Escribe una función para conectar todos los Nodes adyacentes al mismo nivel en un árbol binario. La estructura del Node del árbol binario dado es como la siguiente
C
struct node {
int data;
struct node* left;
struct node* right;
struct node* nextRight;
}
Java
static class node {
int data;
node left;
node right;
node nextRight;
}
// This code is contributed by umadevi9616
Python3
class newnode: def __init__(self,data): self.data = data self.left = None self.right = None self.nextRight = None # this code is contributed by shivanisinghss2110
C#
public class Node
{
public int data;
public Node left;
public Node right;
public Node nextRight;
}
// this code is contributed by shivanisinghss2110
Javascript
class node {
constructor()
{
this.data = 0;
this.left = null;
this.right = null;
this.nextRight = null;
}
}
// This code is contributed by importantly.
Inicialmente, todos los punteros nextRight apuntan a valores basura. Su función debe configurar estos punteros para que apunten a la derecha de cada Node. Solo puede usar espacio adicional constante.
C
// Recursive C program to connect nodes at same level
// using constant extra space
void connectRecur(struct node* p);
struct node *getNextRight(struct node *p);
// Sets the nextRight of root and calls
// connectRecur() for other nodes
void connect (struct node *p)
{
// Set the nextRight for root
p->nextRight = NULL;
// Set the next right for rest of the
// nodes (other than root)
connectRecur(p);
}
/* Set next right of all descendants of p.
This function makes sure that
nextRight of nodes ar level i is set
before level i+1 nodes. */
void connectRecur(struct node* p)
{
// Base case
if (!p)
return;
/* Before setting nextRight of left and right
children, set nextRight of children of other
nodes at same level (because we can access
children of other nodes using p's nextRight only) */
if (p->nextRight != NULL)
connectRecur(p->nextRight);
/* Set the nextRight pointer for p's left child */
if (p->left)
{
if (p->right)
{
p->left->nextRight = p->right;
p->right->nextRight = getNextRight(p);
}
else
p->left->nextRight = getNextRight(p);
/* Recursively call for next level nodes.
Note that we call only for left child.
The call for left child will call for right child */
connectRecur(p->left);
}
/* If left child is NULL then first node of
next level will either be
p->right or getNextRight(p) */
else if (p->right)
{
p->right->nextRight = getNextRight(p);
connectRecur(p->right);
}
else
connectRecur(getNextRight(p));
}
/* This function returns the leftmost child of nodes
at the same level as p. This function is used to
getNExt right of p's right child If right child
of p is NULL then this can also be used for the left child */
struct node *getNextRight(struct node *p)
{
struct node *temp = p->nextRight;
/* Traverse nodes at p's level and find and return
the first node's first child */
while(temp != NULL)
{
if(temp->left != NULL)
return temp->left;
if(temp->right != NULL)
return temp->right;
temp = temp->nextRight;
}
// If all the nodes at p's level are leaf
// nodes then return NULL
return NULL;
}
Java
// Recursive Java program to connect nodes at same level
// using constant extra space
// A binary tree node
class Node
{
int data;
Node left, right, nextRight;
Node(int item)
{
data = item;
left = right = nextRight = null;
}
}
class BinaryTree
{
Node root;
/* Set next right of all descendants of p. This function makes sure that
nextRight of nodes ar level i is set before level i+1 nodes. */
void connectRecur(Node p)
{
// Base case
if (p == null)
return;
/* Before setting nextRight of left and right children, set nextRight
of children of other nodes at same level (because we can access
children of other nodes using p's nextRight only) */
if (p.nextRight != null)
connectRecur(p.nextRight);
/* Set the nextRight pointer for p's left child */
if (p.left != null)
{
if (p.right != null)
{
p.left.nextRight = p.right;
p.right.nextRight = getNextRight(p);
}
else
p.left.nextRight = getNextRight(p);
/* Recursively call for next level nodes. Note that we call only
for left child. The call for left child will call for right child */
connectRecur(p.left);
}
/* If left child is NULL then first node of next level will either be
p->right or getNextRight(p) */
else if (p.right != null)
{
p.right.nextRight = getNextRight(p);
connectRecur(p.right);
}
else
connectRecur(getNextRight(p));
}
/* This function returns the leftmost child of nodes at the same
level as p. This function is used to getNExt right of p's right child
If right child of p is NULL then this can also be used for
the left child */
Node getNextRight(Node p)
{
Node temp = p.nextRight;
/* Traverse nodes at p's level and find and return
the first node's first child */
while (temp != null)
{
if (temp.left != null)
return temp.left;
if (temp.right != null)
return temp.right;
temp = temp.nextRight;
}
// If all the nodes at p's level are leaf nodes then return NULL
return null;
}
/* Driver program to test the above functions */
public static void main(String args[])
{
BinaryTree tree = new BinaryTree();
tree.root = new Node(10);
tree.root.left = new Node(8);
tree.root.right = new Node(2);
tree.root.left.left = new Node(3);
tree.root.right.right = new Node(90);
// Populates nextRight pointer in all nodes
tree.connectRecur(tree.root);
// Let us check the values of nextRight pointers
int a = tree.root.nextRight != null ?
