Postorder traversal of Binary Tree without recursion and without stack

Prerequisite – Inorder/preorder/postorder traversal of tree
Given a binary tree, perform postorder traversal.

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

We have discussed below methods for postorder traversal.
1) Recursive Postorder Traversal.
2) Postorder traversal using Stack.
2) Postorder traversal using two Stacks.

In this method a DFS based solution is discussed. We keep track of visited nodes in a hash table.

C++

// CPP program or postorder traversal 
#include <bits/stdc++.h> 
using namespace std; 
  
/* A binary tree node has data, pointer to left child 
and a pointer to right child */
struct Node { 
    int data; 
    struct Node *left, *right; 
}; 
  
/* Helper function that allocates a new node with the 
given data and NULL left and right pointers. */
void postorder(struct Node* head) 
{ 
    struct Node* temp = head; 
    unordered_set<Node*> visited; 
    while (temp && visited.find(temp) == visited.end()) { 
  
        // Visited left subtree 
        if (temp->left &&  
         visited.find(temp->left) == visited.end()) 
            temp = temp->left; 
  
        // Visited right subtree 
        else if (temp->right &&  
        visited.find(temp->right) == visited.end()) 
            temp = temp->right; 
  
        // Print node 
        else { 
            printf("%d ", temp->data); 
            visited.insert(temp); 
            temp = head; 
        } 
    } 
} 
  
struct Node* newNode(int data) 
{ 
    struct Node* node = new Node; 
    node->data = data; 
    node->left = NULL; 
    node->right = NULL; 
    return (node); 
} 
  
/* Driver program to test above functions*/
int main() 
{ 
    struct Node* root = newNode(8); 
    root->left = newNode(3); 
    root->right = newNode(10); 
    root->left->left = newNode(1); 
    root->left->right = newNode(6); 
    root->left->right->left = newNode(4); 
    root->left->right->right = newNode(7); 
    root->right->right = newNode(14); 
    root->right->right->left = newNode(13); 
    postorder(root); 
    return 0; 
} 

Python

# Python program or postorder traversal  
  
''' A binary tree node has data, pointer to left child  
and a pointer to right child '''
class newNode:  
  
    # Constructor to create a newNode  
    def __init__(self, data):  
        self.data = data  
        self.left = None
        self.right = None
  
''' Helper function that allocates a new node with the  
given data and NULL left and right pointers. '''
def postorder(head): 
      
    temp = head  
    visited = set() 
    while (temp and temp not in visited):  
          
        # Visited left subtree  
        if (temp.left and temp.left not in visited): 
            temp = temp.left  
              
        # Visited right subtree  
        elif (temp.right and temp.right not in visited): 
            temp = temp.right  
          
        # Print node  
        else: 
            print(temp.data, end = " ")  
            visited.add(temp)  
            temp = head  
  
''' Driver program to test above functions'''
if __name__ == '__main__': 
      
    root = newNode(8)  
    root.left = newNode(3)  
    root.right = newNode(10)  
    root.left.left = newNode(1)  
    root.left.right = newNode(6)  
    root.left.right.left = newNode(4)  
    root.left.right.right = newNode(7)  
    root.right.right = newNode(14)  
    root.right.right.left = newNode(13)  
    postorder(root)  
  
# This code is contributed by  
# SHUBHAMSINGH10  

Output:

1 4 7 6 3 13 14 10 8

Alternate Solution:
We can keep visited flag with every node instead of separate hash table.

C++

// CPP program or postorder traversal 
#include <bits/stdc++.h> 
using namespace std; 
  
/* A binary tree node has data, pointer to left child 
and a pointer to right child */
struct Node { 
    int data; 
    struct Node *left, *right; 
    bool visited; 
}; 
  
void postorder(struct Node* head) 
{ 
    struct Node* temp = head; 
    while (temp && temp->visited == false) { 
  
        // Visited left subtree 
        if (temp->left && temp->left->visited == false) 
            temp = temp->left; 
  
        // Visited right subtree 
        else if (temp->right && temp->right->visited == false) 
            temp = temp->right; 
  
        // Print node 
        else { 
            printf("%d ", temp->data); 
            temp->visited = true; 
            temp = head; 
        } 
    } 
} 
  
struct Node* newNode(int data) 
{ 
    struct Node* node = new Node; 
    node->data = data; 
    node->left = NULL; 
    node->right = NULL; 
    node->visited = false; 
    return (node); 
} 
  
/* Driver program to test above functions*/
int main() 
{ 
    struct Node* root = newNode(8); 
    root->left = newNode(3); 
    root->right = newNode(10); 
    root->left->left = newNode(1); 
    root->left->right = newNode(6); 
    root->left->right->left = newNode(4); 
    root->left->right->right = newNode(7); 
    root->right->right = newNode(14); 
    root->right->right->left = newNode(13); 
    postorder(root); 
    return 0; 
} 

Java

// Java program or postorder traversal 
class GFG 
{ 
  
/* A binary tree node has data,  
    pointer to left child 
    and a pointer to right child */
static class Node  
{ 
    int data; 
    Node left, right; 
    boolean visited; 
} 
  
static void postorder( Node head) 
{ 
    Node temp = head; 
    while (temp != null &&  
            temp.visited == false) 
    { 
  
        // Visited left subtree 
        if (temp.left != null &&  
            temp.left.visited == false) 
            temp = temp.left; 
  
        // Visited right subtree 
        else if (temp.right != null &&  
                temp.right.visited == false) 
            temp = temp.right; 
  
