Given a binary tree with parent pointers, find the right sibling of a given node(pointer to the node will be given), if it doesn’t exist return null. Do it in O(1) space and O(n) time?
Examples:
1
/ \
2 3
/ \ \
4 6 5
/ \ \
7 9 8
/ \
10 12
Input : Given above tree with parent pointer and node 10
Output : 12
Approach: The idea is to find out the first right child of the nearest ancestor which is neither the current node nor the parent of the current node, keep track of the level in those while going up. then, iterate through that node first left child, if the left is not there then, right child and if the level becomes 0, then, this is the next right sibling of the given node.
In the above case, if the given node is 7, we will end up with 6 to find the right child which doesn't have any child.
In this case, we need to recursively call for the right sibling with the current level, so that we case reach 8.
Implementation:
// C program to print right sibling of a node
#include <bits/stdc++.h>
// A Binary Tree Node
struct Node {
int data;
Node *left, *right, *parent;
};
// A utility function to create a new Binary
// Tree Node
Node* newNode(int item, Node* parent)
{
Node* temp = new Node;
temp->data = item;
temp->left = temp->right = NULL;
temp->parent = parent;
return temp;
}
// Method to find right sibling
Node* findRightSibling(Node* node, int level)
{
if (node == NULL || node->parent == NULL)
return NULL;
// GET Parent pointer whose right child is not
// a parent or itself of this node. There might
// be case when parent has no right child, but,
// current node is left child of the parent
// (second condition is for that).
while (node->parent->right == node
|| (node->parent->right == NULL
&& node->parent->left == node)) {
if (node->parent == NULL
|| node->parent->parent == NULL)
return NULL;
node = node->parent;
level--;
}
// Move to the required child, where right sibling
// can be present
node = node->parent->right;
if (node == NULL)
return NULL;
// find right sibling in the given subtree(from current
// node), when level will be 0
while (level < 0) {
// Iterate through subtree
if (node->left != NULL)
node = node->left;
else if (node->right != NULL)
node = node->right;
else
// if no child are there, we cannot have right
// sibling in this path
break;
level++;
}
if (level == 0)
return node;
// This is the case when we reach 9 node in the tree,
// where we need to again recursively find the right
// sibling
return findRightSibling(node, level);
}
// Driver Program to test above functions
int main()
{
Node* root = newNode(1, NULL);
root->left = newNode(2, root);
root->right = newNode(3, root);
root->left->left = newNode(4, root->left);
root->left->right = newNode(6, root->left);
root->left->left->left = newNode(7, root->left->left);
root->left->left->left->left = newNode(10, root->left->left->left);
root->left->right->right = newNode(9, root->left->right);
root->right->right = newNode(5, root->right);
root->right->right->right = newNode(8, root->right->right);
root->right->right->right->right = newNode(12, root->right->right->right);
// passing 10
Node* res = findRightSibling(root->left->left->left->left, 0);
if (res == NULL)
printf("No right sibling");
else
printf("%d", res->data);
return 0;
}
// Java program to print right sibling of a node
public class Right_Sibling {
// A Binary Tree Node
static class Node {
int data;
Node left, right, parent;
// Constructor
public Node(int data, Node parent)
{
this.data = data;
left = null;
right = null;
this.parent = parent;
}
};
// Method to find right sibling
static Node findRightSibling(Node node, int level)
{
if (node == null || node.parent == null)
return null;
// GET Parent pointer whose right child is not
// a parent or itself of this node. There might
// be case when parent has no right child, but,
// current node is left child of the parent
// (second condition is for that).
while (node.parent.right == node
|| (node.parent.right == null
&& node.parent.left == node)) {
if (node.parent == null)
return null;
node = node.parent;
level--;
}
// Move to the required child, where right sibling
// can be present
node = node.parent.right;
// find right sibling in the given subtree(from current
// node), when level will be 0
while (level < 0) {
// Iterate through subtree
if (node.left != null)
node = node.left;
else if (node.right != null)
node = node.right;
else
// if no child are there, we cannot have right
// sibling in this path
break;
level++;
}
if (level == 0)
return node;
// This is the case when we reach 9 node in the tree,
// where we need to again recursively find the right
// sibling
return findRightSibling(node, level);
}
// Driver Program to test above functions
public static void main(String args[])
{
Node root = new Node(1, null);
root.left = new Node(2, root);
root.right = new Node(3, root);
root.left.left = new Node(4, root.left);
root.left.right = new Node(6, root.left);
root.left.left.left = new Node(7, root.left.left);
root.left.left.left.left = new Node(10, root.left.left.left);
root.left.right.right = new Node(9, root.left.right);
root.right.right = new Node(5, root.right);
root.right.right.right = new Node(8, root.right.right);
root.right.right.right.right = new Node(12, root.right.right.right);
// passing 10
System.out.println(findRightSibling(root.left.left.left.left, 0).data);
}
}
// This code is contributed by Sumit Ghosh
# Python3 program to print right sibling
# of a node
# A class to create a new Binary
# Tree Node
class newNode:
def __init__(self, item, parent):
self.data = item
self.left = self.right = None
self.parent = parent
# Method to find right sibling
def findRightSibling(node, level):
if (node == None or node.parent == None):
