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[Solved] PostOrder Traversal of Binary Tree Without Recursion (Iterative)

Aniruddha Chaudhari/9162/0
Misc

Traversing elements in postorder traversal with recursion is easy. You can read the postorder traversal in detail where I have explained it with binary tree example and code.

Table of Contents

  • Problem Statement
  • Binary Tree Example
  • Algorithm
  • Python Program for PostOrder Traversal without Recursion (Iterative)

Problem Statement

You have given a binary tree. Implement PostOrder Traversal Without Recursion.

Here we are going to use the iterative method to traverse binary tree elements

(This problem was asked in the OVH cloud interview round. You have to write working code on the whiteboard.)

Binary Tree Example

reorder, inorder and post order traversal

The postorder traversal of the Binary tree is 7 3 1 6 5 4.

Algorithm

Here we are going to use the stack. Each element in the stack will maintain the current node and the flag. The flag will be set to True if the children of the current node are not visited.

In Python, the stack can be implemented using a Python list.

  • Create a binary tree if it is not given
  • Add the root node in the stack (list)
  • While there is an element in the stack (list)
    • Read an element from the stack (node, flag).
    • If the flag is true, set the flag as false. Add right and left children to the stack.
    • Else, print the value of the current node. Pop the node from the stack (Remove the last element from the list).

If you understand the logic and algorithm, you can implement it in many programming languages like C/C++, Java, Python…

Python Program for PostOrder Traversal without Recursion (Iterative)

	
#structure/class for a node
#by defult, left and right pointers are None 
class node:
  def __init__(self, val, left=None, right=None):
    self.left=left
    self.val=val
    self.right=right
  
  
#adding element in the binary tree
root=node(4)
root.left=node(1)
root.right=node(5)
root.left.left=node(7)
root.left.right=node(3)
root.right.right=node(6)
 
#return True if the node has child
#retun False if node has no child
def isChild(node):
    if node.right or node.left:
        return True
    else:
        return False

#print post order traversal
#without recursion
def printPostOrderWithoutRec(node):
    liNode = []
    if node:
        liNode.append([node, isChild(node)])
    else:
        print("Tree is empty.")
        return

    while liNode:
        curNode, bChild = liNode[-1]

        if bChild:
            liNode[-1][1] = 0
            if curNode.right:
                liNode.append([curNode.right, isChild(curNode.right)])
            if curNode.left:
                liNode.append([curNode.left, isChild(curNode.left)])
        else:
            print(curNode.val)
            del liNode[-1]

printPostOrderWithoutRec(root)

Output:

7
3
1
6
5
4

What’s Next?

I hope you understood postorder traversal without recursion. You can practice implementing inorder and preorder traversal without recursion.

If you are preparing for an interview with product-based companies, you have to prepare for a binary tree. If you have any queries, you can comment below.

Binary Treecoding challenge
Aniruddha Chaudhari
I am complete Python Nut, love Linux and vim as an editor. I hold a Master of Computer Science from NIT Trichy. I dabble in C/C++, Java too. I keep sharing my coding knowledge and my own experience on CSEstack.org portal.

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Coding Challenges for Practice

  1. Count Common Factor
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  23. Smallest Subarray with Sum Greater than Given Number
  24. Group Anagrams
  25. Find Duplicates in Array in O(n)
  26. Find Two Unique Numbers from Array in O(n)
  27. Number Patterns & Finding Smallest Number
  28. First Unique Element in a Stream
  29. Flip Equivalent Binary Trees [TeachMint]
  30. Minimum Cost of Merging Files [Amazon]
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  32. Longest Sequence of URLs
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  39. Merge Overlapping Intervals
  40. Longest Balanced Substring
  41. Longest Path in a Weighted Tree
  42. Generate Balanced Parentheses
  43. PostOrder Traversal Without Recursion

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