# 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.

 `// CPP program or postorder traversal ` `#include ` `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 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 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.

 `// CPP program or postorder traversal ` `#include ` `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 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 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# 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 ` `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 parentMap; ` `    ``parentMap.insert(pair(root, nullptr)); ` ` `  `    ``while` `(n) { ` `        ``if` `(n->left && parentMap.find(n->left) == parentMap.end()) { ` `            ``parentMap.insert(pair(n->left, n)); ` `            ``n = n->left; ` `        ``} ` `        ``else` `if` `(n->right && parentMap.find(n->right) == parentMap.end()) { ` `            ``parentMap.insert(pair(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
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :