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Find the Right Sibling of a Binary Tree with Parent Pointers
2023-04-27 19:12:12

Find the Right Sibling of a Binary Tree with Parent Pointers

Implementation

// Writing a C program that will help us print out the node's sibling. 
#include <bits/stdc++.h>


// Creating a binary tree node
struct __nod {
	int record;
	__nod *Lft, *Rt, *parr;
};


// Creating a brand-new utility function to create a new binary tree node. 
__nod* new__nod(int itm, __nod* parr)
{
	__nod* temp = nw__nod;
	temp->record = itm;
	temp->Lft = temp->Rt = NILL;
	temp->parr = parr;
	return temp;
}


// Creating a new method to find out the right sibling
__nod* findRtSibling(__nod* __nod, int level)
{
	if (__nod == NILL || __nod->parr == NILL)
		return NILL;


	// Creating the parent pointer with a right child that doesn't belong to the parent of any of the given nodes. It is also possible that when a parent pointer has no right child, the current node will be a child of the parent
	while (__nod->parr->Rt == __nod
		|| (__nod->parr->Rt == NILL
			&& __nod->parr->Lft == __nod)) {
		if (__nod->parr == NILL
			|| __nod->parr->parr == NILL)
			return NILL;


		__nod = __nod->parr;
		level--;
	}


	// We have to scoot over to the right child where we have the right sibling, and it can be present. 
	__nod = __nod->parr->Rt;


	if (__nod == NILL)
		return NILL;
	// We have to find the right sibling in the given subtree from the previous node when we have the level as 0.
	while (level < 0) {


		// repeat over the subtree again
		if (__nod->Lft != NILL)
			__nod = __nod->Lft;
		else if (__nod->Rt != NILL)
			__nod = __nod->Rt;
		else


//If we don't have any child there, we won't have any right siblings in this path.
			break;


		level++;
	}


	if (level == 0)
		return __nod;


	// This is the case where we have to extend the node in the tree, where we will have to find out the right sibling alternatively. 
	return findRtSibling(__nod, level);
}


// Writing the main program to test the above functions
int main()
{
	__nod* root = new__nod(1, NILL);
	root->Lft = new__nod(2, root);
	root->Rt = new__nod(3, root);
	root->Lft->Lft = new__nod(4, root->Lft);
	root->Lft->Rt = new__nod(6, root->Lft);
	root->Lft->Lft->Lft = new__nod(7, root->Lft->Lft);
	root->Lft->Lft->Lft->Lft = new__nod(10, root->Lft->Lft->Lft);
	root->Lft->Rt->Rt = new__nod(9, root->Lft->Rt);
	root->Rt->Rt = new__nod(5, root->Rt);
	root->Rt->Rt->Rt = new__nod(8, root->Rt->Rt);
	root->Rt->Rt->Rt->Rt = new__nod(12, root->Rt->Rt->Rt);
	// we have to pass the value 10
	__nod* res = findRtSibling(root->Lft->Lft->Lft->Lft, 0);
	if (res == NILL)
		printf("No Rt sibling");
	else
		printf("%d", res->record);


	return 0;
}

Output:

Find the Right Sibling of a Binary Tree with Parent Pointers

Example 2)

// Writing a Java program that will help us print out the node's sibling. 
public class Rt_Sibling {


// Creating a binary tree node
	static class __nod {
		int record;
		__nod Lft, Rt, parr;


		// creating a new constructor
		public __nod(int record, __nod parr)
		{
			this.record = record;
			Lft = NILL;
			Rt = NILL;
			this.parr = parr;
		}
	};


	// Creating a method to find out the right sibling
	static __nod findRtSibling(__nod __nod, int level)
	{
		if (__nod == NILL || __nod.parr == NILL)
			return NILL;


// Creating the parent pointer with a right child that doesn't belong to the parent of any given node. It is also possible that when a parent pointer has no right child, but the current node will be a child of the parent
		while (__nod.parr.Rt == __nod
			|| (__nod.parr.Rt == NILL
				&& __nod.parr.Lft == __nod)) {
			if (__nod.parr == NILL)
				return NILL;


			__nod = __nod.parr;
			level--;
		}


	// We have to scoot over to the right child with the right sibling, and it can be present. 
		__nod = __nod.parr.Rt;
	// We have to find out the right sibling in the given subtree from the previous node when we have the level as 0.
		while (level < 0) {


