Iterative Postorder Traversal | Set 1 (Using Two Stacks)

We have discussed iterative inorder and iterative preorder traversals. In this post, iterative postorder traversal is discussed, which is more complex than the other two traversals (due to its nature of non-tail recursion, there is an extra statement after the final recursive call to itself). Postorder traversal can easily be done using two stacks, though. The idea is to push reverse postorder traversal to a stack. Once we have the reversed postorder traversal in a stack, we can just pop all items one by one from the stack and print them; this order of printing will be in postorder because of the LIFO property of stacks. Now the question is, how to get reversed postorder elements in a stack – the second stack is used for this purpose. For example, in the following tree, we need to get 1, 3, 7, 6, 2, 5, 4 in a stack. If take a closer look at this sequence, we can observe that this sequence is very similar to the preorder traversal. The only difference is that the right child is visited before left child, and therefore the sequence is “root right left” instead of “root left right”. So, we can do something like iterative preorder traversal with the following differences:
a) Instead of printing an item, we push it to a stack.
b) We push the left subtree before the right subtree.

Following is the complete algorithm. After step 2, we get the reverse of a postorder traversal in the second stack. We use the first stack to get the correct order.

1. Push root to first stack.
2. Loop while first stack is not empty
   2.1 Pop a node from first stack and push it to second stack
   2.2 Push left and right children of the popped node to first stack
3. Print contents of second stack

Let us consider the following tree

Following are the steps to print postorder traversal of the above tree using two stacks.

1. Push 1 to first stack.
      First stack: 1
      Second stack: Empty

2. Pop 1 from first stack and push it to second stack. 
   Push left and right children of 1 to first stack
      First stack: 2, 3
      Second stack: 1

3. Pop 3 from first stack and push it to second stack. 
   Push left and right children of 3 to first stack
      First stack: 2, 6, 7
      Second stack: 1, 3

4. Pop 7 from first stack and push it to second stack.
      First stack: 2, 6
      Second stack: 1, 3, 7

5. Pop 6 from first stack and push it to second stack.
      First stack: 2
      Second stack: 1, 3, 7, 6

6. Pop 2 from first stack and push it to second stack. 
   Push left and right children of 2 to first stack
      First stack: 4, 5
      Second stack: 1, 3, 7, 6, 2

7. Pop 5 from first stack and push it to second stack.
      First stack: 4
      Second stack: 1, 3, 7, 6, 2, 5

8. Pop 4 from first stack and push it to second stack.
      First stack: Empty
      Second stack: 1, 3, 7, 6, 2, 5, 4

The algorithm stops here since there are no more items in the first stack. 
Observe that the contents of second stack are in postorder fashion. Print them. 


Following is C implementation of iterative postorder traversal using two stacks.

C

#include <stdio.h>
#include <stdlib.h>
  
// Maximum stack size
#define MAX_SIZE 100
  
// A tree node
struct Node {
    int data;
    struct Node *left, *right;
};
  
// Stack type
struct Stack {
    int size;
    int top;
    struct Node** array;
};
  
// A utility function to create a new tree node
struct Node* newNode(int data)
{
    struct Node* node = (struct Node*)malloc(sizeof(struct Node));
    node->data = data;
    node->left = node->right = NULL;
    return node;
}
  
// A utility function to create a stack of given size
struct Stack* createStack(int size)
{
    struct Stack* stack = (struct Stack*)malloc(sizeof(struct Stack));
    stack->size = size;
    stack->top = -1;
    stack->array = (struct Node**)malloc(stack->size * sizeof(struct Node*));
    return stack;
}
  
// BASIC OPERATIONS OF STACK
int isFull(struct Stack* stack)
{
    return stack->top - 1 == stack->size;
}
  
int isEmpty(struct Stack* stack)
{
    return stack->top == -1;
}
  
void push(struct Stack* stack, struct Node* node)
{
    if (isFull(stack))
        return;
    stack->array[++stack->top] = node;
}
  
struct Node* pop(struct Stack* stack)
{
    if (isEmpty(stack))
        return NULL;
    return stack->array[stack->top--];
}
  