tree.root.nextRight.data : -1;
int b = tree.root.left.nextRight != null ?
tree.root.left.nextRight.data : -1;
int c = tree.root.right.nextRight != null ?
tree.root.right.nextRight.data : -1;
int d = tree.root.left.left.nextRight != null ?
tree.root.left.left.nextRight.data : -1;
int e = tree.root.right.right.nextRight != null ?
tree.root.right.right.nextRight.data : -1;
// Now lets print the values
System.out.println("Following are populated nextRight pointers in "
+ " the tree(-1 is printed if there is no nextRight)");
System.out.println("nextRight of " + tree.root.data + " is " + a);
System.out.println("nextRight of " + tree.root.left.data + " is " + b);
System.out.println("nextRight of " + tree.root.right.data + " is " + c);
System.out.println("nextRight of " + tree.root.left.left.data +
" is " + d);
System.out.println("nextRight of " + tree.root.right.right.data +
" is " + e);
}
}
// This code has been contributed by Mayank Jaiswal
Python3
# Recursive Python program to connect nodes
# at same level using constant extra space
# 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
self.nextRight = None
# Set next right of all descendants of p. This function makes sure that
# nextRight of nodes ar level i is set before level i+1 nodes. */
def connectRecur( p) :
# Base case
if p == None :
return
# Before setting nextRight of left and right children, set nextRight
# of children of other nodes at same level (because we can access
# children of other nodes using p's nextRight only) */
if p.nextRight != None :
connectRecur(p.nextRight)
# Set the nextRight pointer for p's left child */
if p.left != None :
if p.right != None :
p.left.nextRight = p.right
p.right.nextRight = getNextRight(p)
else :
p.left.nextRight = getNextRight(p)
# Recursively call for next level nodes. Note that we call only
# for left child. The call for left child will call for right child */
connectRecur(p.left)
# If left child is NULL then first node of next level will either be
# p->right or getNextRight(p) */
elif p.right != None :
p.right.nextRight = getNextRight(p)
connectRecur(p.right)
else :
connectRecur(getNextRight(p))
# This function returns the leftmost child of
# nodes at the same level as p. This function
# is used to getNExt right of p's right child
# If right child of is None then this can also
# be used for the left child
def getNextRight(p):
temp = p.nextRight
# Traverse nodes at p's level and find
# and return the first node's first child
while (temp != None):
if (temp.left != None):
return temp.left
if (temp.right != None):
return temp.right
temp = temp.nextRight
# If all the nodes at p's level are
# leaf nodes then return None
return None
# Driver Code
if __name__ == '__main__':
# Constructed binary tree is
# 10
# / \
# 8 2
# / \
# 3 90
root = newnode(10)
root.left = newnode(8)
root.right = newnode(2)
root.left.left = newnode(3)
root.right.right = newnode(90)
# Populates nextRight pointer in all nodes
connectRecur(root)
# Let us check the values of nextRight pointers
print("Following are populated nextRight "
"pointers in the tree (-1 is printed "
"if there is no nextRight) \n")
print("nextRight of", root.data,
"is", end = " ")
if root.nextRight:
print(root.nextRight.data)
else:
print(-1)
print("nextRight of", root.left.data,
"is", end = " ")
if root.left.nextRight:
print(root.left.nextRight.data)
else:
print(-1)
print("nextRight of", root.right.data,
"is", end = " ")
if root.right.nextRight:
print(root.right.nextRight.data)
else:
print(-1)
print("nextRight of", root.left.left.data,
"is", end = " ")
if root.left.left.nextRight:
print(root.left.left.nextRight.data)
else:
print(-1)
print("nextRight of", root.right.right.data,
"is", end = " ")
if root.right.right.nextRight:
print(root.right.right.nextRight.data)
else:
print(-1)
