Level order traversal of Binary Tree using Morris Traversal
Given a binary tree, the task is to traverse the Binary Tree in level order fashion.
Examples:
Input: 1 / \ 2 3 Output: 1 2 3 Input: 5 / \ 2 3 \ 6 Output: 5 2 3 6
Approach: The idea is to use Morris Preorder Traversal to traverse the tree in level order traversal.
Observations: There are mainly two observations for the traversal of the tree using Morris preorder traversal. That is –
- In Preorder traversal the left-most nodes of a level are visited first due which it can be used to traverse the tree in level order fashion.
- As we maintain the horizontal distance of the nodes in the top view of the binary tree, Similarly If we maintain the level of the current node and increment or decrement the level accordingly as per the movement, Then the nodes can be traversed easily.
As in the Morris preorder traversal, we connect the right-most node of the left child to its inorder successor to maintain the movement such that we can traverse back to the right child of the parent node after completely exploring the left child of the parent. Therefore, while moving to the rightmost child of the left child we can keep track of the number of the increment in the level to compute the level inorder successor of the child.
Below is the explanation of the approach with the help of example:
Below is the implementation of the above approach:
C++
// C++ implementation of the level // order traversal using Morris traversal #include<bits/stdc++.h> using namespace std; // structure of the node of the binary tree struct Node{ int data; Node* left; Node* right; Node( int data){ this ->data = data; this ->right = NULL; this ->left = NULL; } }; vector<pair<Node*, int >> traversal; // function to traverse the binary // tree in the level order fashion void levelOrderTraversal(Node* root) { // current node is marked as the root node Node* curr = root; int level = 0; // loop to traverse the binary tree until the current node // is not null while (curr != NULL) { // if left child is null, print the // current node data and update the // current pointer to right child. if (curr->left == NULL) { // return the current node with // its level traversal.push_back({curr, level}); curr = curr->right; if (curr != NULL){ level += 1; } else { level -= 1; } } else { // find the inorder predecessor Node* prev = curr->left; int toUp = 0; // loop to find the right most // node of the left child of the current node while (prev->right != NULL && prev->right != curr){ prev = prev->right; toUp += 1; } // If the right child of inorder // predecessor already points to // the current node, update the // current with it's right child if (prev->right == curr){ prev->right = NULL; curr = curr->right; level -= toUp + 1; } // else If right child doesn't // point to the current node, // then print this node's data // and update the right child // pointer with the current node // and update the current with // it's left child else { traversal.push_back({curr, level}); prev->right = curr; curr = curr->left; level += 1; } } } } int main() { // create a binary tree Node* root = new Node(5); root->left = new Node(2); root->right = new Node(3); root->left->right = new Node(6); // traverse the tree in level order traversal levelOrderTraversal(root); // find the height of the tree int h = 0; for ( auto i : traversal){ h = max(h, i.second+1); } // print the date of nodes at each level for ( int i = 0; i<h; i++){ for ( auto j : traversal){ if (j.second == i){ cout<<j.first->data<< " " ; } } cout<<endl; } return 0; } // This code is contributed by Yash Agarwal(yashagarwal2852002) |
Python3
# Python implementation of the Level # order traversal using Morris traversal # Class of the node of the # Binary Tree class Node: def __init__( self , data): self .data = data self .left = None self .right = None # Function to traverse the Binary # tree in the Level Order Fashion def levelOrderTraversal(root): # Current Node is marked as # the Root Node curr = root level = 0 # Loop to traverse the Binary # Tree until the current node # is not Null while curr: # If left child is null, print the # current node data. And, update # the current pointer to right child. if curr.left is None : # Return the current node with # its level yield [curr, level] curr = curr.right if curr: level + = 1 else : level - = 1 else : # Find the inorder predecessor prev = curr.left to_up = 0 # Loop to find the right most # node of the left child of the # current node while prev.right is not None and \ prev.right is not curr: prev = prev.right to_up + = 1 # If the right child of inorder # predecessor already points to # the current node, update the # current with it's right child if prev.right is curr: prev.right = None curr = curr.right level - = to_up + 1 # else If right child doesn't # point to the current node, # then print this node's data # and update the right child # pointer with the current node # and update the current with # it's left child else : yield [curr, level] prev.right = curr curr = curr.left level + = 1 # Driver Code if __name__ = = "__main__" : root = Node( 5 ) root.left = Node( 2 ) root.right = Node( 3 ) root.left.right = Node( 6 ) # Output List to store the # Level Order Traversed Nodes outputData = [[] for i in range ( 100 )] for node, level in levelOrderTraversal(root): outputData[level].append(node.data) h = 0 # Loop to find the height of the # Binary Tree for i in outputData: if i: h + = 1 else : break # Loop to print the Data for i in range (h): print ( ' ' .join( map ( str , outputData[i]))) |
C#
