Flatten binary tree in order of post-order traversal

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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.

Create a dummy node.
Create variable called ‘prev’ and make it point to the dummy node.
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:

C++

// 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;
}

Java

// Java implementation of the approach
class GFG
{
 
// Node of the binary tree
static class node 
{
    int data;
    node left;
    node right;
    node(int data)
    {
        this.data = data;
        left = null;
        right = null;
    }
};
static node prev;
 
// Function to print the flattened
// binary Tree
static void print(node parent)
{
    node curr = parent;
    while (curr != null)
    {
        System.out.print(curr.data + " ");
        curr = curr.right;
    }
}
 
// Function to perform post-order traversal
// recursively
static void postorder(node curr)
{
    // Base case
    if (curr == null)
        return;
    postorder(curr.left);
    postorder(curr.right);
    prev.left = null;
    prev.right = curr;
    prev = curr;
}
 
// Function to flatten the given binary tree
// using post order traversal
static node flatten(node parent)
{
    // Dummy node
    node dummy = new node(-1);
 
    // Pointer to previous element
    prev = dummy;
 
    // Calling post-order traversal
    postorder(parent);
 
    prev.left = null;
    prev.right = null;
    node ret = dummy.right;
 
    // Delete dummy node
    dummy = null;
    return ret;
}
 
// Driver code
public static void main(String[] args)
{
    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));
}
} 
 
// This code is contributed by PrinciRaj1992

Python3

# Python implementation of above algorithm
 
# Utility class to create a node 
class node: 
    def __init__(self, key): 
        self.data = key 
        self.left = self.right = None
 
# Function to print the flattened
# binary Tree
def print_(parent):
 
    curr = parent
    while (curr != None):
        print( curr.data ,end = " ")
        curr = curr.right
 
prev = None
 
# Function to perform post-order traversal
# recursively
def postorder( curr ):
     
    global prev
     
    # Base case
    if (curr == None):
        return
    postorder(curr.left)
    postorder(curr.right)
    prev.left = None
    prev.right = curr
    prev = curr
 
# Function to flatten the given binary tree
# using post order traversal
def flatten(parent):
 
    global prev
     
    # Dummy node
    dummy = node(-1)
 
    # Pointer to previous element
    prev = dummy
 
    # Calling post-order traversal
    postorder(parent)
 
    prev.left = None
    prev.right = None
    ret = dummy.right
 
    return ret
 
# Driver code
root = node(5)
root.left = node(3)
root.right = node(7)
root.left.left = node(2)
root.left.right = node(4)
root.right.left = node(6)
root.right.right = node(8)
 
print_(flatten(root))
 
 
# This code is contributed by Arnab Kundu

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Node of the binary tree
public class node 
{
    public int data;
    public node left;
    public node right;
    public node(int data)
    {
        this.data = data;
        left = null;
        right = null;
    }
};
static node prev;
 
// Function to print the flattened
// binary Tree
static void print(node parent)
{
    node curr = parent;
    while (curr != null)
    {
        Console.Write(curr.data + " ");
        curr = curr.right;
    }
}
 
// Function to perform post-order traversal
// recursively
static void postorder(node curr)
{
    // Base case
    if (curr == null)
        return;
    postorder(curr.left);
    postorder(curr.right);
    prev.left = null;
    prev.right = curr;
    prev = curr;
}
 
// Function to flatten the given binary tree
// using post order traversal
static node flatten(node parent)
{
    // Dummy node
    node dummy = new node(-1);
 
    // Pointer to previous element
    prev = dummy;
 
    // Calling post-order traversal
    postorder(parent);
 
    prev.left = null;
    prev.right = null;
    node ret = dummy.right;
 
    // Delete dummy node
    dummy = null;
    return ret;
}
 
// Driver code
public static void Main(String[] args)
{
    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));
}
}
 
// This code is contributed by Princi Singh

Javascript

<script>
 
// Javascript implementation of the approach
 
// Node of the binary tree
class node 
{
    constructor(data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
};
 
var prev = null;
 
// Function to print the flattened
// binary Tree
function print(parent)
{
    var curr = parent;
    while (curr != null)
    {
        document.write(curr.data + " ");
        curr = curr.right;
    }
}
 
// Function to perform post-order traversal
// recursively
function postorder(curr)
{
     
    // Base case
    if (curr == null)
        return;
         
    postorder(curr.left);
    postorder(curr.right);
    prev.left = null;
    prev.right = curr;
    prev = curr;
}
 
// Function to flatten the given binary tree
// using post order traversal
function flatten(parent)
{
     
    // Dummy node
    var dummy = new node(-1);
 
    // Pointer to previous element
    prev = dummy;
 
    // Calling post-order traversal
    postorder(parent);
 
    prev.left = null;
    prev.right = null;
    var ret = dummy.right;
 
    // Delete dummy node
    dummy = null;
    return ret;
}
 
// Driver code
var 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));
 
// This code is contributed by noob2000
 
</script>

Output:

2 4 3 6 8 7 5

Time complexity: O(N)
Auxiliary Space: O(N). since N extra space has been taken.

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