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 -1
Below is the implementation of the above algorithm:
C++
// C++ program for Print all nodes// between two given levels in// a binary Node#include<bits/stdc++.h>using namespace std; // Class containing left and right// child of current node and key valuestruct Node{ int data; Node *left, *right;};Node* newNode(int x){ Node* temp = new Node(); temp->data = x; temp->left = temp->right = NULL; return temp;} // Iterative function to print all// nodes between two given// levels in a binary Nodevoid printNodes(Node* root, int start, int end){ if (root == NULL) return; // create an empty queue and // enqueue root node queue<Node*> queue; queue.push(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.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(){ 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); /* Constructed binary Node is 1 / \ 2 3 / \ / \ 4 5 6 7 */ int start = 2, end = 3; printNodes(root, start, end);}// THIS CODE IS CONTRIBUTED BY KIRTI AGARWAL |
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 pointers. 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 printNodes(root, start, end): if (root == None): return # create an empty queue and # enqueue root node q = [] q.append(root) # pointer to store current node curr = None # maintains 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 Code if __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 printNodes(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 tree using 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
Time Complexity: O(n)
As we traverse the tree just once.
Auxiliary space: O(b)
Here b is the breadth of the tree. The extra space is used to store the elements of the current level in the queue.
Recursive Approach(Method-2):
Follow the below steps to solve this problem:
1) Create an vector of vector to store the ans in level order fashion.
2) Recursively traverse the whole tree in inorder fashion and keep track the level along with each node.
3) If the level of node lie between the given levels then store them in ans vector respectively to its level number.
4) After traversing the whole tree print the ans vector.
Below is the implementation of above approach:
C++
// C++ program to print all nodes between// two given levels#include <bits/stdc++.h>using namespace std;// A binary tree nodestruct Node{ int data; struct Node* left; struct Node* right;};// utility function to initialize the new nodeNode* newNode(int data){ Node *new_node = new Node(); new_node->data = data; new_node->left = NULL; new_node->right = NULL; return new_node;}// Recursive function to print all nodes between two given// levels in a binary treevoid printNodes(Node* root, int start, int end, vector<vector<int>> &ans, int level){ if(root == NULL) return; printNodes(root->left, start, end, ans, level+1); if(level >= start && level <= end){ ans[level-start].push_back(root->data); } printNodes(root->right, start, end, ans, level+1);}// driver code to test above functionint main(){ 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); int start = 2; int end = 3; // function call vector<vector<int>> ans(end-start+1); printNodes(root, start, end, ans, 1); for(auto i : ans){ for(int j : i){ cout<<j<<" "; } cout<<endl; } return 0;}// THIS CODE IS CONTRIBUTED BY KIRIT AGARWAL(KIRTIAGARWAL23121999) |
Java
// Java program to print all nodes between // two given levels in a binary tree import java.util.*;// A binary tree nodeclass Node { int data; Node left, right; Node(int item) { data = item; left = right = null; }}class BinaryTree { Node root; // utility function to initialize the new node Node newNode(int data) { Node new_node = new Node(data); new_node.left = null; new_node.right = null; return new_node; } // Recursive function to print all nodes between two given // levels in a binary tree void printNodes(Node root, int start, int end, List<List<Integer>> ans, int level) { if (root == null) { return; } printNodes(root.left, start, end, ans, level + 1); if (level >= start && level <= end) { ans.get(level - start).add(root.data); } printNodes(root.right, start, end, ans, level + 1); } // driver code to test above function public static void main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = tree.newNode(1); tree.root.left = tree.newNode(2); tree.root.right = tree.newNode(3); tree.root.left.left = tree.newNode(4); tree.root.left.right = tree.newNode(5); tree.root.right.left = tree.newNode(6); tree.root.right.right = tree.newNode(7); int start = 2; int end = 3; List<List<Integer>> ans = new ArrayList<>(end - start + 1); for (int i = 0; i < end - start + 1; i++) { ans.add(i, new ArrayList<Integer>()); } // function call tree.printNodes(tree.root, start, end, ans, 1); for (List<Integer> i : ans) { for (int j : i) { System.out.print(j + " "); } System.out.println(); } }} |
Python3
# Python program to print all nodes between# two given levelsclass Node: def __init__(self, data): self.data = data self.left = None self.right = None # utility function to get the new nodedef newNode(data): return Node(data)# Recursive function to print all nodes between two given# levels in a binary treedef printNodes(root, start, end, ans, level): if(root is None): return printNodes(root.left, start, end, ans, level+1) if(level >= start and level <= end): ans[level-start].append(root.data) printNodes(root.right, start, end, ans, level+1) # driver code to test above functionroot = 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)start = 2end = 3ans = []for i in range(end-start+1): temp = [] ans.append(temp)# function callprintNodes(root, start, end, ans, 1)for i in range(len(ans)): for j in range (len(ans[i])): print(ans[i][j], end=" ") print("") |
C#
// C# program to print all nodes between two given levels in// a binary treeusing System;using System.Collections.Generic;public class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; }}public class GFG { public Node root; // Utility function to initialize the new node public Node NewNode(int data) { Node newNode = new Node(data); newNode.left = null; newNode.right = null; return newNode; } // Recursive function to print all nodes between two // given levels in a binary tree public void PrintNodes(Node root, int start, int end, List<List<int> > ans, int level) { if (root == null) { return; } PrintNodes(root.left, start, end, ans, level + 1); if (level >= start && level <= end) { ans[level - start].Add(root.data); } PrintNodes(root.right, start, end, ans, level + 1); } static public void Main() { // Code GFG tree = new GFG(); tree.root = tree.NewNode(1); tree.root.left = tree.NewNode(2); tree.root.right = tree.NewNode(3); tree.root.left.left = tree.NewNode(4); tree.root.left.right = tree.NewNode(5); tree.root.right.left = tree.NewNode(6); tree.root.right.right = tree.NewNode(7); int start = 2; int end = 3; List<List<int> > ans = new List<List<int> >(end - start + 1); for (int i = 0; i < end - start + 1; i++) { ans.Add(new List<int>()); } // Function call tree.PrintNodes(tree.root, start, end, ans, 1); for (int i = 0; i < ans.Count; i++) { for (int j = 0; j < ans[i].Count; j++) { Console.Write(ans[i][j] + " "); } Console.WriteLine(); } }}// This code is contributed by lokesh. |
Javascript
// JavaScript Program to print all nodes between// two given levels// a binary tree nodeclass Node{ constructor(data){ this.data = data; this.left = null; this.right = null; }}// utility function to initialize the new nodefunction newNode(data){ return new Node(data);}// Recursive function to print all nodes between two given// levels in a binary treefunction printNodes(root, start, end, ans, level){ if(root == null) return; printNodes(root.left, start, end, ans, level+1); if(level >= start && level <= end){ ans[level-start].push(root.data); } printNodes(root.right, start, end, ans, level+1);} // driver code to test above functionlet 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);let start = 2;let end = 3;let ans = [];for(let i = 0; i<end-start+1; i++){ ans[i] = [];}// function callprintNodes(root, start, end,ans, 1);for(let i = 0; i < ans.length; i++){ for(let j = 0; j < ans[i].length; j++) { console.log(ans[i][j] + " "); } console.log("<br>");}// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002 ) |
Output
2 3 4 5 6 7
Time Complexity: O(N) where N is the number of nodes in given binary tree.
Auxiliary Space : O(N) where N is the number of nodes.
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