Postorder traversal of Binary Tree without recursion and without stack (original) (raw)

Last Updated : 31 Mar, 2023

Given a binary tree, perform postorder traversal.

Prerequisite - Inorder/preorder/postorder traversal of tree

We have discussed the below methods for postorder traversal.

1. Recursive Postorder Traversal.
2. Postorder traversal using Stack.
3. Postorder traversal using two Stacks.

Approach 1

The approach used is based on using an unordered set to keep track of visited nodes and a while loop to traverse the tree. The steps involved in the approach can be expressed mathematically as follows:

• Initialize a pointer temp to the root node of the binary tree.
• Initialize an empty unordered set visited to keep track of visited nodes.
• While temp is not null and temp is not already visited (i.e., temp is not in visited set):
  a. If temp has a left child and the left child is not already visited, set temp to the left child.
  b. Else if temp has a right child and the right child is not already visited, set temp to the right child.
  c. Else, print the data of the current node temp, add temp to visited set, and set temp to the root node.
• Return from the function.

Algorithm

Define a struct Node with integer data, pointer to left child and pointer to right child. Define a helper function called "postorder" which takes a pointer to the head of the tree. Create a pointer "temp" and an unordered set "visited". While "temp" is not NULL and "temp" is not visited before: a. If "temp" has a left child and the left child is not visited before, then set "temp" to its left child and continue the loop. b. If "temp" does not have a left child or the left child is already visited, check if "temp" has a right child and the right child is not visited before. If yes, set "temp" to its right child and continue the loop. c. If "temp" does not have a left child or the left child is already visited, and "temp" does not have a right child or the right child is already visited, then print the data of "temp", insert "temp" into "visited" set, and set "temp" to the head of the tree. Define a function called "newNode" which takes an integer data as input and returns a new Node with the given data, NULL left pointer, and NULL right pointer.

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; }

Java

// JAVA program or postorder traversal import java.util.*;

/* A binary tree node has data, pointer to left child and a pointer to right child */ class Node { int data; Node left, right; Node(int data) { this.data = data; this.left = this.right = null;
} };

class GFG {

Node root;

/* Helper function that allocates a new node with the given data and null left and right pointers. */ void postorder(Node head) { Node temp = root;
HashSet visited = new HashSet<>(); while ((temp != null && !visited.contains(temp))) {

    // Visited left subtree
    if (temp.left != null && 
     !visited.contains(temp.left))
        temp = temp.left;

    // Visited right subtree
    else if (temp.right != null && 
    !visited.contains(temp.right))
        temp = temp.right;

    // Print node
    else 
    {
        System.out.printf("%d ", temp.data);
        visited.add(temp);
        temp = head;
    }
}

}

/* Driver program to test above functions*/ public static void main(String[] args) { GFG gfg = new GFG(); gfg.root = new Node(8); gfg.root.left = new Node(3); gfg.root.right = new Node(10); gfg.root.left.left = new Node(1); gfg.root.left.right = new Node(6); gfg.root.left.right.left = new Node(4); gfg.root.left.right.right = new Node(7); gfg.root.right.right = new Node(14); gfg.root.right.right.left = new Node(13); gfg.postorder(gfg.root); } }

// This code is contributed by Rajput-Ji

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

C#

// C# program or postorder traversal using System; using System.Collections.Generic;

/* A binary tree node has data, pointer to left child and a pointer to right child */ public class Node { public int data; public Node left, right; public Node(int data) { this.data = data; this.left = this.right = null;
} };

class GFG {

Node root;

/* Helper function that allocates a new node with the given data and null left and right pointers. */ void postorder(Node head) { Node temp = root;
HashSet visited = new HashSet(); while ((temp != null && !visited.Contains(temp))) {

  // Visited left subtree
  if (temp.left != null && 
      !visited.Contains(temp.left))
    temp = temp.left;

  // Visited right subtree
  else if (temp.right != null && 
           !visited.Contains(temp.right))
    temp = temp.right;

  // Print node
  else 
  {
    Console.Write(temp.data + " ");
    visited.Add(temp);
    temp = head;
  }
}