# This code is contributed by jana_sayantan.
C#
using System;
// Recursive C# program to connect nodes at same level
// using constant extra space
// A binary tree node
public class Node
{
public int data;
public Node left, right, nextRight;
public Node(int item)
{
data = item;
left = right = nextRight = null;
}
}
public class BinaryTree
{
public Node root;
/* Set next right of all descendants of p. This function makes sure that
nextRight of nodes ar level i is set before level i+1 nodes. */
public virtual void connectRecur(Node p)
{
// Base case
if (p == null)
{
return;
}
/* Before setting nextRight of left and right children, set nextRight
of children of other nodes at same level (because we can access
children of other nodes using p's nextRight only) */
if (p.nextRight != null)
{
connectRecur(p.nextRight);
}
/* Set the nextRight pointer for p's left child */
if (p.left != null)
{
if (p.right != null)
{
p.left.nextRight = p.right;
p.right.nextRight = getNextRight(p);
}
else
{
p.left.nextRight = getNextRight(p);
}
/* Recursively call for next level nodes. Note that we call only
for left child. The call for left child will call for right child */
connectRecur(p.left);
}
/* If left child is NULL then first node of next level will either be
p->right or getNextRight(p) */
else if (p.right != null)
{
p.right.nextRight = getNextRight(p);
connectRecur(p.right);
}
else
{
connectRecur(getNextRight(p));
}
}
/* This function returns the leftmost child of nodes at the same
level as p. This function is used to getNExt right of p's right child
If right child of p is NULL then this can also be used for
the left child */
public virtual Node getNextRight(Node p)
{
Node temp = p.nextRight;
/* Traverse nodes at p's level and find and return
the first node's first child */
while (temp != null)
{
if (temp.left != null)
{
return temp.left;
}
if (temp.right != null)
{
return temp.right;
}
temp = temp.nextRight;
}
// If all the nodes at p's level are leaf nodes then return NULL
return null;
}
/* Driver program to test the above functions */
public static void Main(string[] args)
{
BinaryTree tree = new BinaryTree();
tree.root = new Node(10);
tree.root.left = new Node(8);
tree.root.right = new Node(2);
tree.root.left.left = new Node(3);
tree.root.right.right = new Node(90);
// Populates nextRight pointer in all nodes
tree.connectRecur(tree.root);
// Let us check the values of nextRight pointers
int a = tree.root.nextRight != null ? tree.root.nextRight.data : -1;
int b = tree.root.left.nextRight != null ? tree.root.left.nextRight.data : -1;
int c = tree.root.right.nextRight != null ? tree.root.right.nextRight.data : -1;
int d = tree.root.left.left.nextRight != null ? tree.root.left.left.nextRight.data : -1;
int e = tree.root.right.right.nextRight != null ? tree.root.right.right.nextRight.data : -1;
// Now lets print the values
Console.WriteLine("Following are populated nextRight pointers in the tree(-1 is printed if there is no nextRight)");
Console.WriteLine("nextRight of " + tree.root.data + " is " + a);
Console.WriteLine("nextRight of " + tree.root.left.data + " is " + b);
Console.WriteLine("nextRight of " + tree.root.right.data + " is " + c);
Console.WriteLine("nextRight of " + tree.root.left.left.data + " is " + d);
Console.WriteLine("nextRight of " + tree.root.right.right.data + " is " + e);
}
}
// This code is contributed by Shrikant13
Javascript
<script>
// Recursive javascript program to connect nodes at same level
