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 recreate the Binary Tree from its post-order traversal. This will take O(N) extra space were N is the number of nodes in BST.

A better solution is to simulate post-order traversal of the given binary tree.

  1. Create a dummy node.
  2. Create variable called ‘prev’ and make it point to the dummy node.
  3. Perform post-order traversal and at each step.
    • Set prev -> right = curr
    • Set prev -> left = NULL
    • Set prev = curr

This will improve the space complexity to O(H) in the worst case as post-order traversal takes O(H) extra space.

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Node of the binary tree
struct node {
    int data;
    node* left;
    node* right;
    node(int data)
    {
        this->data = data;
        left = NULL;
        right = NULL;
    }
};
  
// Function to print the flattened
// binary Tree
void print(node* parent)
{
    node* curr = parent;
    while (curr != NULL)
        cout << curr->data << " ", curr = curr->right;
}
  
// Function to perform post-order traversal
// recursively
void postorder(node* curr, node*& prev)
{
    // Base case
    if (curr == NULL)
        return;
    postorder(curr->left, prev);
    postorder(curr->right, prev);
    prev->left = NULL;
    prev->right = curr;
    prev = curr;
}
  
// Function to flatten the given binary tree
// using post order traversal
node* flatten(node* parent)
{
    // Dummy node
    node* dummy = new node(-1);
  
    // Pointer to previous element
    node* prev = dummy;
  
    // Calling post-order traversal
    postorder(parent, prev);
  
    prev->left = NULL;
    prev->right = NULL;
    node* ret = dummy->right;
  
    // Delete dummy node
    delete dummy;
    return ret;
}
  
// Driver code
int main()
{
    node* root = new node(5);
    root->left = new node(3);
    root->right = new node(7);
    root->left->left = new node(2);
    root->left->right = new node(4);
    root->right->left = new node(6);
    root->right->right = new node(8);
  
    print(flatten(root));
  
    return 0;
}

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Output:

2 4 3 6 8 7 5


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