{"id":27638,"date":"2018-04-09T20:17:42","date_gmt":"2018-04-09T14:47:42","guid":{"rendered":"https:\/\/www.wikitechy.com\/technology\/?p=27638"},"modified":"2018-09-14T19:08:19","modified_gmt":"2018-09-14T13:38:19","slug":"python-algorithm-iterative-postorder-traversal-set-2-using-one-stack","status":"publish","type":"post","link":"https:\/\/www.wikitechy.com\/technology\/python-algorithm-iterative-postorder-traversal-set-2-using-one-stack\/","title":{"rendered":"Python Algorithm – Iterative Postorder Traversal | Set 2 (Using One Stack)"},"content":{"rendered":"

We have discussed a simple iterative postorder traversal using two stacks in the previous post. In this post, an approach with only one stack is discussed.<\/span><\/p>\n

The idea is to move down to leftmost node using left pointer. While moving down, push root and root\u2019s right child to stack. Once we reach leftmost node, print it if it doesn\u2019t have a right child. If it has a right child, then change root so that the right child is processed before.<\/p>\n

Following is detailed algorithm.<\/p>\n

1.1 Create an empty stack\r\n2.1 Do following while root is not NULL\r\n    a) Push root's right child and then root to stack.\r\n    b) Set root as root's left child.\r\n2.2 Pop an item from stack and set it as root.\r\n    a) If the popped item has a right child and the right child \r\n       is at top of stack, then remove the right child from stack,\r\n       push the root back and set root as root's right child.\r\n    b) Else print root's data and set root as NULL.\r\n2.3 Repeat steps 2.1 and 2.2 while stack is not empty.<\/pre>\n
Let us consider the following tree<\/p>\n
Following are the steps to print postorder traversal of the above tree using one stack.<\/p>\n
1. Right child of 1 exists. \r\n   Push 3 to stack. Push 1 to stack. Move to left child.\r\n        Stack: 3, 1\r\n\r\n2. Right child of 2 exists. \r\n   Push 5 to stack. Push 2 to stack. Move to left child.\r\n        Stack: 3, 1, 5, 2\r\n\r\n3. Right child of 4 doesn't exist. '\r\n   Push 4 to stack. Move to left child.\r\n        Stack: 3, 1, 5, 2, 4\r\n\r\n4. Current node is NULL. \r\n   Pop 4 from stack. Right child of 4 doesn't exist. \r\n   Print 4. Set current node to NULL.\r\n        Stack: 3, 1, 5, 2\r\n\r\n5. Current node is NULL. \r\n    Pop 2 from stack. Since right child of 2 equals stack top element, \r\n    pop 5 from stack. Now push 2 to stack.     \r\n    Move current node to right child of 2 i.e. 5\r\n        Stack: 3, 1, 2\r\n\r\n6. Right child of 5 doesn't exist. Push 5 to stack. Move to left child.\r\n        Stack: 3, 1, 2, 5\r\n\r\n7. Current node is NULL. Pop 5 from stack. Right child of 5 doesn't exist. \r\n   Print 5. Set current node to NULL.\r\n        Stack: 3, 1, 2\r\n\r\n8. Current node is NULL. Pop 2 from stack. \r\n   Right child of 2 is not equal to stack top element. \r\n   Print 2. Set current node to NULL.\r\n        Stack: 3, 1\r\n\r\n9. Current node is NULL. Pop 1 from stack. \r\n   Since right child of 1 equals stack top element, pop 3 from stack. \r\n   Now push 1 to stack. Move current node to right child of 1 i.e. 3\r\n        Stack: 1\r\n\r\n10. Repeat the same as above steps and Print 6, 7 and 3. \r\n    Pop 1 and Print 1.<\/pre>\n
Python Programming:<\/strong><\/p>\n
 
 <\/div> 
# Python program for iterative postorder traversal# using one stack # Stores the answerans = [] # A Binary tree nodeclass Node:         # Constructor to create a new node    def __init__(self, data):        self.data = data        self.left = None        self.right = None def peek(stack):    if len(stack) > 0:        return stack[-1]    return None# A iterative function to do postorder traversal of # a given binary treedef postOrderIterative(root):             # Check for empty tree    if root is None:        return     stack = []         while(True):                 while (root):             # Push root's right child and then root to stack             if root.right is not None:                stack.append(root.right)             stack.append(root)              # Set root as root's left child             root = root.left                 # Pop an item from stack and set it as root        root = stack.pop()         # If the popped item has a right child and the        # right child is not processed yet, then make sure        # right child is processed before root        if (root.right is not None and            peek(stack) == root.right):            stack.pop() # Remove right child from stack             stack.append(root) # Push root back to stack            root = root.right # change root so that the                              # righ childis processed next         # Else print root's data and set root as None        else:            ans.append(root.data)             root = None         if (len(stack) <= 0):                break # Driver pogram to test above functionroot = Node(1)root.left = Node(2)root.right = Node(3)root.left.left = Node(4)root.left.right = Node(5)root.right.left = Node(6)root.right.right = Node(7) print "Post Order traversal of binary tree is"postOrderIterative(root)print ans # This code is contributed by Nikhil Kumar Singh(nickzuck_007)<\/code><\/pre> <\/div>\n
Output:<\/strong><\/p>\n
Post Order traversal of binary tree is\r\n[4, 5, 2, 6, 7, 3, 1]<\/pre>\n","protected":false},"excerpt":{"rendered":"
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