Print nodes between two given level numbers of a binary tree
Given a binary tree and two level numbers ‘low’ and ‘high’, print nodes from level low to level high.
For example consider the binary tree given in below diagram. Input: Root of below tree, low = 2, high = 4 Output: 8 22 4 12 10 14
A Simple Method is to first write a recursive function that prints nodes of a given level number. Then call a recursive function in a loop from low to high. Time complexity of this method is O(n2)
We can print nodes in O(n) time using queue-based iterative level order traversal. The idea is to do simple queue-based level order traversal. While doing inorder traversal, add a marker node at the end. Whenever we see a marker node, we increase level number. If level number is between low and high, then print nodes.
The following is the implementation of above idea.
Implementation:
C++
// A C++ program to print Nodes level by level between given two levels.#include <bits/stdc++.h>using namespace std;/* A binary tree Node has key, pointer to left and right children */struct Node{ int key; struct Node* left, *right;};/* Given a binary tree, print nodes from level number 'low' to level number 'high'*/void printLevels(Node* root, int low, int high){ queue <Node *> Q; Node *marker = new Node; // Marker node to indicate end of level int level = 1; // Initialize level number // Enqueue the only first level node and marker node for end of level Q.push(root); Q.push(marker); // Simple level order traversal loop while (Q.empty() == false) { // Remove the front item from queue Node *n = Q.front(); Q.pop(); // Check if end of level is reached if (n == marker) { // print a new line and increment level number cout << endl; level++; // Check if marker node was last node in queue or // level number is beyond the given upper limit if (Q.empty() == true || level > high) break; // Enqueue the marker for end of next level Q.push(marker); // If this is marker, then we don't need print it // and enqueue its children continue; } // If level is equal to or greater than given lower level, // print it if (level >= low) cout << n->key << " "; // Enqueue children of non-marker node if (n->left != NULL) Q.push(n->left); if (n->right != NULL) Q.push(n->right); }}/* Helper function that allocates a new Node with the given key and NULL left and right pointers. */Node* newNode(int key){ Node* temp = new Node; temp->key = key; temp->left = temp->right = NULL; return (temp);}/* Driver program to test above functions*/int main(){ // Let us construct the BST shown in the above figure struct Node *root = newNode(20); root->left = newNode(8); root->right = newNode(22); root->left->left = newNode(4); root->left->right = newNode(12); root->left->right->left = newNode(10); root->left->right->right = newNode(14); cout << "Level Order traversal between given two levels is"; printLevels(root, 2, 3); return 0;} |
Java
// Java program to print Nodes level by level between given two levelsimport java.util.LinkedList;import java.util.Queue; /* A binary tree Node has key, pointer to left and right children */class Node { int data; Node left, right; public Node(int item) { data = item; left = right = null; }} class BinaryTree { Node root; /* Given a binary tree, print nodes from level number 'low' to level number 'high'*/ void printLevels(Node node, int low, int high) { Queue<Node> Q = new LinkedList<>(); Node marker = new Node(4); // Marker node to indicate end of level int level = 1; // Initialize level number // Enqueue the only first level node and marker node for end of level Q.add(node); Q.add(marker); // Simple level order traversal loop while (Q.isEmpty() == false) { // Remove the front item from queue Node n = Q.peek(); Q.remove(); // Check if end of level is reached if (n == marker) { // print a new line and increment level number System.out.println(""); level++; // Check if marker node was last node in queue or // level number is beyond the given upper limit if (Q.isEmpty() == true || level > high) break; // Enqueue the marker for end of next level Q.add(marker); // If this is marker, then we don't need print it // and enqueue its children continue; } // If level is equal to or greater than given lower level, // print it if (level >= low) System.out.print( n.data + " "); // Enqueue children of non-marker node if (n.left != null) Q.add(n.left); if (n.right != null) Q.add(n.right); } } // Driver program to test for above functions public static void main(String args[]) { BinaryTree tree = new BinaryTree(); tree.root = new Node(20); tree.root.left = new Node(8); tree.root.right = new Node(22); tree.root.left.left = new Node(4); tree.root.left.right = new Node(12); tree.root.left.right.left = new Node(10); tree.root.left.right.right = new Node(14); System.out.print("Level Order traversal between given two levels is "); tree.printLevels(tree.root, 2, 3); }} // This code has been contributed by Mayank Jaiswal |
Python3
