Reverse Level Order Traversal

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.
 

Example Tree

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++

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// A recursive C++ program to print
// REVERSE level order traversal
#include <bits/stdc++.h>
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 protoypes*/
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

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C

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// A recursive C program to print REVERSE level order traversal
#include <stdio.h>
#include <stdlib.h>
 
/* A binary tree node has data, pointer to left and right child */
struct node
{
    int data;
    struct node* left;
    struct node* right;
};
 
/*Function protoypes*/
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;
}

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Java

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// 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

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Python

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# 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)

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C#

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// 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

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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++

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// A C++ program to print REVERSE level order traversal using stack and queue
// This approach is adopted from following link
#include <bits/stdc++.h>
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 <node *> S;
    queue <node *> 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;
}

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Java

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// 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<Node> S = new Stack();
        Queue<Node> 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

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Python

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# 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)

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C#

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// 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<Node> S = new Stack<Node>();
        Queue<Node> Q = new Queue<Node>();
        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 */

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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.
https://www.youtube.com/watch?v=FfF0FubRtYs
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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