// using constant extra space
// A binary tree node
class Node {
constructor(val) {
this.data = val;
this.left = null;
this.right = null;
this.nextRight = null;
}
}
var root;
/* Set next right of all descendants of p. This function makes sure that
nextRight of nodes ar level i is set before level i+1 nodes. */
function connectRecur(p)
{
// Base case
if (p == null)
return;
/* Before setting nextRight of left and right children, set nextRight
of children of other nodes at same level (because we can access
children of other nodes using p's nextRight only) */
if (p.nextRight != null)
connectRecur(p.nextRight);
/* Set the nextRight pointer for p's left child */
if (p.left != null)
{
if (p.right != null)
{
p.left.nextRight = p.right;
p.right.nextRight = getNextRight(p);
}
else
p.left.nextRight = getNextRight(p);
/* Recursively call for next level nodes. Note that we call only
for left child. The call for left child will call for right child */
connectRecur(p.left);
}
/* If left child is NULL then first node of next level will either be
p->right or getNextRight(p) */
else if (p.right != null)
{
p.right.nextRight = getNextRight(p);
connectRecur(p.right);
}
else
connectRecur(getNextRight(p));
}
/* This function returns the leftmost child of nodes at the same
level as p. This function is used to getNExt right of p's right child
If right child of p is NULL then this can also be used for
the left child */
function getNextRight(p)
{
var temp = p.nextRight;
/* Traverse nodes at p's level and find and return
the first node's first child */
while (temp != null)
{
if (temp.left != null)
return temp.left;
if (temp.right != null)
return temp.right;
temp = temp.nextRight;
}
// If all the nodes at p's level are leaf nodes then return NULL
return null;
}
/* Driver program to test the above functions */
var root = new Node(10);
root.left = new Node(8);
root.right = new Node(2);
root.left.left = new Node(3);
root.right.right = new Node(90);
// Populates nextRight pointer in all nodes
connectRecur(root);
// Let us check the values of nextRight pointers
var a = root.nextRight != null ?
root.nextRight.data : -1;
var b = root.left.nextRight != null ?
root.left.nextRight.data : -1;
var c = root.right.nextRight != null ?
root.right.nextRight.data : -1;
var d = root.left.left.nextRight != null ?
root.left.left.nextRight.data : -1;
var e = root.right.right.nextRight != null ?
root.right.right.nextRight.data : -1;
// Now lets print the values
document.write("Following are populated nextRight pointers in "
+ " the tree(-1 is printed if there is no nextRight)");
document.write("<br/>nextRight of " + root.data + " is " + a);
document.write("<br/>nextRight of " + root.left.data + " is " + b);
document.write("<br/>nextRight of " + root.right.data + " is " + c);
document.write("<br/>nextRight of " + root.left.left.data +
" is " + d);
document.write("<br/>nextRight of " + root.right.right.data +
" is " + e);
// This code is contributed by umadevi9616
</script>
C++
// Iterative CPP program to connect
// nodes at same level using
// constant extra space
#include<bits/stdc++.h>
#include<bits/stdc++.h>
using namespace std;
class node
{
public:
int data;
node* left;
node* right;
node *nextRight;
/* Constructor that allocates a new node with the
given data and NULL left and right pointers. */
node(int data)
{
this->data = data;
this->left = NULL;
this->right = NULL;
this->nextRight = NULL;
}
};
/* This function returns the leftmost
child of nodes at the same level as p.
