Postorder traversal of Binary Tree without recursion and without stack in C++

C++Server Side ProgrammingProgramming

In this problem, we are given a Binary tree. Our task is to print the postorder traversal of the binary tree without using recursion and without stack.

Binary tree is a special type of tree in which each node can have at max 2 child nodes.

Postorder Traversal is a tree traversal technique, in which the first left subtree is traversed than the right subtree and the root is traversed at the end.

postorder traversal of the above tree − 8 4 2 7 9 6

To traverse the tree without using recursion and stack. We will depth-first search based technique and data will be stored in a hash table.

Example

Program to show the implementation of this solution,

 Live Demo

#include <bits/stdc++.h>
using namespace std;
struct Node {
   int data;
   struct Node *left, *right;
};
void postOrderTraversal(struct Node* head) {
   struct Node* temp = head;
   unordered_set<Node*> visited;
   while (temp && visited.find(temp) == visited.end()) {
      if (temp->left &&
         visited.find(temp->left) == visited.end())
         temp = temp->left;
      else if (temp->right &&
         visited.find(temp->right) == visited.end())
         temp = temp->right;
      else {
         cout<<temp->data<<"\t";
         visited.insert(temp);
         temp = head;
      }
   }
}
struct Node* insertNode(int data){
   struct Node* node = new Node;
   node->data = data;
   node->left = NULL;
   node->right = NULL;
   return (node);
}
int main(){
   struct Node* root = insertNode(6);
   root->left = insertNode(2);
   root->right = insertNode(9);
   root->left->left = insertNode(8);
   root->left->right = insertNode(4);
   root->right->left = insertNode(7);
   root->right->left->left = insertNode(13);
   cout<<"Post Order Traversal of the binary tree :\n";
   postOrderTraversal(root);
   return 0;
}

Output

Post Order Traversal of the binary tree :
8    4    2    13    7    9    6

The same solution can be updated and the usage of the hash table can be eliminated. As its work is to store visited nodes. We will add a visited flag to every node it the tree itself to reduce the load on the system, this will make our algorithm better.

A more effective solution is using an unordered map, this will reduce the overhead of trackback to the head.

Example

Program to show the implementation

 Live Demo

#include <bits/stdc++.h>
using namespace std;
struct Node {
   int data;
   struct Node *left, *right;
   bool visited;
};
void postOrderTraversal(Node* root) {
   Node* n = root;
   unordered_map<Node*, Node*> postorder;
   postorder.insert(pair<Node*, Node*>(root, nullptr));
   while (n) {
      if (n->left && postorder.find(n->left) == postorder.end()) {
         postorder.insert(pair<Node*, Node*>(n->left, n));
         n = n->left;
      }
      else if (n->right && postorder.find(n->right) == postorder.end()) {
         postorder.insert(pair<Node*, Node*>(n->right, n));
         n = n->right;
      }
      else {
         cout<<n->data<<"\t";
         n = (postorder.find(n))->second;
      }
   }
}
struct Node* insertNode(int data) {
   struct Node* node = new Node;
   node->data = data;
   node->left = NULL;
   node->right = NULL;
   node->visited = false;
   return (node);
}
int main() {
   struct Node* root = insertNode(6);
   root->left = insertNode(2);
   root->right = insertNode(9);
   root->left->left = insertNode(8);
   root->left->right = insertNode(4);
   root->right->left = insertNode(7);
   root->right->left->left = insertNode(13);
   cout<<"Post Order Traversal of the binary tree :\n";
   postOrderTraversal(root);
   return 0;
}

Output

Post Order Traversal of the binary tree :
8    4    2    13    7    9    6
raja
Published on 17-Apr-2020 07:01:20
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