Print nodes between two given level numbers of a binary tree

• Difficulty Level : Medium
• Last Updated : 14 Jan, 2022

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 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.

C++

 // A C++ program to print Nodes level by level between given two levels.#include 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 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 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 tor 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        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 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.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



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.