This function is used to getNExt right
of p's right child If right child of
is NULL then this can also be used for
the left child */
node *getNextRight(node *p)
{
node *temp = p->nextRight;
/* Traverse nodes at p's level
and find and return the first
node's first child */
while (temp != NULL)
{
if (temp->left != NULL)
return temp->left;
if (temp->right != NULL)
return temp->right;
temp = temp->nextRight;
}
// If all the nodes at p's level
// are leaf nodes then return NULL
return NULL;
}
/* Sets nextRight of all nodes
of a tree with root as p */
void connectRecur(node* p)
{
node *temp;
if (!p)
return;
// Set nextRight for root
p->nextRight = NULL;
// set nextRight of all levels one by one
while (p != NULL)
{
node *q = p;
/* Connect all children nodes of p and
children nodes of all other nodes at
same level as p */
while (q != NULL)
{
// Set the nextRight pointer
// for p's left child
if (q->left)
{
// If q has right child, then
// right child is nextRight of
// p and we also need to set
// nextRight of right child
if (q->right)
q->left->nextRight = q->right;
else
q->left->nextRight = getNextRight(q);
}
if (q->right)
q->right->nextRight = getNextRight(q);
// Set nextRight for other
// nodes in pre order fashion
q = q->nextRight;
}
// start from the first
// node of next level
if (p->left)
p = p->left;
else if (p->right)
p = p->right;
else
p = getNextRight(p);
}
}
/* Driver code*/
int main()
{
/* Constructed binary tree is
10
/ \
8 2
/ \
3 90
*/
node *root = new node(10);
root->left = new node(8);
root->right = new node(2);
root->left->left = new node(3);
root->right->right = new node(90);
// Populates nextRight pointer in all nodes
connectRecur(root);
// Let us check the values of nextRight pointers
cout << "Following are populated nextRight pointers in the tree"
" (-1 is printed if there is no nextRight) \n";
cout << "nextRight of " << root->data << " is "
<< (root->nextRight? root->nextRight->data: -1) <<endl;
cout << "nextRight of " << root->left->data << " is "
<< (root->left->nextRight? root->left->nextRight->data: -1) << endl;
cout << "nextRight of " << root->right->data << " is "
<< (root->right->nextRight? root->right->nextRight->data: -1) << endl;
cout << "nextRight of " << root->left->left->data<< " is "
<< (root->left->left->nextRight? root->left->left->nextRight->data: -1) << endl;
cout << "nextRight of " << root->right->right->data << " is "
<< (root->right->right->nextRight? root->right->right->nextRight->data: -1) << endl;
return 0;
}
// This code is contributed by rathbhupendra
C
// Iterative C program to connect nodes at same level
// using constant extra space
#include <stdio.h>
#include <stdlib.h>
struct node
{
int data;
struct node *left;
struct node *right;
struct node *nextRight;
};
/* This function returns the leftmost child of nodes at the same level as p.
This function is used to getNExt right of p's right child
If right child of is NULL then this can also be used for the left child */
struct node *getNextRight(struct node *p)
{
struct node *temp = p->nextRight;
/* Traverse nodes at p's level and find and return
the first node's first child */
while (temp != NULL)
{
if (temp->left != NULL)
return temp->left;
if (temp->right != NULL)
return temp->right;
temp = temp->nextRight;
}
// If all the nodes at p's level are leaf nodes then return NULL
return NULL;
}
/* Sets nextRight of all nodes of a tree with root as p */
void connect(struct node* p)
{
struct node *temp;
if (!p)
return;
// Set nextRight for root
p->nextRight = NULL;
// set nextRight of all levels one by one
while (p != NULL)
{
struct node *q = p;
/* Connect all children nodes of p and children nodes of all other nodes
at same level as p */
while (q != NULL)
{
// Set the nextRight pointer for p's left child
if (q->left)
{
// If q has right child, then right child is nextRight of
// p and we also need to set nextRight of right child
if (q->right)
q->left->nextRight = q->right;
else
q->left->nextRight = getNextRight(q);
}
if (q->right)
q->right->nextRight = getNextRight(q);
