Reverse Level Order Traversal
We have discussed the level-order traversal of a tree 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.
The 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 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 <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 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 <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 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 nodeclass 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 nodeclass 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 traversaldef reverseLevelOrder(root): h = height(root) for i in reversed(range(1, h+1)): printGivenLevel(root,i)# Print nodes at a given leveldef 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 nodedef 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 functionroot = 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 Codestatic 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
<script> // A recursive JavaScript program to print // reverse level order traversal // A binary tree node class Node { constructor(item) { this.data = item; this.left = null; this.right = null; } } class BinaryTree { constructor() { this.root = null; } /* Function to print REVERSE level order traversal a tree*/ reverseLevelOrder(node) { var h = this.height(node); var i; for ( i = h; i >= 1; i-- // THE ONLY LINE DIFFERENT // FROM NORMAL LEVEL ORDER ) { this.printGivenLevel(node, i); } } /* Print nodes at a given level */ printGivenLevel(node, level) { if (node == null) return; if (level == 1) document.write(node.data + " "); else if (level > 1) { this.printGivenLevel(node.left, level - 1); this.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.*/ height(node) { if (node == null) return 0; else { /* compute the height of each subtree */ var lheight = this.height(node.left); var rheight = this.height(node.right); /* use the larger one */ if (lheight > rheight) return lheight + 1; else return rheight + 1; } } } // Driver Code var 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); document.write( "Level Order traversal " + "of binary tree is : <br>" ); tree.reverseLevelOrder(tree.root); </script> |
Output
Level Order traversal of binary tree is 4 5 2 3 1
Time Complexity: O(n^2)
Auxiliary Space: O(h), where h is the height of the tree, this space is due to the recursive call stack.
METHOD 2 (Using Queue and Stack)
The idea is to use a deque(double-ended queue) to get the reverse level order. A deque allows insertion and deletion at both 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 the 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#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;} |
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<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 |
Python
# Python program to print REVERSE level order traversal using# stack and queuefrom collections import deque# A binary tree nodeclass 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 orderdef 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 functionroot = 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"deq = reverseLevelOrder(root)for key in deq: print (key),# This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
// A recursive C# program to print reverse // level order traversal using stack and queueusing 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 */ |
Javascript
<script>// A recursive javascript program to print// reverse level order traversal// using stack and queue/* A binary tree node has data, pointer to leftand right children */class Node{ constructor(item) { this.data = item; this.left = this.right=null; }}let root;/* Given a binary tree, print its nodes in reverse level order */function reverseLevelOrder(node){ let S = []; let Q = []; Q.push(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.length != 0) { /* Dequeue node and make it root */ node = Q[0]; Q.shift(); S.push(node); /* Enqueue right child */ if (node.right != null) // NOTE: RIGHT CHILD IS ENQUEUED BEFORE LEFT Q.push(node.right); /* Enqueue left child */ if (node.left != null) Q.push(node.left); } // Now pop all items from stack // one by one and print them while (S.length != 0) { node = S.pop(); document.write(node.data + " "); }}// Driver program to test above functions// Let us create trees shown in above diagramroot = new Node(1);root.left = new Node(2);root.right = new Node(3);root.left.left = new Node(4);root.left.right = new Node(5);root.right.left = new Node(6);root.right.right = new Node(7);document.write("Level Order traversal of binary tree is :<br>");reverseLevelOrder(root);// This code is contributed by avanitrachhadiya2155</script> |
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.
Auxiliary Space: O(n), for stack and queue.
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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