Iterative Postorder Traversal | Set 2 (Using One Stack)

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.

The idea is to move down to leftmost node using left pointer. While moving down, push root and root’s right child to stack. Once we reach leftmost node, print it if it doesn’t have a right child. If it has a right child, then change root so that the right child is processed before.

Following is detailed algorithm.

1.1 Create an empty stack
2.1 Do following while root is not NULL
    a) Push root's right child and then root to stack.
    b) Set root as root's left child.
2.2 Pop an item from stack and set it as root.
    a) If the popped item has a right child and the right child 
       is at top of stack, then remove the right child from stack,
       push the root back and set root as root's right child.
    b) Else print root's data and set root as NULL.
2.3 Repeat steps 2.1 and 2.2 while stack is not empty.

Let us consider the following tree

Following are the steps to print postorder traversal of the above tree using one stack.

1. Right child of 1 exists. 
   Push 3 to stack. Push 1 to stack. Move to left child.
        Stack: 3, 1

2. Right child of 2 exists. 
   Push 5 to stack. Push 2 to stack. Move to left child.
        Stack: 3, 1, 5, 2

3. Right child of 4 doesn't exist. '
   Push 4 to stack. Move to left child.
        Stack: 3, 1, 5, 2, 4

4. Current node is NULL. 
   Pop 4 from stack. Right child of 4 doesn't exist. 
   Print 4. Set current node to NULL.
        Stack: 3, 1, 5, 2

5. Current node is NULL. 
    Pop 2 from stack. Since right child of 2 equals stack top element, 
    pop 5 from stack. Now push 2 to stack.     
    Move current node to right child of 2 i.e. 5
        Stack: 3, 1, 2

6. Right child of 5 doesn't exist. Push 5 to stack. Move to left child.
        Stack: 3, 1, 2, 5

7. Current node is NULL. Pop 5 from stack. Right child of 5 doesn't exist. 
   Print 5. Set current node to NULL.
        Stack: 3, 1, 2

8. Current node is NULL. Pop 2 from stack. 
   Right child of 2 is not equal to stack top element. 
   Print 2. Set current node to NULL.
        Stack: 3, 1

9. Current node is NULL. Pop 1 from stack. 
   Since right child of 1 equals stack top element, pop 3 from stack. 
   Now push 1 to stack. Move current node to right child of 1 i.e. 3
        Stack: 1

10. Repeat the same as above steps and Print 6, 7 and 3. 
    Pop 1 and Print 1.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C

// C program for iterative postorder traversal using one stack 
#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--]; 
} 
  
struct Node* peek(struct Stack* stack) 
{ 
    if (isEmpty(stack)) 
        return NULL; 
    return stack->array[stack->top]; 
} 
  
// An iterative function to do postorder traversal of a given binary tree 
void postOrderIterative(struct Node* root) 
{ 
    // Check for empty tree 
    if (root == NULL) 
        return; 
      
    struct Stack* stack = createStack(MAX_SIZE); 
    do
    { 
        // Move to leftmost node 
        while (root) 
        { 
            // Push root's right child and then root to stack. 
            if (root->right) 
                push(stack, root->right); 
            push(stack, root); 
  
            // Set root as root's left child   
            root = root->left; 
        } 
  
        // Pop an item from stack and set it as root     
        root = pop(stack); 
  
        // 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 && peek(stack) == root->right) 
        { 
            pop(stack);  // remove right child from stack 
            push(stack, root);  // push root back to stack 
            root = root->right; // change root so that the right  
                                // child is processed next 
        } 
        else  // Else print root's data and set root as NULL 
        { 
            printf("%d ", root->data); 
            root = NULL; 
        } 
    } while (!isEmpty(stack)); 
} 
  
// 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); 
    printf("Post order traversal of binary tree is :\n"); 
    printf("["); 
    postOrderIterative(root); 
    printf("]"); 
      
  
    return 0; 
} 

Java

// A java program for iterative postorder traversal using stack 
  
import java.util.ArrayList; 
import java.util.Stack; 
   
// A binary tree node 
class Node  
{ 
    int data; 
    Node left, right; 
   
    Node(int item)  
    { 
        data = item; 
        left = right; 
    } 
} 
   
class BinaryTree  
{ 
    Node root; 
    ArrayList<Integer> list = new ArrayList<Integer>(); 
   
