Answer to Question #204552 in Python for Ryukjak

Question

The link below is the photo of the problem. Please create a code without using def

Determine in order, post order and pre order

https://www.bartleby.com/questions-and-answers/52-37-64-24-39-59-73-19-29-56-in-order-pre-order-post-order/100dd94b-74dc-49e0-a700-97b1a7f6afbd

Answer 1

Algorithm Inorder(tree)
   1. Traverse the left subtree, i.e., call Inorder(left-subtree)
   2. Visit the root.
   3. Traverse the right subtree, i.e., call Inorder(right-subtree)

Algorithm Preorder(tree)
   1. Visit the root.
   2. Traverse the left subtree, i.e., call Preorder(left-subtree)
   3. Traverse the right subtree, i.e., call Preorder(right-subtree) 

Algorithm Postorder(tree)
   1. Traverse the left subtree, i.e., call Postorder(left-subtree)
   2. Traverse the right subtree, i.e., call Postorder(right-subtree)
   3. Visit the root.

class Node:
    def __init__(self, key):
        self.left = None
        self.right = None
        self.val = key
 
def printInorder(root):
    if root:
        printInorder(root.left)
        print(root.val, end=' ')
        printInorder(root.right)
 
def printPostorder(root):
    if root:
        printPostorder(root.left)
        printPostorder(root.right)
        print(root.val, end=' ')
 
def printPreorder(root):
     if root:
        print(root.val, end=' ')
        printPreorder(root.left)
        printPreorder(root.right)
 
root = Node(52)
root.left = Node(37)
root.right = Node(64)
root.left.left = Node(24)
root.left.right = Node(39)
root.right.left = Node(59)
root.right.right = Node(73)
root.left.left.left = Node(19)
root.left.left.right = Node(29)
root.right.left.left = Node(56)

print("\nPreorder traversal of binary tree is")
printPreorder(root)
 
print("\nInorder traversal of binary tree is")
printInorder(root)
 
print("\nPostorder traversal of binary tree is")
printPostorder(root)