# Python program to print nodes level by level between # given two levels# A binary tree nodeclass Node: # Constructor to create a new node def __init__(self, key): self.key = key self.left = None self.right = None # Given a binary tree, print nodes form level number 'low'# to level number 'high'def printLevels(root, low, high): Q = [] marker = Node(11114) # Marker node to indicate end of level level = 1 # Initialize level number # Enqueue the only first level node and marker node for # end of level Q.append(root) Q.append(marker) #print Q # Simple level order traversal loop while(len(Q) >0): # Remove the front item from queue n = Q[0] Q.pop(0) #print Q # Check if end of level is reached if n == marker: # print a new line and increment level number print() level += 1 # Check if marker node was last node in queue # or level number is beyond the given upper limit if len(Q) == 0 or level > high: break # Enqueue the marker for end of next level Q.append(marker) # If this is marker, then we don't need print it # and enqueue its children continue if level >= low: print (n.key,end=" ") # Enqueue children of non-marker node if n.left is not None: Q.append(n.left) Q.append(n.right)# Driver program to test the above functionroot = Node(20)root.left = Node(8)root.right = Node(22)root.left.left = Node(4)root.left.right = Node(12)root.left.right.left = Node(10)root.left.right.right = Node(14)print ("Level Order Traversal between given two levels is",printLevels(root,2,3))# This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
using System;using System.Collections.Generic;// c# program to print Nodes level by level between given two levels /* A binary tree Node has key, pointer to left and right children */public class Node{ public int data; public Node left, right; public Node(int item) { data = item; left = right = null; }}public class BinaryTree{ public Node root; /* Given a binary tree, print nodes from level number 'low' to level number 'high'*/ public virtual void printLevels(Node node, int low, int high) { LinkedList<Node> Q = new LinkedList<Node>(); Node marker = new Node(4); // Marker node to indicate end of level int level = 1; // Initialize level number // Enqueue the only first level node and marker node for end of level Q.AddLast(node); Q.AddLast(marker); // Simple level order traversal loop while (Q.Count > 0) { // Remove the front item from queue Node n = Q.First.Value; Q.RemoveFirst(); // Check if end of level is reached if (n == marker) { // print a new line and increment level number Console.WriteLine(""); level++; // Check if marker node was last node in queue or // level number is beyond the given upper limit if (Q.Count == 0 || level > high) { break; } // Enqueue the marker for end of next level Q.AddLast(marker); // If this is marker, then we don't need print it // and enqueue its children continue; } // If level is equal to or greater than given lower level, // print it if (level >= low) { Console.Write(n.data + " "); } // Enqueue children of non-marker node if (n.left != null) { Q.AddLast(n.left); } if (n.right != null) { Q.AddLast(n.right); } } } // Driver program to test for above functions public static void Main(string[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(20); tree.root.left = new Node(8); tree.root.right = new Node(22); tree.root.left.left = new Node(4); tree.root.left.right = new Node(12); tree.root.left.right.left = new Node(10); tree.root.left.right.right = new Node(14); Console.Write("Level Order traversal between given two levels is "); tree.printLevels(tree.root, 2, 3); }}// This code is contributed by Shrikant13 |
Javascript
<script> // JavaScript program to print Nodes // level by level between given two levels /* A binary tree Node has key, pointer to left and right children */class Node{ constructor(item) { this.data = item; this.left = null; this.right = null; }}var root = null;/* Given a binary tree, print nodesfrom level number 'low' to level number 'high'*/function printLevels(node, low, high){ var Q = []; var marker = new Node(4); // Marker node to indicate end of level var level = 1; // Initialize level number // Enqueue the only first level node and // marker node for end of level Q.push(node); Q.push(marker); // Simple level order traversal loop while (Q.length > 0) { // Remove the front item from queue var n = Q[0]; Q.shift(); // Check if end of level is reached if (n == marker) { // print a new line and increment level number document.write("<br>"); level++; // Check if marker node was last node in queue or // level number is beyond the given upper limit if (Q.length == 0 || level > high) { break; } // Enqueue the marker for end of next level Q.push(marker); // If this is marker, then we don't need print it // and enqueue its children continue; } // If level is equal to or greater than given lower level, // print it if (level >= low) { document.write(n.data + " "); } // Enqueue children of non-marker node if (n.left != null) { Q.push(n.left); } if (n.right != null) { Q.push(n.right); } }}// Driver program to test for above functions root = new Node(20);root.left = new Node(8);root.right = new Node(22);root.left.left = new Node(4);root.left.right = new Node(12);root.left.right.left = new Node(10);root.left.right.right = new Node(14);document.write("Level Order traversal between given two levels is ");printLevels(root, 2, 3);</script> |
Output
Level Order traversal between given two levels is 8 22 4 12
Time complexity of above method is O(n) as it does a simple level order traversal.
Auxiliary Space: O(N) where N is the number of nodes in the binary tree due to queue.
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