// C# program for the above approach using System; using System.Collections.Generic; // class to represent tree node public class Node{ public int data; public Node left, right; public Node( int item){ data = item; left = null ; right = null ; } } // class to print desired output public class BinaryTree{ Node root; List<Tuple<Node, int >> traversal = new List<Tuple<Node, int >>(); void levelOrderTraversal(){ // current node is marked as the root node Node curr = root; int level = 0; // loop to traverse the binary tree until the current node while (curr != null ){ // if left child is null, print the // current node data and update the // current pointer to right child. if (curr.left == null ){ // return the current node with // its level traversal.Add( new Tuple<Node, int >(curr, level)); curr = curr.right; if (curr != null ){ level += 1; } else { level -= 1; } } else { // find the inorder predecessor Node prev = curr.left; int toUp = 0; // loop to find the right most // node of the left child of the current node while (prev.right != null && prev.right != curr){ prev = prev.right; toUp += 1; } // If the right child of inorder // predecessor already points to // the current node, update the // current with it's right child if (prev.right == curr){ prev.right = null ; curr = curr.right; level -= toUp + 1; } // else If right child doesn't // point to the current node, // then print this node's data // and update the right child // pointer with the current node // and update the current with // it's left child else { traversal.Add( new Tuple<Node, int >(curr, level)); prev.right = curr; curr = curr.left; level += 1; } } } } public static void Main(){ /* creating a binary tree and entering the nodes */ BinaryTree tree = new BinaryTree(); tree.root = new Node(5); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.right = new Node(6); // traverse the tree in level order traversal tree.levelOrderTraversal(); // find the height of the tree int h = 0; foreach ( var i in tree.traversal){ h = Math.Max(h, i.Item2 + 1); } // print the date of nodes at each level for ( int i = 0; i<h; i++){ foreach ( var j in tree.traversal){ if (j.Item2 == i){ Console.Write(j.Item1.data + " " ); } } Console.WriteLine( " " ); } } } // THIS CODE IS CONTRIBUTED BY KIRIT AGARWAL(KIRITAGARWAL23121999) |
Javascript
// JavaScript code for the above approach class Node { constructor(data) { this .data = data; this .left = null ; this .right = null ; } } // Function to traverse the Binary // tree in the Level Order Fashion function * levelOrderTraversal(root) { // Current Node is marked as // the Root Node let curr = root; let level = 0; // Loop to traverse the Binary // Tree until the current node // is not Null while (curr) { // If left child is null, print the // current node data. And, update // the current pointer to right child. if (!curr.left) { // Return the current node with // its level yield [curr, level]; curr = curr.right; if (curr) { level += 1; } else { level -= 1; } } else { // Find the inorder predecessor let prev = curr.left; let toUp = 0; // Loop to find the right most // node of the left child of the // current node while (prev.right && prev.right !== curr) { prev = prev.right; toUp += 1; } // If the right child of inorder // predecessor already points to // the current node, update the // current with it's right child if (prev.right === curr) { prev.right = null ; curr = curr.right; level -= toUp + 1; } // else If right child doesn't // point to the current node, // then print this node's data // and update the right child // pointer with the current node // and update the current with // it's left child else { yield [curr, level]; prev.right = curr; curr = curr.left; level += 1; } } } } // Create a binary tree const root = new Node(5); root.left = new Node(2); root.right = new Node(3); root.left.right = new Node(6); // Traverse the tree in level order const traversal = [...levelOrderTraversal(root)]; // Find the height of the tree let h = 0; for (const [node, level] of traversal) { h = Math.max(h, level + 1); } // Print the data of the nodes at each level for (let i = 0; i < h; i++) { for (const [node, level] of traversal) { if (level === i) { console.log(node.data+ " " ); } } console.log( "<br>" ); } // This code is contributed by Potta Lokesh. |
Java
// JAVA implementation of the level // order traversal using Morris traversal import java.util.*; // structure of the node of the binary tree class Node{ int data; Node left; Node right; public Node( int data) { this .data=data; } } public class Main { static ArrayList<Object[]> traversal; // function to traverse the binary // tree in the level order fashion public static void levelOrderTraversal(Node root) { // current node is marked as the root node Node curr=root; int level= 0 ; // loop to traverse the binary tree until the current node // is not null while (curr!= null ) { // if left child is null, print the // current node data and update the // current pointer to right child. if (curr.left== null ) { // return the current node with // its level traversal.add( new Object[]{curr, new Integer(level)}); curr=curr.right; if (curr!= null ) { level++; } else level--; } else { // find the inorder predecessor Node prev=curr.left; int toUp= 0 ; // loop to find the right most // node of the left child of the current node while (prev.right!= null && prev.right!=curr) { prev=prev.right; toUp++; } // If the right child of inorder // predecessor already points to // the current node, update the // current with it's right child if (prev.right==curr) { prev.right= null ; curr=curr.right; level-=toUp+ 1 ; } // else If right child doesn't // point to the current node, // then print this node's data // and update the right child // pointer with the current node // and update the current with // it's left child else { traversal.add( new Object[]{curr, new Integer(level)}); prev.right=curr; curr=curr.left; level++; } } } } public static void main(String[] args) { // create a binary tree Node root= new Node( 5 ); root.left= new Node( 2 ); root.right= new Node( 3 ); root.left.right= new Node( 6 ); traversal= new ArrayList<>(); // traverse the tree in level order traversal levelOrderTraversal(root); // find the height of the tree int h= 0 ; for (Object[] i:traversal) { h=Math.max(h,( int )i[ 1 ]+ 1 ); } // print the date of nodes at each level for ( int i = 0 ; i<h; i++){ for (Object[] j : traversal){ if (( int )j[ 1 ] == i){ System.out.print(((Node)j[ 0 ]).data+ " " ); } } System.out.println(); } } } |
5 2 3 6
Performance Analysis:
- Time Complexity: As in the above approach, every node is touched at max twice due to which the time complexity is O(N), where N is the number of nodes.
- Auxiliary Space: As in the above approach, there is no extra space used due to which auxiliary space used will be O(1)
Contact Us