Flatten binary tree in order of post-order traversal
Last Updated :
08 Aug, 2022
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++
#include <bits/stdc++.h>
using namespace std;
struct node {
int data;
node* left;
node* right;
node( int data)
{
this ->data = data;
left = NULL;
right = NULL;
}
};
void print(node* parent)
{
node* curr = parent;
while (curr != NULL)
cout << curr->data << " " , curr = curr->right;
}
void postorder(node* curr, node*& prev)
{
if (curr == NULL)
return ;
postorder(curr->left, prev);
postorder(curr->right, prev);
prev->left = NULL;
prev->right = curr;
prev = curr;
}
node* flatten(node* parent)
{
node* dummy = new node(-1);
node* prev = dummy;
postorder(parent, prev);
prev->left = NULL;
prev->right = NULL;
node* ret = dummy->right;
delete dummy;
return ret;
}
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
class GFG
{
static class node
{
int data;
node left;
node right;
node( int data)
{
this .data = data;
left = null ;
right = null ;
}
};
static node prev;
static void print(node parent)
{
node curr = parent;
while (curr != null )
{
System.out.print(curr.data + " " );
curr = curr.right;
}
}
static void postorder(node curr)
{
if (curr == null )
return ;
postorder(curr.left);
postorder(curr.right);
prev.left = null ;
prev.right = curr;
prev = curr;
}
static node flatten(node parent)
{
node dummy = new node(- 1 );
prev = dummy;
postorder(parent);
prev.left = null ;
prev.right = null ;
node ret = dummy.right;
dummy = null ;
return ret;
}
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));
}
}
|
Python3
class node:
def __init__( self , key):
self .data = key
self .left = self .right = None
def print_(parent):
curr = parent
while (curr ! = None ):
print ( curr.data ,end = " " )
curr = curr.right
prev = None
def postorder( curr ):
global prev
if (curr = = None ):
return
postorder(curr.left)
postorder(curr.right)
prev.left = None
prev.right = curr
prev = curr
def flatten(parent):
global prev
dummy = node( - 1 )
prev = dummy
postorder(parent)
prev.left = None
prev.right = None
ret = dummy.right
return ret
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))
|
C#
using System;
class GFG
{
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;
static void print(node parent)
{
node curr = parent;
while (curr != null )
{
Console.Write(curr.data + " " );
curr = curr.right;
}
}
static void postorder(node curr)
{
if (curr == null )
return ;
postorder(curr.left);
postorder(curr.right);
prev.left = null ;
prev.right = curr;
prev = curr;
}
static node flatten(node parent)
{
node dummy = new node(-1);
prev = dummy;
postorder(parent);
prev.left = null ;
prev.right = null ;
node ret = dummy.right;
dummy = null ;
return ret;
}
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));
}
}
|
Javascript
<script>
class node
{
constructor(data)
{
this .data = data;
this .left = null ;
this .right = null ;
}
};
var prev = null ;
function print(parent)
{
var curr = parent;
while (curr != null )
{
document.write(curr.data + " " );
curr = curr.right;
}
}
function postorder(curr)
{
if (curr == null )
return ;
postorder(curr.left);
postorder(curr.right);
prev.left = null ;
prev.right = curr;
prev = curr;
}
function flatten(parent)
{
var dummy = new node(-1);
prev = dummy;
postorder(parent);
prev.left = null ;
prev.right = null ;
var ret = dummy.right;
dummy = null ;
return ret;
}
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));
</script>
|
Time complexity: O(N)
Auxiliary Space: O(N). since N extra space has been taken.
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