Dado un árbol binario, compruebe si todas las hojas están al mismo nivel o no.
12
/ \
5 7
/ \
3 1
Leaves are at same level
12
/ \
5 7
/
3
Leaves are Not at same level
12
/
5
/ \
3 9
/ /
1 2
Leaves are at same level
La idea es encontrar primero el nivel de la hoja más a la izquierda y almacenarlo en una variable leafLevel. Luego compare el nivel de todas las demás hojas con leafLevel, si es el mismo, devuelva verdadero, de lo contrario, devuelva falso. Atravesamos el árbol binario dado en forma de preorden. Se pasa un argumento leaflevel a todas las llamadas. El valor de leafLevel se inicializa como 0 para indicar que aún no se ha visto la primera hoja. El valor se actualiza cuando encontramos la primera hoja. El nivel de las hojas subsiguientes (en preorden) se compara con leafLevel.
C++
// C++ program to check if all leaves
// are at same level
#include <bits/stdc++.h>
using namespace std;
// A binary tree node
struct Node
{
int data;
struct Node *left, *right;
};
// A utility function to allocate
// a new tree node
struct Node* newNode(int data)
{
struct Node* node = (struct Node*) malloc(sizeof(struct Node));
node->data = data;
node->left = node->right = NULL;
return node;
}
/* Recursive function which checks whether
all leaves are at same level */
bool checkUtil(struct Node *root,
int level, int *leafLevel)
{
// Base case
if (root == NULL) return true;
// If a leaf node is encountered
if (root->left == NULL &&
root->right == NULL)
{
// When a leaf node is found
// first time
if (*leafLevel == 0)
{
*leafLevel = level; // Set first found leaf's level
return true;
}
// If this is not first leaf node, compare
// its level with first leaf's level
return (level == *leafLevel);
}
// If this node is not leaf, recursively
// check left and right subtrees
return checkUtil(root->left, level + 1, leafLevel) &&
checkUtil(root->right, level + 1, leafLevel);
}
/* The main function to check
if all leafs are at same level.
It mainly uses checkUtil() */
bool check(struct Node *root)
{
int level = 0, leafLevel = 0;
return checkUtil(root, level, &leafLevel);
}
// Driver Code
int main()
{
// Let us create tree shown in third example
struct Node *root = newNode(12);
root->left = newNode(5);
root->left->left = newNode(3);
root->left->right = newNode(9);
root->left->left->left = newNode(1);
root->left->right->left = newNode(1);
if (check(root))
cout << "Leaves are at same level\n";
else
cout << "Leaves are not at same level\n";
getchar();
return 0;
}
// This code is contributed
// by Akanksha Rai
C
// C program to check if all leaves are at same level
#include <stdio.h>
#include <stdlib.h>
// A binary tree node
struct Node
{
int data;
struct Node *left, *right;
};
// A utility function to allocate a new tree node
struct Node* newNode(int data)
{
struct Node* node = (struct Node*) malloc(sizeof(struct Node));
node->data = data;
node->left = node->right = NULL;
return node;
}
/* Recursive function which checks whether all leaves are at same level */
bool checkUtil(struct Node *root, int level, int *leafLevel)
{
// Base case
if (root == NULL) return true;
// If a leaf node is encountered
if (root->left == NULL && root->right == NULL)
{
// When a leaf node is found first time
if (*leafLevel == 0)
{
*leafLevel = level; // Set first found leaf's level
return true;
}
// If this is not first leaf node, compare its level with
// first leaf's level
return (level == *leafLevel);
}
// If this node is not leaf, recursively check left and right subtrees
return checkUtil(root->left, level+1, leafLevel) &&
checkUtil(root->right, level+1, leafLevel);
}
/* The main function to check if all leafs are at same level.
It mainly uses checkUtil() */
bool check(struct Node *root)
{
int level = 0, leafLevel = 0;
return checkUtil(root, level, &leafLevel);
}
// Driver program to test above function
int main()
{
// Let us create tree shown in thirdt example
struct Node *root = newNode(12);
root->left = newNode(5);
root->left->left = newNode(3);
root->left->right = newNode(9);
root->left->left->left = newNode(1);
root->left->right->left = newNode(1);
if (check(root))
printf("Leaves are at same level\n");
else
printf("Leaves are not at same level\n");
getchar();
return 0;
}
Java
// Java program to check if all leaves are at same level
// A binary tree node
class Node
{
int data;
Node left, right;
Node(int item)
{
data = item;
left = right = null;
}
}
class Leaf
{
int leaflevel=0;
}
class BinaryTree
{
Node root;
Leaf mylevel = new Leaf();
/* Recursive function which checks whether all leaves are at same
level */
boolean checkUtil(Node node, int level, Leaf leafLevel)
{
// Base case
if (node == null)
return true;
// If a leaf node is encountered
if (node.left == null && node.right == null)
{
// When a leaf node is found first time
if (leafLevel.leaflevel == 0)
{
// Set first found leaf's level
leafLevel.leaflevel = level;
return true;
}
// If this is not first leaf node, compare its level with
// first leaf's level
return (level == leafLevel.leaflevel);
}
// If this node is not leaf, recursively check left and right
// subtrees
return checkUtil(node.left, level + 1, leafLevel)
&& checkUtil(node.right, level + 1, leafLevel);
}
/* The main function to check if all leafs are at same level.
It mainly uses checkUtil() */
boolean check(Node node)
{
int level = 0;
return checkUtil(node, level, mylevel);
}
public static void main(String args[])
{
// Let us create the tree as shown in the example
BinaryTree tree = new BinaryTree();
tree.root = new Node(12);
tree.root.left = new Node(5);
tree.root.left.left = new Node(3);
tree.root.left.right = new Node(9);
tree.root.left.left.left = new Node(1);
tree.root.left.right.left = new Node(1);
if (tree.check(tree.root))
System.out.println("Leaves are at same level");
else
System.out.println("Leaves are not at same level");
}
}
// This code has been contributed by Mayank Jaiswal
Python
# Python program to check if all leaves are at same level # A binary tree node class Node: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None # Recursive function which check whether all leaves are at # same level def checkUtil(root, level): # Base Case if root is None: return True # If a tree node is encountered if root.left is None and root.right is None: # When a leaf node is found first time if check.leafLevel == 0 : check.leafLevel = level # Set first leaf found return True # If this is not first leaf node, compare its level # with first leaf's level return level == check.leafLevel # If this is not first leaf node, compare its level # with first leaf's level return (checkUtil(root.left, level+1)and checkUtil(root.right, level+1)) def check(root): level = 0 check.leafLevel = 0 return (checkUtil(root, level)) # Driver program to test above function root = Node(12) root.left = Node(5) root.left.left = Node(3) root.left.right = Node(9) root.left.left.left = Node(1) root.left.right.left = Node(2) if(check(root)): print "Leaves are at same level" else: print "Leaves are not at same level" # This code is contributed by Nikhil Kumar Singh(nickzuck_007)
C#
// C# program to check if all leaves
// are at same level
using System;
// A binary tree node
public class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
public class Leaf
{
public int leaflevel = 0;
}
class GFG
{
public Node root;
public Leaf mylevel = new Leaf();
/* Recursive function which checks
whether all leaves are at same level */
public virtual bool checkUtil(Node node, int level,
Leaf leafLevel)
{
// Base case
if (node == null)
{
return true;
}
// If a leaf node is encountered
if (node.left == null && node.right == null)
{
// When a leaf node is found first time
if (leafLevel.leaflevel == 0)
{
// Set first found leaf's level
leafLevel.leaflevel = level;
return true;
}
// If this is not first leaf node,
// compare its level with first leaf's level
return (level == leafLevel.leaflevel);
}
// If this node is not leaf, recursively
// check left and right subtrees
return checkUtil(node.left, level + 1, leafLevel) &&
checkUtil(node.right, level + 1, leafLevel);
}
/* The main function to check if all leafs
are at same level. It mainly uses checkUtil() */
public virtual bool check(Node node)
{
int level = 0;
return checkUtil(node, level, mylevel);
}
// Driver Code
public static void Main(string[] args)
{
// Let us create the tree as shown in the example
GFG tree = new GFG();
tree.root = new Node(12);
tree.root.left = new Node(5);
tree.root.left.left = new Node(3);
tree.root.left.right = new Node(9);
tree.root.left.left.left = new Node(1);
tree.root.left.right.left = new Node(1);
if (tree.check(tree.root))
{
Console.WriteLine("Leaves are at same level");
}
else
{
Console.WriteLine("Leaves are not at same level");
}
}
}
// This code is contributed by Shrikant13
Javascript
<script>
// Javascript program to check if all
// leaves are at same level
// A binary tree node
class Node
{
constructor(item)
{
this.data = item;
this.left = this.right = null;
}
}
class Leaf
{
leaflevel = 0;
}
let root;
let mylevel = new Leaf();
// Recursive function which checks
// whether all leaves are at same level
function checkUtil(node, level, leafLevel)
{
// Base case
if (node == null)
return true;
// If a leaf node is encountered
if (node.left == null && node.right == null)
{
// When a leaf node is found first time
if (leafLevel.leaflevel == 0)
{
// Set first found leaf's level
leafLevel.leaflevel = level;