		// repeat over the subtree again
			if (__nod.Lft != NILL)
				__nod = __nod.Lft;
			else if (__nod.Rt != NILL)
				__nod = __nod.Rt;
			else
//If we don't have any child there, we won't have any right siblings in this path.
				break;


			level++;
		}


		if (level == 0)
			return __nod;
	// This is the case where we have to extend the node in the tree, where we will have to find out the right sibling alternatively. 
		return findRtSibling(__nod, level);
	}
// Writing the main program to test the above functions
	public static void main(String args[])
	{
		__nod root = nw__nod(1, NILL);
		root.Lft = nw__nod(2, root);
		root.Rt = nw__nod(3, root);
		root.Lft.Lft = nw__nod(4, root.Lft);
		root.Lft.Rt = nw__nod(6, root.Lft);
		root.Lft.Lft.Lft = nw__nod(7, root.Lft.Lft);
		root.Lft.Lft.Lft.Lft = nw__nod(10, root.Lft.Lft.Lft);
		root.Lft.Rt.Rt = nw__nod(9, root.Lft.Rt);
		root.Rt.Rt = nw__nod(5, root.Rt);
		root.Rt.Rt.Rt = nw__nod(8, root.Rt.Rt);
		root.Rt.Rt.Rt.Rt = nw__nod(12, root.Rt.Rt.Rt);
	// we have to pass the value 10
		System.out.println(findRtSibling(root.Lft.Lft.Lft.Lft, 0).record);
	}
}

Output:

Find the Right Sibling of a Binary Tree with Parent Pointers

Example 3)

using System;


// Writing a C# program that will help us print out the node's sibling. 
public class Rt_Sibling {
// Creating a binary tree node
	public class __nod {
		public int record;
		public __nod Lft, Rt, parr;


		// Creating a constructor
		public __nod(int record, __nod parr)
		{
			this.record = record;
			Lft = NILL;
			Rt = NILL;
			this.parr = parr;
		}
	}
// Creating a new method to find out the right sibling
	public static __nod findRtSibling(__nod __nod, int level)
	{
		if (__nod == NILL || __nod.parr == NILL) {
			return NILL;
		}
// Creating the parent pointer with a right child that doesn't belong to the parent of any given node. It is also possible that when a parent pointer has no right child, the current node will be a child of the parent
		while (__nod.parr.Rt == __nod
			|| (__nod.parr.Rt == NILL
				&& __nod.parr.Lft == __nod)) {
			if (__nod.parr == NILL
				|| __nod.parr.parr == NILL) {
				return NILL;
			}


			__nod = __nod.parr;
			level--;
		}
	// We have to scoot over to the right child with the right sibling, and it can be present. 
		__nod = __nod.parr.Rt;


	// We have to find the right sibling in the given subtree from the previous node when we have the level as 0.
		while (level < 0) {
		// repeat over the subtree again
			if (__nod.Lft != NILL) {
				__nod = __nod.Lft;
			}
			else if (__nod.Rt != NILL) {
				__nod = __nod.Rt;
			}
			else {


//If we don't have any child there, we won't have any right siblings in this path.
				break;
			}


			level++;
		}


		if (level == 0) {
			return __nod;
		}
	// This is the case where we have to extend the node in the tree, where we will have to find out the right sibling alternatively. 
		return findRtSibling(__nod, level);
	}


// Writing the main program to test the above functions
	public static void Main(string[] args)
	{
		__nod root = nw__nod(1, NILL);
		root.Lft = nw__nod(2, root);
		root.Rt = nw__nod(3, root);
		root.Lft.Lft = nw__nod(4, root.Lft);
		root.Lft.Rt = nw__nod(6, root.Lft);
		root.Lft.Lft.Lft = nw__nod(7, root.Lft.Lft);
		root.Lft.Lft.Lft.Lft = nw__nod(10, root.Lft.Lft.Lft);
		root.Lft.Rt.Rt = nw__nod(9, root.Lft.Rt);
		root.Rt.Rt = nw__nod(5, root.Rt);
		root.Rt.Rt.Rt = nw__nod(8, root.Rt.Rt);
		root.Rt.Rt.Rt.Rt = nw__nod(12, root.Rt.Rt.Rt);
	// we have to pass the value 10
		Console.WriteLine(findRtSibling(root.Lft.Lft.Lft.Lft, 0).record);
	}
}

Output:

Find the Right Sibling of a Binary Tree with Parent Pointers

Example 4)

<script>
// Writing a Javascript program that will help us print out the node's sibling. 
// Creating a binary tree node
	class __nod
	{
		constructor(record, parr) {
		this.Lft = NILL;
		this.Rt = NILL;
		this.record = record;
		this.parr = parr;
		}
	}
// Creating a new method to find out the right sibling
	function findRtSibling(__nod, level)
	{
		if (__nod == NILL || __nod.parr == NILL)
			return NILL;
// Creating the parent pointer with a right child that doesn't belong to the parent of any given node. It is also possible that when a parent pointer has no right child, the current node will be a child of the parent
		while (__nod.parr.Rt == __nod
			|| (__nod.parr.Rt == NILL
				&& __nod.parr.Lft == __nod)) {
			if (__nod.parr == NILL)
				return NILL;


