Dado un árbol binario que contiene n Nodes. El problema es encontrar e imprimir el mayor valor presente en cada nivel.
Ejemplos:
Input :
1
/ \
2 3
Output : 1 3
Input :
4
/ \
9 2
/ \ \
3 5 7
Output : 4 9 7
Enfoque: En la publicación anterior , se ha discutido un método recursivo. En esta publicación se ha discutido un método iterativo. La idea es realizar un recorrido de orden de nivel iterativo del árbol binario usando la cola. Mientras se recorre, mantenga la variable max que almacena el elemento máximo del nivel actual del árbol que se está procesando. Cuando el nivel esté completamente recorrido, imprima ese valor máximo .
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 tree
struct Node {
int data;
Node *left, *right;
};
// function to get a new node
Node* 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 Tree
void 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 code
int 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 Tree
import 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 tree INT_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 values def 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 Code if __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 Tree
using 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>
Publicación traducida automáticamente
Artículo escrito por ayushjauhari14 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA