Print all nodes between two given levels in Binary Tree
Given a binary tree, print all nodes between two given levels in a binary tree. Print the nodes level wise, i.e., the nodes for any level should be printed from left to right.
In the above tree, if the starting level is 2 and the ending level is 3 then the solution should print:
2 3 4 5 6 7
Note: Level number starts with 1. That is, the root node is at level 1.
Prerequisite: Level order Traversal.
The idea is to do level order traversal of the tree using a queue and keep track of the current level. If the current level lies between the starting and ending level then print the nodes at that level.
Algorithm:
levelordertraverse (root, startLevel, endLevel)
q -> empty queue
q.enqueue (root)
level -> 0
while (not q.isEmpty())
size -> q.size()
level = level + 1
while (size)
node -> q.dequeue()
if (level between startLevel and endevel)
print (node)
if(node.leftnull)
q.enqueue (node. left)
if(node.leftnull)
q.enqueue(node.right)
size =size -1Below is the implementation of the above algorithm:
C++
// C++ program for Print all nodes// between two given levels in// a binary tree#include<bits/stdc++.h>using namespace std;// Class containing left and right// child of current node and key valuestruct tree{ int data; tree *left, *right;};struct tree* newNode(int x){ tree* temp = new tree; temp->data = x; temp->left = temp->right = NULL;}// Iterative function to print all// nodes between two given// levels in a binary treevoid printNodes(tree* root, int start, int end){ if (root == NULL) { return; } // create an empty queue and // enqueue root node queue<tree*> queue ; queue.push(root); // pointer to store current node tree *curr = NULL; // maintains level of current node int level = 0; // run till queue is not empty while (!queue.empty()) { // increment level by 1 level++; // calculate number of nodes in // current level int size = queue.size(); // process every node of current level // and enqueue their non-empty left // and right child to queue while (size != 0) { curr = queue.front(); queue.pop(); // print the node if its level is // between given levels if (level >= start && level <= end) { cout << curr->data << " "; } if (curr->left != NULL) { queue.push(curr->left); } if (curr->right != NULL) { queue.push(curr->right); } size--; } if (level >= start && level <= end) { cout << ("\n"); }; }}// Driver Codeint main(){ tree *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); /* Constructed binary tree is 1 / \ 2 3 / \ / \ 4 5 6 7 */ int start = 2, end = 3; printNodes(root, start, end);}// This code is contributed by Rajput-Ji |
Java
// Java program for Print all nodes// between two given levels in// a binary treeimport java.util.LinkedList;import java.util.Queue;public class BinaryTree { // Class containing left and right // child of current node and key value static class Node { int data; Node left, right; public Node(int item) { data = item; left = right = null; } } // Root of the Binary Tree Node root; public BinaryTree() { root = null; } // Iterative function to print all // nodes between two given // levels in a binary tree void printNodes(Node root, int start, int end) { if (root == null) { return; } // create an empty queue and // enqueue root node Queue<Node> queue = new LinkedList<Node>(); queue.add(root); // pointer to store current node Node curr = null; // maintains level of current node int level = 0; // run till queue is not empty while (!queue.isEmpty()) { // increment level by 1 level++; // calculate number of nodes in // current level int size = queue.size(); // process every node of current level // and enqueue their non-empty left // and right child to queue while (size != 0) { curr = queue.poll(); // print the node if its level is // between given levels if (level >= start && level <= end) { System.out.print(curr.data + " "); } if (curr.left != null) { queue.add(curr.left); } if (curr.right != null) { queue.add(curr.right); } size--; } if (level >= start && level <= end) { System.out.println(""); }; } } // Driver Code public static void main(String args[]) { BinaryTree tree = new BinaryTree(); 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); /* Constructed binary tree is 1 / \ 2 3 / \ / \ 4 5 6 7 */ int start = 2, end = 3; tree.printNodes(tree.root, start, end); }} |
Python3