// An iterative function to do post order traversal of a given binary tree
void postOrderIterative(struct Node* root)
{
    if (root == NULL)
        return;
  
    // Create two stacks
    struct Stack* s1 = createStack(MAX_SIZE);
    struct Stack* s2 = createStack(MAX_SIZE);
  
    // push root to first stack
    push(s1, root);
    struct Node* node;
  
    // Run while first stack is not empty
    while (!isEmpty(s1)) {
        // Pop an item from s1 and push it to s2
        node = pop(s1);
        push(s2, node);
  
        // Push left and right children of removed item to s1
        if (node->left)
            push(s1, node->left);
        if (node->right)
            push(s1, node->right);
    }
  
    // Print all elements of second stack
    while (!isEmpty(s2)) {
        node = pop(s2);
        printf("%d ", node->data);
    }
}
  
// Driver program to test above functions
int main()
{
    // Let us construct the tree shown in above figure
    struct Node* root = NULL;
    root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
  
    postOrderIterative(root);
  
    return 0;
}

Java

// Java program for iterative post
// order using two stacks
  
import java.util.*;
public class IterativePostorder {
  
    static class node {
        int data;
        node left, right;
  
        public node(int data)
        {
            this.data = data;
        }
    }
  
    // Two stacks as used in explanation
    static Stack<node> s1, s2;
  
    static void postOrderIterative(node root)
    {
        // Create two stacks
        s1 = new Stack<>();
        s2 = new Stack<>();
  
        if (root == null)
            return;
  
        // push root to first stack
        s1.push(root);
  
        // Run while first stack is not empty
        while (!s1.isEmpty()) {
            // Pop an item from s1 and push it to s2
            node temp = s1.pop();
            s2.push(temp);
  
            // Push left and right children of
            // removed item to s1
            if (temp.left != null)
                s1.push(temp.left);
            if (temp.right != null)
                s1.push(temp.right);
        }
  
        // Print all elements of second stack
        while (!s2.isEmpty()) {
            node temp = s2.pop();
            System.out.print(temp.data + " ");
        }
    }
  
    public static void main(String[] args)
    {
        // Let us construct the tree
        // shown in above figure
  
        node root = null;
        root = new node(1);
        root.left = new node(2);
        root.right = new node(3);
        root.left.left = new node(4);
        root.left.right = new node(5);
        root.right.left = new node(6);
        root.right.right = new node(7);
  
        postOrderIterative(root);
    }
}
  
// This code is contributed by Rishabh Mahrsee

Python

# Python program for iterative postorder traversal using
# two stacks
  
# A binary tree node
class Node:
      
    # Constructor to create a new node
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
  
# An iterative function to do postorder traversal of a
# given binary tree
def postOrderIterative(root): 
  
    if root is None:
        return          
      
    # Create two stacks 
    s1 = []
    s2 = []
      
    # Push root to first stack
    s1.append(root)
      
    # Run while first stack is not empty
    while s1:
          
        # Pop an item from s1 and append it to s2
        node = s1.pop()
        s2.append(node)
      
        # Push left and right children of removed item to s1
        if node.left:
            s1.append(node.left)
        if node.right:
            s1.append(node.right)
  
        # Print all elements of second stack
    while s2:
        node = s2.pop()
        print node.data,
  
# Driver program to test above function
root = 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)
postOrderIterative(root)


Output:

4 5 2 6 7 3 1

Following is an overview of the above post.
Iterative preorder traversal can be easily implemented using two stacks. The first stack is used to get the reverse postorder traversal. The steps to get a reverse postorder are similar to iterative preorder.

You may also like to see a method which uses only one stack.

This article is compiled by Aashish Barnwal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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Improved By : IshitaTripathi



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