// Set nextRight for other nodes in pre order fashion
q = q->nextRight;
}
// start from the first node of next level
if (p->left)
p = p->left;
else if (p->right)
p = p->right;
else
p = getNextRight(p);
}
}
/* UTILITY FUNCTIONS */
/* 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 = (struct node*)
malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
node->nextRight = NULL;
return(node);
}
/* Driver program to test above functions*/
int main()
{
/* Constructed binary tree is
10
/ \
8 2
/ \
3 90
*/
struct node *root = newnode(10);
root->left = newnode(8);
root->right = newnode(2);
root->left->left = newnode(3);
root->right->right = newnode(90);
// Populates nextRight pointer in all nodes
connect(root);
// Let us check the values of nextRight pointers
printf("Following are populated nextRight pointers in the tree "
"(-1 is printed if there is no nextRight) \n");
printf("nextRight of %d is %d \n", root->data,
root->nextRight? root->nextRight->data: -1);
printf("nextRight of %d is %d \n", root->left->data,
root->left->nextRight? root->left->nextRight->data: -1);
printf("nextRight of %d is %d \n", root->right->data,
root->right->nextRight? root->right->nextRight->data: -1);
printf("nextRight of %d is %d \n", root->left->left->data,
root->left->left->nextRight? root->left->left->nextRight->data: -1);
printf("nextRight of %d is %d \n", root->right->right->data,
root->right->right->nextRight? root->right->right->nextRight->data: -1);
getchar();
return 0;
}
Java
// Iterative Java program to connect nodes at same level
// using constant extra space
// A binary tree node
class Node
{
int data;
Node left, right, nextRight;
Node(int item)
{
data = item;
left = right = nextRight = null;
}
}
class BinaryTree
{
Node root;
/* This function returns the leftmost child of nodes at the same level
as p. This function is used to getNExt right of p's right child
If right child of is NULL then this can also be used for the
left child */
Node getNextRight(Node p)
{
Node temp = p.nextRight;
/* Traverse nodes at p's level and find and return
the first node's first child */
while (temp != null)
{
if (temp.left != null)
return temp.left;
if (temp.right != null)
return temp.right;
temp = temp.nextRight;
}
// If all the nodes at p's level are leaf nodes then return NULL
return null;
}
/* Sets nextRight of all nodes of a tree with root as p */
void connect(Node p) {
Node temp = null;
if (p == null)
return;
// Set nextRight for root
p.nextRight = null;
// set nextRight of all levels one by one
while (p != null)
{
Node q = p;
/* Connect all children nodes of p and children nodes of all other
nodes at same level as p */
while (q != null)
{
// Set the nextRight pointer for p's left child
if (q.left != null)
{
// If q has right child, then right child is nextRight of
// p and we also need to set nextRight of right child
if (q.right != null)
q.left.nextRight = q.right;
else
q.left.nextRight = getNextRight(q);
}
if (q.right != null)
q.right.nextRight = getNextRight(q);
// Set nextRight for other nodes in pre order fashion
q = q.nextRight;
}
// start from the first node of next level
if (p.left != null)
p = p.left;
else if (p.right != null)
p = p.right;
else
p = getNextRight(p);
}
}
/* Driver program to test above functions */
public static void main(String args[])
{
/* Constructed binary tree is
10
/ \
8 2
/ \
3 90
*/
BinaryTree tree = new BinaryTree();
tree.root = new Node(10);
tree.root.left = new Node(8);
tree.root.right = new Node(2);
tree.root.left.left = new Node(3);
tree.root.right.right = new Node(90);
// Populates nextRight pointer in all nodes
tree.connect(tree.root);
// Let us check the values of nextRight pointers
int a = tree.root.nextRight != null ?
tree.root.nextRight.data : -1;
int b = tree.root.left.nextRight != null ?
tree.root.left.nextRight.data : -1;
int c = tree.root.right.nextRight != null ?
tree.root.right.nextRight.data : -1;
int d = tree.root.left.left.nextRight != null ?
tree.root.left.left.nextRight.data : -1;
int e = tree.root.right.right.nextRight != null ?
