Given a binary tree containing n nodes. The problem is to find and print the largest value present in each level.
Examples:
Input :
1
/ \
2 3
Output : 1 3
Input :
4
/ \
9 2
/ \ \
3 5 7
Output : 4 9 7
Approach: In the previous post, a recursive method have been discussed. In this post an iterative method has been discussed. The idea is to perform iterative level order traversal of the binary tree using queue. While traversing keep max variable which stores the maximum element of the current level of the tree being processed. When the level is completely traversed, print that max value.
Implementation:
C++
// C++ implementation to print largest// value in each level of Binary Tree#include <bits/stdc++.h>using namespace std;// structure of a node of binary treestruct Node { int data; Node *left, *right;};// function to get a new nodeNode* newNode(int data){ // allocate space Node* temp = new Node; // put in the data temp->data = data; temp->left = temp->right = NULL; return temp;}// function to print largest value// in each level of Binary Treevoid largestValueInEachLevel(Node* root){ // if tree is empty if (!root) return; queue<Node*> q; int nc, max; // push root to the queue 'q' q.push(root); while (1) { // node count for the current level nc = q.size(); // if true then all the nodes of // the tree have been traversed if (nc == 0) break; // maximum element for the current // level max = INT_MIN; while (nc--) { // get the front element from 'q' Node* front = q.front(); // remove front element from 'q' q.pop(); // if true, then update 'max' if (max < front->data) max = front->data; // if left child exists if (front->left) q.push(front->left); // if right child exists if (front->right) q.push(front->right); } // print maximum element of // current level cout << max << " "; }}// Driver codeint main(){ /* Construct a Binary Tree 4 / \ 9 2 / \ \ 3 5 7 */ Node* root = NULL; root = newNode(4); root->left = newNode(9); root->right = newNode(2); root->left->left = newNode(3); root->left->right = newNode(5); root->right->right = newNode(7); // Function call largestValueInEachLevel(root); return 0;} |
Java
// Java implementation to print largest// value in each level of Binary Treeimport java.util.*;class GfG { // structure of a node of binary tree static class Node { int data; Node left = null; Node right = null; } // function to get a new node static Node newNode(int val) { // allocate space Node temp = new Node(); // put in the data temp.data = val; temp.left = null; temp.right = null; return temp; } // function to print largest value // in each level of Binary Tree static void largestValueInEachLevel(Node root) { // if tree is empty if (root == null) return; Queue<Node> q = new LinkedList<Node>(); int nc, max; // push root to the queue 'q' q.add(root); while (true) { // node count for the current level nc = q.size(); // if true then all the nodes of // the tree have been traversed if (nc == 0) break; // maximum element for the current // level max = Integer.MIN_VALUE; while (nc != 0) { // get the front element from 'q' Node front = q.peek(); // remove front element from 'q' q.remove(); // if true, then update 'max' if (max < front.data) max = front.data; // if left child exists if (front.left != null) q.add(front.left); // if right child exists if (front.right != null) q.add(front.right); nc--; } // print maximum element of // current level System.out.println(max + " "); } } // Driver code public static void main(String[] args) { /* Construct a Binary Tree 4 / \ 9 2 / \ \ 3 5 7 */ Node root = null; root = newNode(4); root.left = newNode(9); root.right = newNode(2); root.left.left = newNode(3); root.left.right = newNode(5); root.right.right = newNode(7); // Function call largestValueInEachLevel(root); }} |
Python3
# Python program to print largest value# on each level of binary treeINT_MIN = -2147483648# Helper function that allocates a new# node with the given data and None left# and right pointers.class newNode: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None# function to find largest valuesdef largestValueInEachLevel(root): if (not root): return q = [] nc = 10 max = 0 q.append(root) while (1): # node count for the current level nc = len(q) # if true then all the nodes of # the tree have been traversed if (nc == 0): break # maximum element for the current # level max = INT_MIN while (nc): # get the front element from 'q' front = q[0] # remove front element from 'q' q = q[1:] # if true, then update 'max' if (max < front.data): max = front.data # if left child exists if (front.left): q.append(front.left) # if right child exists if (front.right != None): q.append(front.right) nc -= 1 # print maximum element of # current level print(max, end=" ")# Driver Codeif __name__ == '__main__': """ Let us construct the following Tree 4 / \ 9 2 / \ \ 3 5 7 """ root = newNode(4) root.left = newNode(9) root.right = newNode(2) root.left.left = newNode(3) root.left.right = newNode(5) root.right.right = newNode(7) # Function call largestValueInEachLevel(root)# This code is contributed# Shubham Singh(SHUBHAMSINGH10) |
C#
