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Postorder traversal of Binary Tree without recursion and without stack

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  • Difficulty Level : Hard
  • Last Updated : 21 Jun, 2022

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

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;
}

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<Node> 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<Node> visited = new HashSet<Node>();
    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




<script>
 
// JavaScript program or postorder traversal
 
/* A binary tree node has data, pointer to left child
and a pointer to right child */
class Node
{
    constructor(data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
};
 
 
var root = null;
 
  /* Helper function that allocates a new node with the
given data and null left and right pointers. */
  function postorder(head)
  {
    var temp = root;   
    var visited = new Set();
    while ((temp != null  && !visited.has(temp)))
    {
 
      // Visited left subtree
      if (temp.left != null &&
          !visited.has(temp.left))
        temp = temp.left;
 
      // Visited right subtree
      else if (temp.right != null &&
               !visited.has(temp.right))
        temp = temp.right;
 
      // Print node
      else
      {
        document.write(temp.data + " ");
        visited.add(temp);
        temp = head;
      }
    }
  }
 
/* Driver code*/
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);
 
</script>

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

Javascript




<script>
 
    // JavaScript program or postorder traversal
     
    /* A binary tree node has data,
    pointer to left child
    and a pointer to right child */
    class Node
    {
        constructor() {
               this.data;
            this.left;
            this.right;
            this.visited;
        }
    }
     
    function postorder(head)
    {
        let 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
            {
                document.write(temp.data + " ");
                temp.visited = true;
                temp = head;
            }
        }
    }
 
    function newNode(data)
    {
        let node = new Node();
        node.data = data;
        node.left = null;
        node.right = null;
        node.visited = false;
        return (node);
    }
     
    let 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);
   
</script>

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).
 

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;
}

Output: 
 

1 4 7 6 3 13 14 10 8 

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

 


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