return None
# GET Parent pointer whose right child is not
# a parent or itself of this node. There might
# be case when parent has no right child, but,
# current node is left child of the parent
# (second condition is for that).
while (node.parent.right == node or
(node.parent.right == None and
node.parent.left == node)):
if (node.parent == None):
return None
node = node.parent
level -= 1
# Move to the required child, where
# right sibling can be present
node = node.parent.right
# find right sibling in the given subtree
# (from current node), when level will be 0
while (level < 0):
# Iterate through subtree
if (node.left != None):
node = node.left
else if (node.right != None):
node = node.right
else:
# if no child are there, we cannot
# have right sibling in this path
break
level += 1
if (level == 0):
return node
# This is the case when we reach 9 node
# in the tree, where we need to again
# recursively find the right sibling
return findRightSibling(node, level)
# Driver Code
if __name__ == '__main__':
root = newNode(1, None)
root.left = newNode(2, root)
root.right = newNode(3, root)
root.left.left = newNode(4, root.left)
root.left.right = newNode(6, root.left)
root.left.left.left = newNode(7, root.left.left)
root.left.left.left.left = newNode(10, root.left.left.left)
root.left.right.right = newNode(9, root.left.right)
root.right.right = newNode(5, root.right)
root.right.right.right = newNode(8, root.right.right)
root.right.right.right.right = newNode(12, root.right.right.right)
# passing 10
res = findRightSibling(root.left.left.left.left, 0)
if (res == None):
print("No right sibling")
else:
print(res.data)
# This code is contributed by PranchalK
using System;
// C# program to print right sibling of a node
public class Right_Sibling {
// A Binary Tree Node
public class Node {
public int data;
public Node left, right, parent;
// Constructor
public Node(int data, Node parent)
{
this.data = data;
left = null;
right = null;
this.parent = parent;
}
}
// Method to find right sibling
public static Node findRightSibling(Node node, int level)
{
if (node == null || node.parent == null) {
return null;
}
// GET Parent pointer whose right child is not
// a parent or itself of this node. There might
// be case when parent has no right child, but,
// current node is left child of the parent
// (second condition is for that).
while (node.parent.right == node
|| (node.parent.right == null
&& node.parent.left == node)) {
if (node.parent == null
|| node.parent.parent == null) {
return null;
}
node = node.parent;
level--;
}
// Move to the required child, where right sibling
// can be present
node = node.parent.right;
// find right sibling in the given subtree(from current
// node), when level will be 0
while (level < 0) {
// Iterate through subtree
if (node.left != null) {
node = node.left;
}
else if (node.right != null) {
node = node.right;
}
else {
// if no child are there, we cannot have right
// sibling in this path
break;
}
level++;
}
if (level == 0) {
return node;
}
// This is the case when we reach 9 node in the tree,
// where we need to again recursively find the right
// sibling
return findRightSibling(node, level);
}
// Driver Program to test above functions
public static void Main(string[] args)
{
Node root = new Node(1, null);
root.left = new Node(2, root);
root.right = new Node(3, root);
root.left.left = new Node(4, root.left);
root.left.right = new Node(6, root.left);
root.left.left.left = new Node(7, root.left.left);
root.left.left.left.left = new Node(10, root.left.left.left);
root.left.right.right = new Node(9, root.left.right);
root.right.right = new Node(5, root.right);
root.right.right.right = new Node(8, root.right.right);
root.right.right.right.right = new Node(12, root.right.right.right);
// passing 10
Console.WriteLine(findRightSibling(root.left.left.left.left, 0).data);
}
}
// This code is contributed by Shrikant13
<script>
// Javascript program to print right sibling of a node
// A Binary Tree Node
class Node
{
constructor(data, parent) {
this.left = null;
this.right = null;
this.data = data;
this.parent = parent;
}
}
// Method to find right sibling
function findRightSibling(node, level)
{
if (node == null || node.parent == null)
return null;
// GET Parent pointer whose right child is not
// a parent or itself of this node. There might
// be case when parent has no right child, but,
// current node is left child of the parent
// (second condition is for that).
while (node.parent.right == node
|| (node.parent.right == null
&& node.parent.left == node)) {
if (node.parent == null)
return null;
node = node.parent;
level--;
}
// Move to the required child, where right sibling
// can be present
node = node.parent.right;
// find right sibling in the given subtree(from current
// node), when level will be 0
while (level < 0) {
// Iterate through subtree
if (node.left != null)
node = node.left;
else if (node.right != null)
node = node.right;
else
// if no child are there, we cannot have right
// sibling in this path
break;
level++;
}
if (level == 0)
return node;
// This is the case when we reach 9 node in the tree,
// where we need to again recursively find the right
// sibling
return findRightSibling(node, level);
}
let root = new Node(1, null);
root.left = new Node(2, root);
root.right = new Node(3, root);
root.left.left = new Node(4, root.left);
root.left.right = new Node(6, root.left);
root.left.left.left = new Node(7, root.left.left);
root.left.left.left.left = new Node(10, root.left.left.left);
root.left.right.right = new Node(9, root.left.right);
root.right.right = new Node(5, root.right);
root.right.right.right = new Node(8, root.right.right);
root.right.right.right.right = new Node(12, root.right.right.right);
// passing 10
document.write(findRightSibling(root.left.left.left.left, 0).data);
// This code is contributed by divyesh072019.
</script>
Output
12
Time Complexity: O(N)
Auxiliary Space: O(1)