}

/* Driver code*/ public static void Main(String[] args) { GFG gfg = new GFG(); gfg.root = new Node(8); gfg.root.left = new Node(3); gfg.root.right = new Node(10); gfg.root.left.left = new Node(1); gfg.root.left.right = new Node(6); gfg.root.left.right.left = new Node(4); gfg.root.left.right.right = new Node(7); gfg.root.right.right = new Node(14); gfg.root.right.right.left = new Node(13); gfg.postorder(gfg.root); } }

// This code is contributed by Rajput-Ji

JavaScript

`

Output:

1 4 7 6 3 13 14 10 8

Time complexity: O(N) where N is no of nodes in a binary tree

Auxiliary Space: O(n) since using unordered_set

Alternate Solution:

We can keep the visited flag with every node instead of a 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

JavaScript

`

Output:

1 4 7 6 3 13 14 10 8

Time complexity: O(n2) in worst case we move pointer back to head after visiting every node.

Auxiliary Space: O(1)

Alternate solution using unordered_map in which we do not have to move pointer back to head, so time complexity is O(n).

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(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; }

Java

import java.util.HashMap; import java.util.Map;

class Node { int data; Node left, right; boolean visited; }

public class Tree { static Map<Node, Node> parentMap = new HashMap<>();

static void postorder(Node root) {
    Node n = root;
    parentMap.put(root, null);

    while (n != null) {
        if (n.left != null && !parentMap.containsKey(n.left)) {
            parentMap.put(n.left, n);
            n = n.left;
        } else if (n.right != null && !parentMap.containsKey(n.right)) {
            parentMap.put(n.right, n);
            n = n.right;
        } else {
            System.out.print(n.data + " ");
            n = parentMap.get(n);
        }
    }
}

static Node newNode(int data) {
    Node node = new Node();
    node.data = data;
    node.left = null;
    node.right = null;
    node.visited = false;
    return node;
}

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);
}

}

Python3

Python3 code for the above approach

A binary tree node class

class Node: def init(self, data): self.data = data self.left = None self.right = None self.visited = False

def postorder(root): n = root parent_map = {} parent_map[root] = None

while n:
    if n.left and n.left not in parent_map:
        parent_map[n.left] = n
        n = n.left
    elif n.right and n.right not in parent_map:
        parent_map[n.right] = n
        n = n.right
    else:
        print(n.data, end = " ")
        n = parent_map[n]

Driver code

if name == 'main': root = Node(8) root.left = Node(3) root.right = Node(10) root.left.left = Node(1) root.left.right = Node(6) root.left.right.left = Node(4) root.left.right.right = Node(7) root.right.right = Node(14) root.right.right.left = Node(13) postorder(root)

C#

// C# program or postorder traversal using System; using System.Collections.Generic;

/* 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; }

class Tree { static Dictionary<Node, Node> parentMap = new Dictionary<Node, Node>();

static void postorder(Node root) { Node n = root; parentMap[root] = null;

while (n != null) {
  if (n.left != null && !parentMap.ContainsKey(n.left)) {
    parentMap[n.left] = n;
    n = n.left;
  } else if (n.right != null && !parentMap.ContainsKey(n.right)) {
    parentMap[n.right] = n;
    n = n.right;
  } else {
    Console.Write(n.data + " ");
    n = parentMap[n];
  }
}

}

static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = null; node.right = null; node.visited = false; return node; }

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); } }

JavaScript

class Node { constructor(data) { this.data = data; this.left = null; this.right = null; this.visited = false; } }

function postorder(root) { let n = root; const parentMap = new Map(); parentMap.set(root, null);

while (n) { if (n.left && !parentMap.has(n.left)) { parentMap.set(n.left, n); n = n.left; } else if (n.right && !parentMap.has(n.right)) { parentMap.set(n.right, n); n = n.right; } else { console.log(n.data + " "); n = parentMap.get(n); } } }

// Test const root = new Node(8); root.left = new Node(3); root.right = new Node(10); root.left.left = new Node(1); root.left.right = new Node(6); root.left.right.left = new Node(4); root.left.right.right = new Node(7); root.right.right = new Node(14); root.right.right.left = new Node(13); postorder(root);

// This code is contributed by divyansh2212

`

Output:

1 4 7 6 3 13 14 10 8

Time complexity: O(n) where n is no of nodes in a binary tree

Auxiliary Space: O(n) since using unordered_map