# Python3 program for Print all nodes# between two given levels in# a binary tree# Helper function that allocates a new# node with the given data and None# left and right poers. class newNode: # Construct to create a new node def __init__(self, key): self.data = key self.left = None self.right = None# Iterative function to print all# nodes between two given# levels in a binary treedef prNodes(root, start, end): if (root == None): return # create an empty queue and # enqueue root node q = [] q.append(root) # poer to ste current node curr = None # maains level of current node level = 0 # run till queue is not empty while (len(q)): # increment level by 1 level += 1 # calculate number of nodes in # current level size = len(q) # process every node of current level # and enqueue their non-empty left # and right child to queue while (size != 0) : curr = q[0] q.pop(0) # print the node if its level is # between given levels if (level >= start and level <= end) : print(curr.data, end = " ") if (curr.left != None) : q.append(curr.left) if (curr.right != None) : q.append(curr.right) size -= 1 if (level >= start and level <= end) : print("") # Driver Codeif __name__ == '__main__': """ Let us create Binary Tree shown in above example """ root = newNode(1) root.left = newNode(2) root.left.left = newNode(4) root.left.right = newNode(5) root.right = newNode(3) root.right.right = newNode(7) root.right.left = newNode(6) start = 2 end = 3 prNodes(root, start, end) # This code is contributed by# Shubham Singh(SHUBHAMSINGH10) |
C#
// C# program for Print all nodes// between two given levels in// a binary treeusing System;using System.Collections.Generic;public class BinaryTree{ // Class containing left and right // child of current node and key value public class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; } } // Root of the Binary Tree Node root; public BinaryTree() { root = null; } // Iterative function to print all // nodes between two given // levels in a binary tree void printNodes(Node root, int start, int end) { if (root == null) { return; } // create an empty queue and // enqueue root node Queue<Node> queue = new Queue<Node>(); queue.Enqueue(root); // pointer to store current node Node curr = null; // maintains level of current node int level = 0; // run till queue is not empty while (queue.Count >0) { // increment level by 1 level++; // calculate number of nodes in // current level int size = queue.Count; // process every node of current level // and enqueue their non-empty left // and right child to queue while (size != 0) { curr = queue.Peek(); queue.Dequeue(); // print the node if its level is // between given levels if (level >= start && level <= end) { Console.Write(curr.data + " "); } if (curr.left != null) { queue.Enqueue(curr.left); } if (curr.right != null) { queue.Enqueue(curr.right); } size--; } if (level >= start && level <= end) { Console.WriteLine(""); }; } } // Driver Code public static void Main(String []args) { BinaryTree tree = new BinaryTree(); 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); /* Constructed binary tree is 1 / \ 2 3 / \ / \ 4 5 6 7 */ int start = 2, end = 3; tree.printNodes(tree.root, start, end); }}// This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript program for Print all nodes // between two given levels in // a binary tree class Node { constructor(data) { this.left = null; this.right = null; this.data = data; } } // Root of the Binary Tree let root = null; // Iterative function to print all // nodes between two given // levels in a binary tree function printNodes(root, start, end) { if (root == null) { return; } // create an empty queue and // enqueue root node let queue = []; queue.push(root); // pointer to store current node let curr = null; // maintains level of current node let level = 0; // run till queue is not empty while (queue.length > 0) { // increment level by 1 level++; // calculate number of nodes in // current level let size = queue.length; // process every node of current level // and enqueue their non-empty left // and right child to queue while (size != 0) { curr = queue[0]; queue.shift(); // print the node if its level is // between given levels if (level >= start && level <= end) { document.write(curr.data + " "); } if (curr.left != null) { queue.push(curr.left); } if (curr.right != null) { queue.push(curr.right); } size--; } if (level >= start && level <= end) { document.write("</br>"); } } } let tree = new Node(0); 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); /* Constructed binary tree is 1 / \ 2 3 / \ / \ 4 5 6 7 */ let start = 2, end = 3; printNodes(tree.root, start, end); // This code is contributed by decode2207.</script> |
Output
2 3 4 5 6 7
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