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.

Hackerrank Tree: Postorder Traversal problem solution


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;
}