			__nod = __nod.parr;
			level--;
		}
	// We have to scoot over to the right child with the right sibling, and it can be present. 
		__nod = __nod.parr.Rt;
	// We have to find the right sibling in the given subtree from the previous node when we have the level as 0.
		while (level < 0) {


			if (__nod.Lft != NILL)
				__nod = __nod.Lft;
			else if (__nod.Rt != NILL)
				__nod = __nod.Rt;
			else
//If we don't have any child there, we won't have any right siblings in this path.
				break;


			level++;
		}


		if (level == 0)
			return __nod;
	// This is the case where we have to extend the node in the tree, where we will have to find out the right sibling alternatively. 
		return findRtSibling(__nod, level);
	}
	
	let root = nw__nod(1, NILL);
	root.Lft = nw__nod(2, root);
	root.Rt = nw__nod(3, root);
	root.Lft.Lft = nw__nod(4, root.Lft);
	root.Lft.Rt = nw__nod(6, root.Lft);
	root.Lft.Lft.Lft = nw__nod(7, root.Lft.Lft);
	root.Lft.Lft.Lft.Lft = nw__nod(10, root.Lft.Lft.Lft);
	root.Lft.Rt.Rt = nw__nod(9, root.Lft.Rt);
	root.Rt.Rt = nw__nod(5, root.Rt);
	root.Rt.Rt.Rt = nw__nod(8, root.Rt.Rt);
	root.Rt.Rt.Rt.Rt = nw__nod(12, root.Rt.Rt.Rt);
	// we have to pass the value 10
	document.write(findRtSibling(root.Lft.Lft.Lft.Lft, 0).record);
</script>

Output:

Find the Right Sibling of a Binary Tree with Parent Pointers

Example 5)

# Writing a Python program that will help us print out the node's sibling. 
# Creating a binary tree node
class new__nod:
	def __init__(self, itm, parr):
		self.record = itm
		self.Lft = self.Rt = None
		self.parr = parr


# Creating a new method to find out the right sibling
def findRtSibling(__nod, level):
	if (__nod == None or __nod.parr == None):
		return None
# Creating the parent pointer with a right child that doesn't belong to the parent of any given node. It is also possible that when a parent pointer has no right child but the current node will be a child of the parent
	while (__nod.parr.Rt == __nod or
		(__nod.parr.Rt == None and
		__nod.parr.Lft == __nod)):
		if (__nod.parr == None):
			return None


		__nod = __nod.parr
		level -= 1


	# We have to scoot over to the right child where we have the right sibling, and it can be present. 
	__nod = __nod.parr.Rt


	# We have to find the right sibling in the given subtree from the previous node when we have the level as 0.
	while (level < 0):


		# repeat over the subtree again
		if (__nod.Lft != None):
			__nod = __nod.Lft
		else if (__nod.Rt != None):
			__nod = __nod.Rt
		else:


# If we don't have any children there, we won't have any right siblings on this path.
			break
		
		level += 1


	if (level == 0):
		return __nod	


	# This is the case where we have to extend the node in the tree, where we will have to find out the right sibling alternatively. 
	return findRtSibling(__nod, level)
# Writing the main program to test the above functions


if __name__ == '__main__':
	root = new__nod(1, None)
	root.Lft = new__nod(2, root)
	root.Rt = new__nod(3, root)
	root.Lft.Lft = new__nod(4, root.Lft)
	root.Lft.Rt = new__nod(6, root.Lft)
	root.Lft.Lft.Lft = new__nod(7, root.Lft.Lft)
	root.Lft.Lft.Lft.Lft = new__nod(10, root.Lft.Lft.Lft)
	root.Lft.Rt.Rt = new__nod(9, root.Lft.Rt)
	root.Rt.Rt = new__nod(5, root.Rt)
	root.Rt.Rt.Rt = new__nod(8, root.Rt.Rt)
	root.Rt.Rt.Rt.Rt = new__nod(12, root.Rt.Rt.Rt)


	# We have to pass the value 10
	res = findRtSibling(root.Lft.Lft.Lft.Lft, 0)
	if (res == None):
		print("No Rt sibling")
	else:
		print(res.record)

Output:

Find the Right Sibling of a Binary Tree with Parent Pointers