tree.root.right.right.nextRight.data : -1;
// Now lets print the values
System.out.println("Following are populated nextRight pointers in "
+ " the tree(-1 is printed if there is no nextRight)");
System.out.println("nextRight of " + tree.root.data + " is " + a);
System.out.println("nextRight of " + tree.root.left.data
+ " is " + b);
System.out.println("nextRight of " + tree.root.right.data +
" is " + c);
System.out.println("nextRight of " + tree.root.left.left.data +
" is " + d);
System.out.println("nextRight of " + tree.root.right.right.data +
" is " + e);
}
}
// This code has been contributed by Mayank Jaiswal
Python3
# Iterative Python program to connect nodes
# at same level using constant extra space
# 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
self.nextRight = None
# This function returns the leftmost child of
# nodes at the same level as p. This function
# is used to getNExt right of p's right child
# If right child of is None then this can also
# be used for the left child
def getNextRight(p):
temp = p.nextRight
# Traverse nodes at p's level and find
# and return the first node's first child
while (temp != None):
if (temp.left != None):
return temp.left
if (temp.right != None):
return temp.right
temp = temp.nextRight
# If all the nodes at p's level are
# leaf nodes then return None
return None
# Sets nextRight of all nodes of a tree
# with root as p
def connect(p):
temp = None
if (not p):
return
# Set nextRight for root
p.nextRight = None
# set nextRight of all levels one by one
while (p != None):
q = p
# Connect all children nodes of p and
# children nodes of all other nodes
# at same level as p
while (q != None):
# Set the nextRight pointer for
# p's left child
if (q.left):
# If q has right child, then right
# child is nextRight of p and we
# also need to set nextRight of
# right child
if (q.right):
q.left.nextRight = q.right
else:
q.left.nextRight = getNextRight(q)
if (q.right):
q.right.nextRight = getNextRight(q)
# Set nextRight for other nodes in
# pre order fashion
q = q.nextRight
# start from the first node
# of next level
if (p.left):
p = p.left
elif (p.right):
p = p.right
else:
p = getNextRight(p)
# Driver Code
if __name__ == '__main__':
# Constructed binary tree is
# 10
# / \
# 8 2
# / \
# 3 90
root = newnode(10)
root.left = newnode(8)
root.right = newnode(2)
root.left.left = newnode(3)
root.right.right = newnode(90)
# Populates nextRight pointer in all nodes
connect(root)
# Let us check the values of nextRight pointers
print("Following are populated nextRight "
"pointers in the tree (-1 is printed "
"if there is no nextRight) \n")
print("nextRight of", root.data,
"is", end = " ")
if root.nextRight:
print(root.nextRight.data)
else:
print(-1)
print("nextRight of", root.left.data,
"is", end = " ")
if root.left.nextRight:
print(root.left.nextRight.data)
else:
print(-1)
print("nextRight of", root.right.data,
"is", end = " ")
if root.right.nextRight:
print(root.right.nextRight.data)
else:
print(-1)
print("nextRight of", root.left.left.data,
"is", end = " ")
if root.left.left.nextRight:
print(root.left.left.nextRight.data)
else:
print(-1)
print("nextRight of", root.right.right.data,
"is", end = " ")
if root.right.right.nextRight:
print(root.right.right.nextRight.data)
else:
print(-1)
# This code is contributed by PranchalK
C#
using System;
// Iterative c# program to connect nodes at same level
// using constant extra space
// A binary tree node
public class Node
{
public int data;
public Node left, right, nextRight;
public Node(int item)
{
data = item;
left = right = nextRight = null;
}
}
public class BinaryTree
{
public Node root;
/* This function returns the leftmost child of nodes at the same level
as p. This function is used to getNExt right of p's right child
If right child of is NULL then this can also be used for the
left child */
public virtual Node getNextRight(Node p)
{
Node temp = p.nextRight;
/* Traverse nodes at p's level and find and return
the first node's first child */
while (temp != null)
{
if (temp.left != null)
{
return temp.left;
}
if (temp.right != null)
{
return temp.right;
}
temp = temp.nextRight;
}
// If all the nodes at p's level are leaf nodes then return NULL
return null;
}
/* Sets nextRight of all nodes of a tree with root as p */
public virtual void connect(Node p)
{
Node temp = null;
if (p == null)
{
return;
}
// Set nextRight for root
p.nextRight = null;
// set nextRight of all levels one by one
while (p != null)
{
Node q = p;
/* Connect all children nodes of p and children nodes of all other
nodes at same level as p */
while (q != null)
{
// Set the nextRight pointer for p's left child
if (q.left != null)
{
// If q has right child, then right child is nextRight of
// p and we also need to set nextRight of right child
