# Hackerrank Tree: Postorder Traversal problem solution

In this HackerRank Tree: Postorder Traversal problem we have given a pointer to the root node of a binary tree. and we need to print the values of trees in postorder in a single line.

## Problem solution in Python programming.

```class Node:
def __init__(self, info):
self.info = info
self.left = None
self.right = None
self.level = None

def __str__(self):
return str(self.info)

class BinarySearchTree:
def __init__(self):
self.root = None

def create(self, val):
if self.root == None:
self.root = Node(val)
else:
current = self.root

while True:
if val < current.info:
if current.left:
current = current.left
else:
current.left = Node(val)
break
elif val > current.info:
if current.right:
current = current.right
else:
current.right = Node(val)
break
else:
break

def postOrder(root):
if root.left:
postOrder(root.left)

if root.right:
postOrder(root.right)

print(root.info, end=" ")

tree = BinarySearchTree()
t = int(input())

arr = list(map(int, input().split()))

for i in range(t):
tree.create(arr[i])

postOrder(tree.root)```

## Problem solution in Java Programming.

```import java.util.*;
import java.io.*;

class Node {
Node left;
Node right;
int data;

Node(int data) {
this.data = data;
left = null;
right = null;
}
}

class Solution {

public static void postOrder(Node root) {
Node t = root;
Deque<Node> stack = new ArrayDeque<Node>();
stack.push(root);
while(!stack.isEmpty() && root!=null){
root = stack.peek();
//nodes without children should be printed
if( (root.left==null&&root.right==null)
|| (t==root.left||t==root.right) ){//or nodes whose children have already been printed
System.out.print(root.data+" ");
stack.pop();
t = root;
}else{
if(root.right!=null) stack.push(root.right);
if(root.left!=null) stack.push(root.left);
}
}
}

public static Node insert(Node root, int data) {
if(root == null) {
return new Node(data);
} else {
Node cur;
if(data <= root.data) {
cur = insert(root.left, data);
root.left = cur;
} else {
cur = insert(root.right, data);
root.right = cur;
}
return root;
}
}

public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
Node root = null;
while(t-- > 0) {
int data = scan.nextInt();
root = insert(root, data);
}
scan.close();
postOrder(root);
}
}
```

### Problem solution in C++ programming.

```#include <bits/stdc++.h>

using namespace std;

class Node {
public:
int data;
Node *left;
Node *right;
Node(int d) {
data = d;
left = NULL;
right = NULL;
}
};

class Solution {
public:
Node* insert(Node* root, int data) {
if(root == NULL) {
return new Node(data);
} else {
Node* cur;
if(data <= root->data) {
cur = insert(root->left, data);
root->left = cur;
} else {
cur = insert(root->right, data);
root->right = cur;
}

return root;
}
}

void postOrder(Node *root) {
if(root == NULL)
return;

postOrder(root->left);
postOrder(root->right);
cout << root->data << " ";
}

}; //End of Solution

int main() {

Solution myTree;
Node* root = NULL;

int t;
int data;

std::cin >> t;

while(t-- > 0) {
std::cin >> data;
root = myTree.insert(root, data);
}

myTree.postOrder(root);
return 0;
}```

### Problem solution in C programming.

```#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

struct node {

int data;
struct node *left;
struct node *right;

};

struct node* insert( struct node* root, int data ) {

if(root == NULL) {

struct node* node = (struct node*)malloc(sizeof(struct node));

node->data = data;

node->left = NULL;
node->right = NULL;
return node;

} else {

struct node* cur;

if(data <= root->data) {
cur = insert(root->left, data);
root->left = cur;
} else {
cur = insert(root->right, data);
root->right = cur;
}

return root;
}
}

void postOrder( struct node *root) {
if (root != NULL) {
postOrder(root->left);
postOrder(root->right);
printf("%d ", root->data);
}
}

int main() {

struct node* root = NULL;

int t;
int data;

scanf("%d", &t);

while(t-- > 0) {
scanf("%d", &data);
root = insert(root, data);
}

postOrder(root);
return 0;
}```