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# Find parent of given node in a Binary Tree with given postorder traversal

• Difficulty Level : Hard
• Last Updated : 15 Mar, 2023

Given two integers N and K where N denotes the height of a binary tree, the task is to find the parent of the node with value K in a binary tree whose postorder traversal is first

natural numbers

For N = 3, the Tree will be -

7
/   \
3     6
/   \  /  \
1     2 4   5

Examples:

Input: N = 4, K = 5
Output:
Explanation:
Parent of the node 5 is 6. As shown in the tree above.
Input: N = 5, K = 3
Output:
Explanation:
Parent of the node 3 is 7. As shown in the tree above.

Naive Approach: A simple approach is to build the tree according to the following pattern and then traverse the whole tree to find the parent of a given node.
Efficient Approach: The idea is to use a binary search to find the parent of the node. As we know the binary Tree of Height N has

nodes. Therefore, the search space for the binary search will be 1 to

Now each node has children value either

or

Therefore, parents of such nodes can be found easily.

Algorithm:

Step 1: Start
Step 2: Create a function called “findParent” that has two inputs: height and node. This function returns a number that represents                the binary tree’s parent node for the specified node.
a. Set the initial values of the two variables “start” and “end” to 1 and 2height – 1, respectively.
b. Verify whether the specified node is the binary tree’s root node. Return -1 if the specified node is the root node because                   the root node has no parents.
c. Start a loop that keeps going until the specified node is discovered.
1. In the loop, as we ascend to the parent node, deduct 1 from the “end” variable.
2. Get the average of the “start” and “end” variables to determine the center node of the current range of nodes.
3. Verify whether the range’s final node or the intermediate node is the same as the specified node. Return the parent                          node that is equal to (end + 1) if either of the conditions is true.
4. Verify that the provided node is lower than the center node. If so, change the value of the “end” variable to the middle                      node.
5. If not, change the “start” variable such that it now represents the middle node.
6. If the specified node is not located, repeat steps 5â€“9 until it is.
7. Lastly, give the supplied node’s parent node back.
Step 3: End

Below is the implementation of the above approach:

## C++

 // C++ implementation to find the // parent of the given node K in // a binary tree whose post-order // traversal is N natural numbers   #include  using namespace std;   // Function to find the parent // of the given node int findParent(int height, int node) {     int start = 1;     int end = pow(2, height) - 1;       // Condition to check whether     // the given node is a root node.     // if it is then return -1 because     // root node has no parent     if (end == node)         return -1;       // Loop till we found     // the given node     while (node >= 1) {         end = end - 1;           // Finding the middle node of the         // tree because at every level         // tree parent is         // divided into two halves         int mid = start                   + (end - start)                         / 2;           // if the node is found return         // the parent always the child         // nodes of every node         // is node/2 or (node-1)         if (mid == node || end == node) {             return (end + 1);         }           // if the node to be found         // is greater than the mid         // search for left subtree else         // search in right subtree         else if (node < mid) {             end = mid;         }         else {             start = mid;         }     } }   // Driver Code int main() {     int height = 4;     int node = 6;       int k = findParent(height, node);     cout << k;       return 0; }

## Java

 // Java implementation to find the // parent of the given node K in // a binary tree whose post-order // traversal is N natural numbers import java.util.*; class GFG{   // Function to find the parent // of the given node static int findParent(int height,                        int node) {   int start = 1;   int end = (int)Math.pow(2, height) - 1;     // Condition to check whether   // the given node is a root node.   // if it is then return -1 because   // root node has no parent   if (end == node)     return -1;     // Loop till we found   // the given node   while (node >= 1)    {     end = end - 1;       // Finding the middle node of the     // tree because at every level     // tree parent is     // divided into two halves     int mid = start + (end - start) / 2;       // if the node is found return     // the parent always the child     // nodes of every node     // is node*/2 or (node-1)     if (mid == node || end == node)      {       return (end + 1);     }       // if the node to be found     // is greater than the mid     // search for left subtree else     // search in right subtree     else if (node < mid)      {       end = mid;     }     else     {       start = mid;     }   }   return -1; }   // Driver Code public static void main(String[] args) {   int height = 4;   int node = 6;   int k = findParent(height, node);   System.out.print(k); } }   // This code is contributed by gauravrajput1

## Python3

 # Python implementation to find the # parent of the given node   import math    # Function to find the parent  # of the given node def findParent(height, node):       start = 1     end = pow(2, height) - 1       # Check whether the given node      # is a root node.if it is then      # return -1 because root      # node has no parent     if (end == node):         return -1       # Loop till we found      # the given node     while(node >= 1):           end = end - 1           # Find the middle node of the          # tree because at every level          # tree parent is divided          # into two halves         mid = start + (end - start)//2           # if the node is found          # return the parent         # always the child nodes of every         # node is node / 2 or (node-1)          if(mid == node or end == node):             return (end + 1)                   # if the node to be found is greater         # than the mid search for left         # subtree else search in right subtree         elif (node < mid):             end = mid           else:             start = mid   # Driver code if __name__ == "__main__":     height = 4     node = 6           # Function Call     k = findParent(height, node)     print(k)

## C#

 // C# implementation to find the // parent of the given node K in // a binary tree whose post-order // traversal is N natural numbers using System; class GFG{   // Function to find the parent // of the given node static int findParent(int height,                        int node) {   int start = 1;   int end = (int)Math.Pow(2, height) - 1;     // Condition to check whether   // the given node is a root node.   // if it is then return -1 because   // root node has no parent   if (end == node)     return -1;     // Loop till we found   // the given node   while (node >= 1)    {     end = end - 1;       // Finding the middle node of the     // tree because at every level     // tree parent is     // divided into two halves     int mid = start + (end - start) / 2;       // if the node is found return     // the parent always the child     // nodes of every node     // is node*/2 or (node-1)     if (mid == node || end == node)      {       return (end + 1);     }       // if the node to be found     // is greater than the mid     // search for left subtree else     // search in right subtree     else if (node < mid)      {       end = mid;     }     else     {       start = mid;     }   }   return -1; }   // Driver Code public static void Main(String[] args) {   int height = 4;   int node = 6;   int k = findParent(height, node);   Console.Write(k); } }   // This code is contributed by Princi Singh

## Javascript

 

Output:

7

Time complexity: O(log n) where n is no of nodes in binary tree

Auxiliary Space: O(1)

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