if (q.right != null)
{
q.left.nextRight = q.right;
}
else
{
q.left.nextRight = getNextRight(q);
}
}
if (q.right != null)
{
q.right.nextRight = getNextRight(q);
}
// Set nextRight for other nodes in pre order fashion
q = q.nextRight;
}
// start from the first node of next level
if (p.left != null)
{
p = p.left;
}
else if (p.right != null)
{
p = p.right;
}
else
{
p = getNextRight(p);
}
}
}
/* Driver program to test above functions */
public static void Main(string[] args)
{
/* Constructed binary tree is
10
/ \
8 2
/ \
3 90
*/
BinaryTree tree = new BinaryTree();
tree.root = new Node(10);
tree.root.left = new Node(8);
tree.root.right = new Node(2);
tree.root.left.left = new Node(3);
tree.root.right.right = new Node(90);
// Populates nextRight pointer in all nodes
tree.connect(tree.root);
// Let us check the values of nextRight pointers
int a = tree.root.nextRight != null ? tree.root.nextRight.data : -1;
int b = tree.root.left.nextRight != null ? tree.root.left.nextRight.data : -1;
int c = tree.root.right.nextRight != null ? tree.root.right.nextRight.data : -1;
int d = tree.root.left.left.nextRight != null ? tree.root.left.left.nextRight.data : -1;
int e = tree.root.right.right.nextRight != null ? tree.root.right.right.nextRight.data : -1;
// Now lets print the values
Console.WriteLine("Following are populated nextRight pointers in " + " the tree(-1 is printed if there is no nextRight)");
Console.WriteLine("nextRight of " + tree.root.data + " is " + a);
Console.WriteLine("nextRight of " + tree.root.left.data + " is " + b);
Console.WriteLine("nextRight of " + tree.root.right.data + " is " + c);
Console.WriteLine("nextRight of " + tree.root.left.left.data + " is " + d);
Console.WriteLine("nextRight of " + tree.root.right.right.data + " is " + e);
}
}
// This code is contributed by Shrikant13
Javascript
<script>
// Iterative Javascript program to connect nodes at same level
// using constant extra space
class Node
{
constructor(item) {
this.data = item;
this.left = null;
this.right = null;
this.nextRight = null;
}
}
let root;
/* This function returns the leftmost
child of nodes at the same level
as p. This function is used to getNExt right of p's right child
If right child of is NULL then this can also be used for the
left child */
function getNextRight(p)
{
let temp = p.nextRight;
/* Traverse nodes at p's level and find and return
the first node's first child */
while (temp != null)
{
if (temp.left != null)
return temp.left;
if (temp.right != null)
return temp.right;
temp = temp.nextRight;
}
// If all the nodes at p's level are
// leaf nodes then return NULL
return null;
}
/* Sets nextRight of all nodes of a tree with root as p */
function connect(p) {
let temp = null;
if (p == null)
return;
// Set nextRight for root
p.nextRight = null;
// set nextRight of all levels one by one
while (p != null)
{
let q = p;
/* Connect all children nodes of p and
children nodes of all other
nodes at same level as p */
while (q != null)
{
// Set the nextRight pointer for p's left child
if (q.left != null)
{
// If q has right child, then
// right child is nextRight of
// p and we also need to set
// nextRight of right child
if (q.right != null)
q.left.nextRight = q.right;
else
q.left.nextRight = getNextRight(q);
}
if (q.right != null)
q.right.nextRight = getNextRight(q);
// Set nextRight for other nodes
// in pre order fashion
q = q.nextRight;
}
// start from the first node of next level
if (p.left != null)
p = p.left;
else if (p.right != null)
p = p.right;
else
p = getNextRight(p);
}
}
/* Constructed binary tree is
10
/ \
8 2
/ \
3 90
*/
root = new Node(10);
root.left = new Node(8);
root.right = new Node(2);
root.left.left = new Node(3);
root.right.right = new Node(90);
// Populates nextRight pointer in all nodes
connect(root);
// Let us check the values of nextRight pointers
let a = root.nextRight != null ?
root.nextRight.data : -1;
let b = root.left.nextRight != null ?
root.left.nextRight.data : -1;
let c = root.right.nextRight != null ?
root.right.nextRight.data : -1;
let d = root.left.left.nextRight != null ?
root.left.left.nextRight.data : -1;
let e = root.right.right.nextRight != null ?
root.right.right.nextRight.data : -1;
// Now lets print the values
document.write("Following are populated nextRight pointers in "
+ " the tree(-1 is printed if there is no nextRight)" +
"</br>");
document.write("nextRight of " + root.data + " is " + a +
"</br>");
document.write("nextRight of " + root.left.data
+ " is " + b + "</br>");
document.write("nextRight of " + root.right.data +
" is " + c + "</br>");
document.write("nextRight of " + root.left.left.data +
" is " + d + "</br>");
document.write("nextRight of " + root.right.right.data +
" is " + e + "</br>");
</script>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA