Postorder traversal of Binary Tree without recursion and without stack
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.
2) 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 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 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# 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
<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
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 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
<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: 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;
}
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 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# 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);
}
}
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
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Given a binary tree, the task is to find the postorder traversal of the tree without using recursion.Examples:Input: Output: 4 5 2 3 1Explanation: Postorder traversal (Left->Right->Root) of the tree is 4 5 2 3 1.Input: Output: 10 7 1 6 10 6 5 8 Explanation: Postorder traversal (Left->Right-
10 min read
Flatten binary tree in order of post-order traversal
Given a binary tree, the task is to flatten it in order of its post-order traversal. In the flattened binary tree, the left node of all the nodes must be NULL. Examples: Input: 5 / \ 3 7 / \ / \ 2 4 6 8 Output: 2 4 3 6 8 7 5 Input: 1 \ 2 \ 3 \ 4 \ 5 Output: 5 4 3 2 1 A simple approach will be to rec
6 min read
Modify a binary tree to get preorder traversal using right pointers only
Given a binary tree. The task is to modify it in such a way that after modification preorder traversal of it can get only with the right pointers. During modification, we can use right as well as left pointers. Examples: Input : Output : Explanation: The preorder traversal of given binary tree is 10
12 min read