// C# implementation to print largest// value in each level of Binary Treeusing System;using System.Collections.Generic;class GfG{ // structure of a node of binary tree class Node { public int data; public Node left = null; public Node right = null; } // function to get a new node static Node newNode(int val) { // allocate space Node temp = new Node(); // put in the data temp.data = val; temp.left = null; temp.right = null; return temp; } // function to print largest value // in each level of Binary Tree static void largestValueInEachLevel(Node root) { // if tree is empty if (root == null) return; Queue<Node> q = new Queue<Node>(); int nc, max; // push root to the queue 'q' q.Enqueue(root); while (true) { // node count for the current level nc = q.Count; // if true then all the nodes of // the tree have been traversed if (nc == 0) break; // maximum element for the current // level max = int.MinValue; while (nc != 0) { // get the front element from 'q' Node front = q.Peek(); // remove front element from 'q' q.Dequeue(); // if true, then update 'max' if (max < front.data) max = front.data; // if left child exists if (front.left != null) q.Enqueue(front.left); // if right child exists if (front.right != null) q.Enqueue(front.right); nc--; } // print maximum element of // current level Console.Write(max + " "); } } // Driver code public static void Main(String[] args) { /* Construct a Binary Tree 4 / \ 9 2 / \ \ 3 5 7 */ Node root = null; root = newNode(4); root.left = newNode(9); root.right = newNode(2); root.left.left = newNode(3); root.left.right = newNode(5); root.right.right = newNode(7); // Function call largestValueInEachLevel(root); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script> // JavaScript implementation to print largest // value in each level of Binary Tree // structure of a node of binary tree class Node { constructor(data) { this.left = null; this.right = null; this.data = data; } } // function to get a new node function newNode(val) { // allocate space let temp = new Node(val); return temp; } // function to print largest value // in each level of Binary Tree function largestValueInEachLevel(root) { // if tree is empty if (root == null) return; let q = []; let nc, max; // push root to the queue 'q' q.push(root); while (true) { // node count for the current level nc = q.length; // if true then all the nodes of // the tree have been traversed if (nc == 0) break; // maximum element for the current // level max = Number.MIN_VALUE; while (nc != 0) { // get the front element from 'q' let front = q[0]; // remove front element from 'q' q.shift(); // if true, then update 'max' if (max < front.data) max = front.data; // if left child exists if (front.left != null) q.push(front.left); // if right child exists if (front.right != null) q.push(front.right); nc--; } // print maximum element of // current level document.write(max + " "); } } /* Construct a Binary Tree 4 / \ 9 2 / \ \ 3 5 7 */ let root = null; root = newNode(4); root.left = newNode(9); root.right = newNode(2); root.left.left = newNode(3); root.left.right = newNode(5); root.right.right = newNode(7); // Function call largestValueInEachLevel(root); </script> |
Output
4 9 7
Complexity Analysis:
- Time Complexity: O(N) where N is the total number of nodes in the tree.
In level order traversal, every node of the tree is processed once and hence the complexity due to the level order traversal is O(N) if there are total N nodes in the tree. Also, while processing every node, we are maintaining the maximum element at each level, however, this does not affect the overall time complexity. Therefore, the time complexity is O(N). - Auxiliary Space: O(w) where w is the maximum width of the tree.
In level order traversal, a queue is maintained whose maximum size at any moment can go upto the maximum width of the binary tree.
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