    // An iterative function to do postorder traversal  
    // of a given binary tree 
    ArrayList<Integer> postOrderIterative(Node node)  
    { 
        Stack<Node> S = new Stack<Node>(); 
           
        // Check for empty tree 
        if (node == null) 
            return list; 
        S.push(node); 
        Node prev = null; 
        while (!S.isEmpty())  
        { 
            Node current = S.peek(); 
   
            /* go down the tree in search of a leaf an if so process it  
            and pop stack otherwise move down */
            if (prev == null || prev.left == current ||  
                                        prev.right == current)  
            { 
                if (current.left != null) 
                    S.push(current.left); 
                else if (current.right != null) 
                    S.push(current.right); 
                else
                { 
                    S.pop(); 
                    list.add(current.data); 
                } 
   
                /* go up the tree from left node, if the child is right  
                   push it onto stack otherwise process parent and pop  
                   stack */
            }  
            else if (current.left == prev)  
            { 
                if (current.right != null) 
                    S.push(current.right); 
                else 
                { 
                    S.pop(); 
                    list.add(current.data); 
                } 
                   
                /* go up the tree from right node and after coming back 
                 from right node process parent and pop stack */
            }  
            else if (current.right == prev)  
            { 
                S.pop(); 
                list.add(current.data); 
            } 
   
            prev = current; 
        } 
   
        return list; 
    } 
   
    // Driver program to test above functions 
    public static void main(String args[])  
    { 
     BinaryTree tree = new BinaryTree(); 
   
        // Let us create trees shown in above diagram 
        tree.root = new Node(1); 
        tree.root.left = new Node(2); 
        tree.root.right = new Node(3); 
        tree.root.left.left = new Node(4); 
        tree.root.left.right = new Node(5); 
        tree.root.right.left = new Node(6); 
        tree.root.right.right = new Node(7); 
   
        ArrayList<Integer> mylist = tree.postOrderIterative(tree.root); 
           
        System.out.println("Post order traversal of binary tree is :"); 
        System.out.println(mylist); 
    } 
} 
   
// This code has been contributed by Mayank Jaiswal 

Python

# Python program for iterative postorder traversal 
# using one stack 
  
# Stores the answer 
ans = [] 
  
# 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
  
def peek(stack): 
    if len(stack) > 0: 
        return stack[-1] 
    return None
# A iterative function to do postorder traversal of  
# a given binary tree 
def 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 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) 
  
print "Post Order traversal of binary tree is"
postOrderIterative(root) 
print ans 
  
# This code is contributed by Nikhil Kumar Singh(nickzuck_007) 

Output:

Post Order traversal of binary tree is
[4, 5, 2, 6, 7, 3, 1]

Method 2:
Push directly root node two times while traversing to the left. While poping if you find stack top() is same as root then go for root->right else print root.

// Simple Java program to print PostOrder Traversal(Iterative) 
import java.util.Stack; 
  
// A binary tree node 
class Node  
{ 
    int data; 
    Node left, right; 
  
    Node(int item)  
    { 
        data = item; 
        left = right; 
    } 
} 
  
// create a postorder class 
class PostOrder 
{  
    Node root; 
      
    // An iterative function to do postorder traversal  
    // of a given binary tree 
    private void postOrderIterative(Node root) { 
        Stack<Node> stack = new Stack<>(); 
        while(true) { 
            while(root != null) { 
                stack.push(root); 
                stack.push(root); 
                root = root.left; 
            } 
              
            // Check for empty stack 
            if(stack.empty()) return; 
            root = stack.pop(); 
              
            if(!stack.empty() && stack.peek() == root) root = root.right; 
              
            else { 
                  
                System.out.print(root.data + " "); root = null; 
            } 
        } 
    } 
      
    // Driver program to test above functions 
    public static void main(String args[])  
    { 
    PostOrder tree = new PostOrder(); 
          
        // Let us create trees shown in above diagram 
        tree.root = new Node(1); 
        tree.root.left = new Node(2); 
        tree.root.right = new Node(3); 
        tree.root.left.left = new Node(4); 
        tree.root.left.right = new Node(5); 
        tree.root.right.left = new Node(6); 
        tree.root.right.right = new Node(7); 
        System.out.println("Post order traversal of binary tree is :"); 
        tree.postOrderIterative(tree.root); 
    } 
} 

Output:

Post Order traversal of binary tree is:
4, 5, 2, 6, 7, 3, 1

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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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C

Java

Python

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