return true;
}
// If this is not first leaf node,
// compare its level with first leaf's level
return (level == leafLevel.leaflevel);
}
// If this node is not leaf, recursively
// check left and right subtrees
return checkUtil(node.left, level + 1, leafLevel) &&
checkUtil(node.right, level + 1, leafLevel);
}
// The main function to check if all
// leafs are at same level. It mainly
// uses checkUtil()
function check(node)
{
let level = 0;
return checkUtil(node, level, mylevel);
}
// Driver code
// Let us create the tree as shown in the example
root = new Node(12);
root.left = new Node(5);
root.left.left = new Node(3);
root.left.right = new Node(9);
root.left.left.left = new Node(1);
root.left.right.left = new Node(1);
if (check(root))
document.write("Leaves are at same level");
else
document.write("Leaves are not at same level");
// This code is contributed by rag2127
</script>
C++
// C++ program to check if all leaf nodes are at
// same level of binary tree
#include <bits/stdc++.h>
using namespace std;
// tree node
struct Node {
int data;
Node *left, *right;
};
// returns a new tree Node
Node* newNode(int data)
{
Node* temp = new Node();
temp->data = data;
temp->left = temp->right = NULL;
return temp;
}
// return true if all leaf nodes are
// at same level, else false
int checkLevelLeafNode(Node* root)
{
if (!root)
return 1;
// create a queue for level order traversal
queue<Node*> q;
q.push(root);
int result = INT_MAX;
int level = 0;
// traverse until the queue is empty
while (!q.empty()) {
int size = q.size();
level += 1;
// traverse for complete level
while(size > 0){
Node* temp = q.front();
q.pop();
// check for left child
if (temp->left) {
q.push(temp->left);
// if its leaf node
if(!temp->left->right && !temp->left->left){
// if it's first leaf node, then update result
if (result == INT_MAX)
result = level;
// if it's not first leaf node, then compare
// the level with level of previous leaf node
else if (result != level)
return 0;
}
}
// check for right child
if (temp->right){
q.push(temp->right);
// if it's leaf node
if (!temp->right->left && !temp->right->right)
// if it's first leaf node till now,
// then update the result
if (result == INT_MAX)
result = level;
// if it is not the first leaf node,
// then compare the level with level
// of previous leaf node
else if(result != level)
return 0;
}
size -= 1;
}
}
return 1;
}
// driver program
int main()
{
// construct a tree
Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->right = newNode(4);
root->right->left = newNode(5);
root->right->right = newNode(6);
int result = checkLevelLeafNode(root);
if (result)
cout << "All leaf nodes are at same level\n";
else
cout << "Leaf nodes not at same level\n";
return 0;
}
Java
// Java program to check if all leaf nodes are at
// same level of binary tree
import java.util.*;
// User defined node class
class Node {
int data;
Node left, right;
// Constructor to create a new tree node
Node(int key) {
int data = key;
left = right = null;
}
}
class GFG {
// return true if all leaf nodes are
// at same level, else false
static boolean checkLevelLeafNode(Node root)
{
if (root == null)
return true;
// create a queue for level order traversal
Queue<Node> q = new LinkedList<>();
q.add(root);
int result = Integer.MAX_VALUE;
int level = 0;
// traverse until the queue is empty
while (q.size() != 0) {
int size = q.size();
level++;
// traverse for complete level
while (size > 0) {
Node temp = q.remove();
// check for left child
if (temp.left != null) {
q.add(temp.left);
// if its leaf node
if (temp.left.left == null && temp.left.right == null) {
// if it's first leaf node, then update result
if (result == Integer.MAX_VALUE)
result = level;
// if it's not first leaf node, then compare
// the level with level of previous leaf node.
else if (result != level)
return false;
}
}
// check for right child
if (temp.right != null) {
q.add(temp.right);
// if its leaf node
if (temp.right.left == null && temp.right.right == null) {
// if it's first leaf node, then update result
if (result == Integer.MAX_VALUE)
result = level;
// if it's not first leaf node, then compare
// the level with level of previous leaf node.
