelOrder(root);` `    ``return` `0;``}`

## Java

 `// A recursive java program to print reverse level order traversal``// using stack and queue`` ` `import` `java.util.LinkedList;``import` `java.util.Queue;``import` `java.util.Stack;`` ` `/* A binary tree node has data, pointer to left and right children */``class` `Node``{``    ``int` `data;``    ``Node left, right;`` ` `    ``Node(``int` `item)``    ``{``        ``data = item;``        ``left = right;``    ``}``}`` ` `class` `BinaryTree``{``    ``Node root;`` ` `    ``/* Given a binary tree, print its nodes in reverse level order */``    ``void` `reverseLevelOrder(Node node)``    ``{``        ``Stack S = ``new` `Stack();``        ``Queue Q = ``new` `LinkedList();``        ``Q.add(node);`` ` `        ``// Do something like normal level order traversal order.Following``        ``// are the differences with normal level order traversal``        ``// 1) Instead of printing a node, we push the node to stack``        ``// 2) Right subtree is visited before left subtree``        ``while` `(Q.isEmpty() == ``false``)``        ``{``            ``/* Dequeue node and make it root */``            ``node = Q.peek();``            ``Q.remove();``            ``S.push(node);`` ` `            ``/* Enqueue right child */``            ``if` `(node.right != ``null``)``                ``// NOTE: RIGHT CHILD IS ENQUEUED BEFORE LEFT``                ``Q.add(node.right);``                ` `            ``/* Enqueue left child */``            ``if` `(node.left != ``null``)``                ``Q.add(node.left);``        ``}`` ` `        ``// Now pop all items from stack one by one and print them``        ``while` `(S.empty() == ``false``)``        ``{``            ``node = S.peek();``            ``System.out.print(node.data + ``" "``);``            ``S.pop();``        ``}``    ``}`` ` `    ``// Driver program to test above functions``    ``public` `static` `void` `main(String args[])``    ``{``        ``BinaryTree tree = ``new` `BinaryTree();`` ` `        ``// Let us create trees shown in above diagram``        ``tree.root = ``new` `Node(``1``);``        ``tree.root.left = ``new` `Node(``2``);``        ``tree.root.right = ``new` `Node(``3``);``        ``tree.root.left.left = ``new` `Node(``4``);``        ``tree.root.left.right = ``new` `Node(``5``);``        ``tree.root.right.left = ``new` `Node(``6``);``        ``tree.root.right.right = ``new` `Node(``7``);`` ` `        ``System.out.println(``"Level Order traversal of binary tree is :"``);``        ``tree.reverseLevelOrder(tree.root);`` ` `    ``}``}`` ` `// This code has been contributed by Mayank Jaiswal`

## Python

 `# Python program to print REVERSE level order traversal using``# stack and queue` `from` `collections ``import` `deque``# A binary tree node``class` `Node:` `    ``# Constructor to create a new node``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.left ``=` `None``        ``self``.right ``=` `None`  `# Given a binary tree, print its nodes in reverse level order`  `def` `reverseLevelOrder(root):``  ``# we can use a double ended queue which provides O(1) insert at the beginning``  ``# using the appendleft method``  ``# we do the regular level order traversal but instead of processing the``  ``# left child first we process the right child first and the we process the left child``  ``# of the current Node``  ``# we can do this One pass reduce the space usage not in terms of complexity but intuitively``  ` `    ``q ``=` `deque()``    ``q.append(root)``    ``ans ``=` `deque()``    ``while` `q:``        ``node ``=` `q.popleft()``        ``if` `node ``is` `None``:``            ``continue``        ` `        ``ans.appendleft(node.data)``        ` `        ``if` `node.right:``            ``q.append(node.right)``            ` `        ``if` `node.left:``            ``q.append(node.left)``            ` `    ``return` `ans` `# Driver program to test above function``root ``=` `Node(``1``)``root.left ``=` `Node(``2``)``root.right ``=` `Node(``3``)``root.left.left ``=` `Node(``4``)``root.left.right ``=` `Node(``5``)``root.right.left ``=` `Node(``6``)``root.right.right ``=` `Node(``7``)` `print` `"Level Order traversal of binary tree is"``reverseLevelOrder(root)` `# This code is contributed by Nikhil Kumar Singh(nickzuck_007)`

## C#

 `// A recursive C# program to print reverse``// level order traversal using stack and queue``using` `System.Collections.Generic;``using` `System;` `/* A binary tree node has data,``pointer to left and right children */``public` `class` `Node``{``    ``public` `int` `data;``    ``public` `Node left, right;` `    ``public` `Node(``int` `item)``    ``{``        ``data = item;``        ``left = right;``    ``}``}` `public` `class` `BinaryTree``{``    ``Node root;` `    ``/* Given a binary tree, print its``    ``nodes in reverse level order */``    ``void` `reverseLevelOrder(Node node)``    ``{``        ``Stack S = ``new` `Stack();``        ``Queue Q = ``new` `Queue();``        ``Q.Enqueue(node);` `        ``// Do something like normal level``        ``// order traversal order.Following``        ``// are the differences with normal``        ``// level order traversal``        ``// 1) Instead of printing a node, we push the node to stack``        ``// 2) Right subtree is visited before left subtree``        ``while` `(Q.Count>0)``        ``{``            ``/* Dequeue node and make it root */``            ``node = Q.Peek();``            ``Q.Dequeue();``            ``S.Push(node);` `            ``/* Enqueue right child */``            ``if` `(node.right != ``null``)``                ``// NOTE: RIGHT CHILD IS ENQUEUED BEFORE LEFT``                ``Q.Enqueue(node.right);``                ` `            ``/* Enqueue left child */``            ``if` `(node.left != ``null``)``                ``Q.Enqueue(node.left);``        ``}` `        ``// Now pop all items from stack``        ``// one by one and print them``        ``while` `(S.Count>0)``        ``{``            ``node = S.Peek();``            ``Console.Write(node.data + ``" "``);``            ``S.Pop();``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``BinaryTree tree = ``new` `BinaryTree();` `        ``// Let us create trees shown in above diagram``        ``tree.root = ``new` `Node(1);``        ``tree.root.left = ``new` `Node(2);``        ``tree.root.right = ``new` `Node(3);``        ``tree.root.left.left = ``new` `Node(4);``        ``tree.root.left.right = ``new` `Node(5);``        ``tree.root.right.left = ``new` `Node(6);``        ``tree.root.right.right = ``new` `Node(7);` `        ``Console.WriteLine(``"Level Order traversal of binary tree is :"``);``        ``tree.reverseLevelOrder(tree.root);` `    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## Javascript

 ``

Output:

```Level Order traversal of binary tree is
4 5 6 7 2 3 1```

Time Complexity: O(n) where n is the number of nodes in the binary tree.

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Please write comments if you find any bug in the above programs/algorithms or other ways to solve the same problem.

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