        // Print node 
        else 
        { 
            System.out.printf("%d ", temp.data); 
            temp.visited = true; 
            temp = head; 
        } 
    } 
} 
  
static Node newNode(int data) 
{ 
    Node node = new Node(); 
    node.data = data; 
    node.left = null; 
    node.right = null; 
    node.visited = false; 
    return (node); 
} 
  
/* Driver code*/
public static void main(String []args) 
{ 
    Node root = newNode(8); 
    root.left = newNode(3); 
    root.right = newNode(10); 
    root.left.left = newNode(1); 
    root.left.right = newNode(6); 
    root.left.right.left = newNode(4); 
    root.left.right.right = newNode(7); 
    root.right.right = newNode(14); 
    root.right.right.left = newNode(13); 
    postorder(root); 
} 
} 
  
// This code is contributed by Arnab Kundu 

Python3

"""Python3 program or postorder traversal """
  
# A Binary Tree Node  
# Utility function to create a  
# new tree node  
class newNode:  
  
    # Constructor to create a newNode  
    def __init__(self, data):  
        self.data = data  
        self.left = None
        self.right = None
        self.visited = False
  
def postorder(head) : 
  
    temp = head  
    while (temp and temp.visited == False): 
  
        # Visited left subtree  
        if (temp.left and 
            temp.left.visited == False): 
            temp = temp.left  
  
        # Visited right subtree  
        elif (temp.right and 
              temp.right.visited == False):  
            temp = temp.right  
  
        # Print node  
        else: 
            print(temp.data, end = " ")  
            temp.visited = True
            temp = head 
                          
# Driver Code 
if __name__ == '__main__': 
  
    root = newNode(8)  
    root.left = newNode(3)  
    root.right = newNode(10)  
    root.left.left = newNode(1)  
    root.left.right = newNode(6)  
    root.left.right.left = newNode(4)  
    root.left.right.right = newNode(7)  
    root.right.right = newNode(14)  
    root.right.right.left = newNode(13)  
    postorder(root) 
  
# This code is contributed by  
# SHUBHAMSINGH10 

C#

// C# program or postorder traversal 
using System; 
  
class GFG 
{ 
  
/* A binary tree node has data,  
    pointer to left child 
    and a pointer to right child */
class Node  
{ 
    public int data; 
    public Node left, right; 
    public bool visited; 
} 
  
static void postorder( Node head) 
{ 
    Node temp = head; 
    while (temp != null &&  
            temp.visited == false) 
    { 
  
        // Visited left subtree 
        if (temp.left != null &&  
            temp.left.visited == false) 
            temp = temp.left; 
  
        // Visited right subtree 
        else if (temp.right != null &&  
                temp.right.visited == false) 
            temp = temp.right; 
  
        // Print node 
        else
        { 
            Console.Write("{0} ", temp.data); 
            temp.visited = true; 
            temp = head; 
        } 
    } 
} 
  
static Node newNode(int data) 
{ 
    Node node = new Node(); 
    node.data = data; 
    node.left = null; 
    node.right = null; 
    node.visited = false; 
    return (node); 
} 
  
/* Driver code*/
public static void Main(String []args) 
{ 
    Node root = newNode(8); 
    root.left = newNode(3); 
    root.right = newNode(10); 
    root.left.left = newNode(1); 
    root.left.right = newNode(6); 
    root.left.right.left = newNode(4); 
    root.left.right.right = newNode(7); 
    root.right.right = newNode(14); 
    root.right.right.left = newNode(13); 
    postorder(root); 
} 
} 
  
// This code is contributed by 29AjayKumar 

Output:

1 4 7 6 3 13 14 10 8

Time complexity of above solution is O(n²) in worst case we move pointer back to head after visiting every node.
Alternate solution using unordered_map in which we do not have to move pointer back to head, so time complexity is O(n).

// CPP program or postorder traversal 
#include <bits/stdc++.h> 
using namespace std; 
  
/* A binary tree node has data, pointer to left child 
and a pointer to right child */
struct Node { 
    int data; 
    struct Node *left, *right; 
    bool visited; 
}; 
  
void postorder(Node* root) 
{ 
    Node* n = root; 
    unordered_map<Node*, Node*> parentMap; 
    parentMap.insert(pair<Node*, Node*>(root, nullptr)); 
  
    while (n) { 
        if (n->left && parentMap.find(n->left) == parentMap.end()) { 
            parentMap.insert(pair<Node*, Node*>(n->left, n)); 
            n = n->left; 
        } 
        else if (n->right && parentMap.find(n->right) == parentMap.end()) { 
            parentMap.insert(pair<Node*, Node*>(n->right, n)); 
            n = n->right; 
        } 
        else { 
            cout << n->data << " "; 
            n = (parentMap.find(n))->second; 
        } 
    } 
} 
struct Node* newNode(int data) 
{ 
    struct Node* node = new Node; 
    node->data = data; 
    node->left = NULL; 
    node->right = NULL; 
    node->visited = false; 
    return (node); 
} 
  
/* Driver program to test above functions*/
int main() 
{ 
    struct Node* root = newNode(8); 
    root->left = newNode(3); 
    root->right = newNode(10); 
    root->left->left = newNode(1); 
    root->left->right = newNode(6); 
    root->left->right->left = newNode(4); 
    root->left->right->right = newNode(7); 
    root->right->right = newNode(14); 
    root->right->right->left = newNode(13); 
    postorder(root); 
    return 0; 
} 

Output:

1 4 7 6 3 13 14 10 8

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My Personal Notes

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

C++

Python

C++

Java

Python3

C#

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