else if (result != level)
return false;
}
}
size--;
}
}
return true;
}
// Driver code
public static void main(String args[])
{
// construct a tree
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.right = new Node(4);
root.right.left = new Node(5);
root.right.right = new Node(6);
boolean result = checkLevelLeafNode(root);
if (result == true)
System.out.println("All leaf nodes are at same level");
else
System.out.println("Leaf nodes not at same level");
}
}
// This code is contributed by rachana soma
Python3
# Python3 program to check if all leaf nodes
# are at same level of binary tree
INT_MAX = 2**31
INT_MIN = -2**31
# Tree Node
# returns a new tree Node
class newNode:
def __init__(self, data):
self.data = data
self.left = self.right = None
# return true if all leaf nodes are
# at same level, else false
def checkLevelLeafNode(root) :
if (not root) :
return 1
# create a queue for level
# order traversal
q = []
q.append(root)
result = INT_MAX
level = 0
# traverse until the queue is empty
while (len(q)):
size = len(q)
level += 1
# traverse for complete level
while(size > 0 or len(q)):
temp = q[0]
q.pop(0)
# check for left child
if (temp.left) :
q.append(temp.left)
# if its leaf node
if(not temp.left.right and
not temp.left.left):
# if it's first leaf node,
# then update result
if (result == INT_MAX):
result = level
# if it's not first leaf node,
# then compare the level with
# level of previous leaf node
elif (result != level):
return 0
# check for right child
if (temp.right) :
q.append(temp.right)
# if it's leaf node
if (not temp.right.left and
not temp.right.right):
# if it's first leaf node till now,
# then update the result
if (result == INT_MAX):
result = level
# if it is not the first leaf node,
# then compare the level with level
# of previous leaf node
elif(result != level):
return 0
size -= 1
return 1
# Driver Code
if __name__ == '__main__':
# construct a tree
root = newNode(1)
root.left = newNode(2)
root.right = newNode(3)
root.left.right = newNode(4)
root.right.left = newNode(5)
root.right.right = newNode(6)
result = checkLevelLeafNode(root)
if (result) :
print("All leaf nodes are at same level")
else:
print("Leaf nodes not at same level")
# This code is contributed by SHUBHAMSINGH10
C#
// C# program to check if all leaf nodes are at
// same level of binary tree
using System;
using System.Collections.Generic;
// User defined node class
public class Node
{
public int data;
public Node left, right;
// Constructor to create a new tree node
public Node(int key)
{
int data = key;
left = right = null;
}
}
public class GFG
{
// return true if all leaf nodes are
// at same level, else false
static bool checkLevelLeafNode(Node root)
{
if (root == null)
return true;
// create a queue for level order traversal
Queue<Node> q = new Queue<Node>();
q.Enqueue(root);
int result = int.MaxValue;
int level = 0;
// traverse until the queue is empty
while (q.Count != 0)
{
int size = q.Count;
level++;
// traverse for complete level
while (size > 0)
{
Node temp = q.Dequeue();
// check for left child
if (temp.left != null)
{
q.Enqueue(temp.left);
// if its leaf node
if (temp.left.left != null &&
temp.left.right != null)
{
// if it's first leaf node, then update result
if (result == int.MaxValue)
result = level;
// if it's not first leaf node, then compare
// the level with level of previous leaf node.
else if (result != level)
return false;
}
}
// check for right child
if (temp.right != null)
{
q.Enqueue(temp.right);
// if its leaf node
if (temp.right.left != null &&
temp.right.right != null)
{
// if it's first leaf node, then update result
if (result == int.MaxValue)
result = level;
// if it's not first leaf node, then compare
// the level with level of previous leaf node.
else if (result != level)
return false;
}
}
size--;
}
}
return true;
}
// Driver code
public static void Main(String []args)
{
// construct a tree
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.right = new Node(4);
root.right.left = new Node(5);
root.right.right = new Node(6);
bool result = checkLevelLeafNode(root);
if (result == true)
Console.WriteLine("All leaf nodes are at same level");
else
Console.WriteLine("Leaf nodes not at same level");
}
}
// This code has been contributed by 29AjayKumar
Javascript
<script>
// Javascript program to check if all
// leaf nodes are at same level of binary tree
// User defined node class
class Node
{
// Constructor to create a new tree node
constructor(key)
{
this.data = key;
this.left = this.right = null;
}
}
// Return true if all leaf nodes are
// at same level, else false
function checkLevelLeafNode(root)
{
if (root == null)
return true;
// Create a queue for level
// order traversal
let q = [];
q.push(root);
let result = Number.MAX_VALUE;
let level = 0;
// Traverse until the queue is empty
while (q.length != 0)
{
let size = q.length;
level++;
// traverse for complete level
while (size > 0)
{
let temp = q.shift();
// check for left child
if (temp.left != null)
{
q.push(temp.left);
// if its leaf node
if (temp.left.left == null &&
temp.left.right == null)
{
// If it's first leaf node,
// then update result
if (result == Number.MAX_VALUE)
result = level;
// If it's not first leaf node,
// then compare the level with
// level of previous leaf node.
else if (result != level)
return false;
}
}
// Check for right child
if (temp.right != null)
{
q.push(temp.right);
// If its leaf node
if (temp.right.left == null &&
temp.right.right == null)
{
// If it's first leaf node, then
// update result
if (result == Number.MAX_VALUE)
result = level;
// If it's not first leaf node,
// then compare the level with
// level of previous leaf node.
else if (result != level)
return false;
}
}
size--;
}
}
return true;
}
// Driver code
// construct a tree
let root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.right = new Node(4);
root.right.left = new Node(5);
root.right.right = new Node(6);
let result = checkLevelLeafNode(root);
if (result == true)
document.write("All leaf nodes are at same level");
else
document.write("Leaf nodes not at same level");
// This code is contributed by avanitrachhadiya2155
</script>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA