Display postOrderBefore(v) and postOrderAfter(v) in perfect binary tree, without using tree traversal methods(inOrder, PostOrder and preOrder)

Question

For example, Take n nodes from user and target element v. input = [1,2,3,4,5,6,7] where 1 is root. 2 & 3 are left and right child of 1. 4 & 5 are left & right child of 2. 6 & 7 are left & right child of 3.

if user give input v as 5 -> program should display 4 and 2 as output. Since, postOrder of the given binary tree will be -> [4,5,2,6,7,3,1]. Here postOrderBefore is 4 and postOrderAfter is 2 when v = 5.

Write a pseudo-code to find out before and after element of v. Also, calculate running time of code.

My code:

class Node:
    def __init__(self, key):
        self.left = None
        self.right = None
        self.val = key

# A function to do postorder tree traversal
def printPostorder(root):
 
    if root:
 
        # First recur on left child
        printPostorder(root.left)
 
        # the recur on right child
        printPostorder(root.right)
 
        # now print the data of node
        print(root.val),
        
print("Enter number of nodes: ")
n = int(input())
list = []

print("Enter nodes: ")

for i in range(0, n):
    ele = int(input())
    list.append(ele)

for i in range(0, n):    
     print("\n",list[i])

for i in range(0, n):
    root = Node(list[i])
    root.left = Node(list[i+1])
    root.right = Node(list[i+2])
    root.left.left = Node(list[i+3])
    root.left.right = Node(list[i+4])
    root.right.left = Node(list[i+5])
    root.right.right = Node(list[i+6])
    break
 
print ("\nPostorder traversal of binary tree is")
printPostorder(root)

Here n is 7.

Answer 1

postorderBefore(v):
    if v.right! = None:
      return v.right.data
    else:
      node = v.parent
      if node.right = v:
        return node.left.data
      else:
        while node!= None AND v == node.left:
           v = node
           node = node.parent
    if node == None:
       return node.data
    else 
       return node.left.data