Level with maximum number of nodes
Find the level in a binary tree which has the maximum number of nodes. The root is at level 0.
Examples:
Input:Output : 2 Explanation:
Input:
Output:1 Explanation
Approach: It is known that in level order traversal of binary tree with queue, at any time our queue contains all elements of a particular level. So find level with maximum number of nodes in queue.
BFS traversal is an algorithm for traversing or searching tree or graphs . It starts at the tree root , and explores all of the neighbor nodes at the present depth prior to moving on to the nodes at the next depth level.
So at any point the queue of BFS will contain elements of adjacent layers. So this makes the algorithm perfect for this problem.
Algorithm:
- Create the tree, a queue to store the nodes and insert the root in the queue. Create variables level=0,count =0 and level_no=-1
- The implementation will be slightly different, all the elements of same level will be removed in a single iteration.
- Run a loop while size of queue is greater than 0. Get the size of queue (size) and store it. If size is greater than count then update count = size and level_no = level.
- Now run a loop size times, and pop one node from the queue and insert its childrens (if present).
- Increment level.
Implementation:
C++
// C++ implementation to find the level // having maximum number of Nodes #include <bits/stdc++.h> using namespace std; /* A binary tree Node has data, pointer to left child and a pointer to right child */struct Node { int data; struct Node* left; struct Node* right; }; /* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct Node* newNode(int data) { struct Node* node = new Node; node->data = data; node->left = NULL; node->right = NULL; return(node); } // function to find the level // having maximum number of Nodes int maxNodeLevel(Node *root) { if (root == NULL) return -1; queue<Node *> q; q.push(root); // Current level int level = 0; // Maximum Nodes at same level int max = INT_MIN; // Level having maximum Nodes int level_no = 0; while (1) { // Count Nodes in a level int NodeCount = q.size(); if (NodeCount == 0) break; // If it is maximum till now // Update level_no to current level if (NodeCount > max) { max = NodeCount; level_no = level; } // Pop complete current level while (NodeCount > 0) { Node *Node = q.front(); q.pop(); if (Node->left != NULL) q.push(Node->left); if (Node->right != NULL) q.push(Node->right); NodeCount--; } // Increment for next level level++; } return level_no; } // Driver program to test above int main() { // binary tree formation struct Node *root = newNode(2); /* 2 */ root->left = newNode(1); /* / \ */ root->right = newNode(3); /* 1 3 */ root->left->left = newNode(4); /* / \ \ */ root->left->right = newNode(6); /* 4 6 8 */ root->right->right = newNode(8); /* / */ root->left->right->left = newNode(5);/* 5 */ printf("Level having maximum number of Nodes : %d", maxNodeLevel(root)); return 0; } |
Java
// Java implementation to find the level // having maximum number of Nodes import java.util.*; class GfG { /* A binary tree Node has data, pointer to left child and a pointer to right child */static class Node { int data; Node left; Node right; } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = null; node.right = null; return(node); } // function to find the level // having maximum number of Nodes static int maxNodeLevel(Node root) { if (root == null) return -1; Queue<Node> q = new LinkedList<Node> (); q.add(root); // Current level int level = 0; // Maximum Nodes at same level int max = Integer.MIN_VALUE; // Level having maximum Nodes int level_no = 0; while (true) { // Count Nodes in a level int NodeCount = q.size(); if (NodeCount == 0) break; // If it is maximum till now // Update level_no to current level if (NodeCount > max) { max = NodeCount; level_no = level; } // Pop complete current level while (NodeCount > 0) { Node Node = q.peek(); q.remove(); if (Node.left != null) q.add(Node.left); if (Node.right != null) q.add(Node.right); NodeCount--; } // Increment for next level level++; } return level_no; } // Driver program to test above public static void main(String[] args) { // binary tree formation Node root = newNode(2); /* 2 */ root.left = newNode(1); /* / \ */ root.right = newNode(3); /* 1 3 */ root.left.left = newNode(4); /* / \ \ */ root.left.right = newNode(6); /* 4 6 8 */ root.right.right = newNode(8); /* / */ root.left.right.left = newNode(5);/* 5 */ System.out.println("Level having maximum number of Nodes : " + maxNodeLevel(root)); } } |
Python3
# Python3 implementation to find the # level having Maximum number of Nodes # Importing Queue from queue import Queue # Helper class that allocates a new # node with the given data and None # left and right pointers. class newNode: def __init__(self, data): self.data = data self.left = None self.right = None # function to find the level # having Maximum number of Nodes def maxNodeLevel(root): if (root == None): return -1 q = Queue() q.put(root) # Current level level = 0 # Maximum Nodes at same level Max = -999999999999 # Level having Maximum Nodes level_no = 0 while (1): # Count Nodes in a level NodeCount = q.qsize() if (NodeCount == 0): break # If it is Maximum till now # Update level_no to current level if (NodeCount > Max): Max = NodeCount level_no = level # Pop complete current level while (NodeCount > 0): Node = q.queue[0] q.get() if (Node.left != None): q.put(Node.left) if (Node.right != None): q.put(Node.right) NodeCount -= 1 # Increment for next level level += 1 return level_no # Driver Code if __name__ == '__main__': # binary tree formation root = newNode(2) # 2 root.left = newNode(1) # / \ root.right = newNode(3) # 1 3 root.left.left = newNode(4) # / \ \ root.left.right = newNode(6) # 4 6 8 root.right.right = newNode(8) # / root.left.right.left = newNode(5)# 5 print("Level having Maximum number of Nodes : ", maxNodeLevel(root)) # This code is contributed by Pranchalk |
C#
using System; using System.Collections.Generic; // C# implementation to find the level // having maximum number of Nodes public class GfG { /* A binary tree Node has data, pointer to left child and a pointer to right child */public class Node { public int data; public Node left; public Node right; } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */public static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = null; node.right = null; return (node); } // function to find the level // having maximum number of Nodes public static int maxNodeLevel(Node root) { if (root == null) { return -1; } LinkedList<Node> q = new LinkedList<Node> (); q.AddLast(root); // Current level int level = 0; // Maximum Nodes at same level int max = int.MinValue; // Level having maximum Nodes int level_no = 0; while (true) { // Count Nodes in a level int NodeCount = q.Count; if (NodeCount == 0) { break; } // If it is maximum till now // Update level_no to current level if (NodeCount > max) { max = NodeCount; level_no = level; } // Pop complete current level while (NodeCount > 0) { Node Node = q.First.Value; q.RemoveFirst(); if (Node.left != null) { q.AddLast(Node.left); } if (Node.right != null) { q.AddLast(Node.right); } NodeCount--; } // Increment for next level level++; } return level_no; } // Driver program to test above public static void Main(string[] args) { // binary tree formation Node root = newNode(2); // 2 root.left = newNode(1); // / \ root.right = newNode(3); // 1 3 root.left.left = newNode(4); // / \ \ root.left.right = newNode(6); // 4 6 8 root.right.right = newNode(8); // / root.left.right.left = newNode(5); // 5 Console.WriteLine("Level having maximum number of Nodes : " + maxNodeLevel(root)); } } // This code is contributed by Shrikant13 |
Output:
Level having maximum number of nodes : 2
Complexity Analysis:
- Time Complexity : O(n).
In BFS traversal every node is visited only once, So Time Complexity is O(n). - Space Complexity: O(n).
The space is required to store the nodes in a queue.
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