Given a Binary Tree, the task is to flatten it in order of ZigZag traversal of the tree. In the flattened binary tree, the left node of all the nodes must be NULL.
Examples:
Input: 1 / \ 5 2 / \ / \ 6 4 9 3 Output: 1 2 5 6 4 9 3 Input: 1 \ 2 \ 3 \ 4 \ 5 Output: 1 2 3 4 5
Approach: We will solve this problem by simulating the ZigZag traversal of Binary Tree.
Algorithm:
- Create two stacks, “c_lev” and “n_lev” and to store the nodes of current and next level Binary tree.
- Create a variable “prev” and initialise it by parent node.
- Push right and left children of parent in the c_lev stack.
- Apply ZigZag traversal. Lets say “curr” is top most element in “c_lev”. Then,
- If ‘curr’ is NULL, continue.
- Else push curr->left and curr->right on the stack “n_lev” in appropriate order. If we are performing left to right traversal then curr->left is pushed first else curr->right is pushed first.
- Set prev = curr.
Below is the implementation of the above approach:
CPP
// 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 flatten Binary tree void flatten(node* parent)
{ // Queue to store node
// for BFS
stack<node *> c_lev, n_lev;
c_lev.push(parent->left);
c_lev.push(parent->right);
bool lev = 1;
node* prev = parent;
// Code for BFS
while (c_lev.size()) {
// Size of queue
while (c_lev.size()) {
// Front most node in
// queue
node* curr = c_lev.top();
c_lev.pop();
// Base case
if (curr == NULL)
continue ;
prev->right = curr;
prev->left = NULL;
prev = curr;
// Pushing new elements
// in queue
if (!lev)
n_lev.push(curr->left);
n_lev.push(curr->right);
if (lev)
n_lev.push(curr->left);
}
lev = (!lev);
c_lev = n_lev;
while (n_lev.size())
n_lev.pop();
}
prev->left = NULL;
prev->right = NULL;
} // Function to print flattened // binary tree void print(node* parent)
{ node* curr = parent;
while (curr != NULL)
cout << curr->data << " " , curr = curr->right;
} // Driver code int main()
{ node* root = new node(1);
root->left = new node(5);
root->right = new node(2);
root->left->left = new node(6);
root->left->right = new node(4);
root->right->left = new node(9);
root->right->right = new node(3);
// Calling required functions
flatten(root);
print(root);
return 0;
} |
Java
// Java implementation of the approach import java.io.*;
import java.util.*;
class GFG {
static class node {
int data;
node left, right;
node( int data)
{
this .data = data;
left = right = null ;
}
}
// Function to flatten Binary tree
static void flatten(node parent)
{
// Stack to store node for BFS
Stack<node> c_lev = new Stack<>();
Stack<node> n_lev = new Stack<>();
c_lev.push(parent.left);
c_lev.push(parent.right);
boolean lev = true ;
node prev = parent;
// Code for BFS
while (c_lev.size() != 0 ) {
// Size
while (c_lev.size() != 0 ) {
node curr = c_lev.peek();
c_lev.pop();
if (curr == null ) {
continue ;
}
prev.right = curr;
prev.left = null ;
prev = curr;
if (!lev) {
n_lev.push(curr.left);
}
n_lev.push(curr.right);
if (lev) {
n_lev.push(curr.left);
}
}
lev = (!lev);
c_lev = (Stack)n_lev.clone();
while (n_lev.size() != 0 ) {
n_lev.pop();
}
}
prev.left = null ;
prev.right = null ;
}
static void print(node parent)
{
node curr = parent;
while (curr != null ) {
System.out.print(curr.data + " " );
curr = curr.right;
}
}
public static void main(String[] args)
{
node root = new node( 1 );
root.left = new node( 5 );
root.right = new node( 2 );
root.left.left = new node( 6 );
root.left.right = new node( 4 );
root.right.left = new node( 9 );
root.right.right = new node( 3 );
// Calling required functions
flatten(root);
print(root);
}
} // This code is contributed by lokeshmvs21. |
Python3
# python implementation of the approach # node of the binary tree class node:
# A utility function to create a new node
def __init__( self , key):
self .data = key
self .left = None
self .right = None
# Function to flatten Binary Tree def flatten(parent):
# queue to store for bfs
c_lev = []
n_lev = []
c_lev.append(parent.left)
c_lev.append(parent.right)
lev = True
prev = parent
# code for bfs
while ( len (c_lev) > 0 ):
# size of queue
while ( len (c_lev) > 0 ):
# front most node in queue
curr = c_lev.pop()
# base case
if (curr = = None ):
continue
prev.right = curr
prev.left = None
prev = curr
# pushing new elements in queue
if (lev = = False ):
n_lev.append(curr.left)
n_lev.append(curr.right)
if (lev = = True ):
n_lev.append(curr.left)
if lev is True :
lev = False
else :
lev = True
c_lev = n_lev.copy()
while ( len (n_lev) > 0 ):
n_lev.pop()
prev.left = None
prev.right = None
# function to print flattened binary tree def printTree(parent):
curr = parent
while (curr is not None ):
print (curr.data, end = " " )
curr = curr.right
# driver code root = node( 1 )
root.left = node( 5 )
root.right = node( 2 )
root.left.left = node( 6 )
root.left.right = node( 4 )
root.right.left = node( 9 )
root.right.right = node( 3 )
# calling required function flatten(root) printTree(root) # This code is contributed by Yash Agarwal(yashagarwal2852002) |
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 ;
}
}; // Function to flatten Binary tree function flatten()
{ // Queue to store node
// for BFS
var c_lev = [], n_lev = [];
c_lev.push(parent.left);
c_lev.push(parent.right);
var lev = 1;
var prev = parent;
// Code for BFS
while (c_lev.length!=0) {
// Size of queue
while (c_lev.length!=0) {
// Front most node in
// queue
var curr = c_lev[c_lev.length-1];
c_lev.pop();
// Base case
if (curr == null )
continue ;
prev.right = curr;
prev.left = null ;
prev = curr;
// Pushing new elements
// in queue
if (!lev)
n_lev.push(curr.left);
n_lev.push(curr.right);
if (lev)
n_lev.push(curr.left);
}
lev = lev==0?1:0;
c_lev = n_lev;
n_lev = [];
}
prev.left = null ;
prev.right = null ;
} // Function to print flattened // binary tree function print()
{ var curr = parent;
while (curr != null )
{
document.write(curr.data + " " );
curr = curr.right;
}
} // Driver code var parent = new node(1);
parent.left = new node(5);
parent.right = new node(2);
parent.left.left = new node(6);
parent.left.right = new node(4);
parent.right.left = new node(9);
parent.right.right = new node(3);
// Calling required functions flatten(); print(); // This code is contributed by rrrtnx. </script> |
C#
using System;
using System.Collections.Generic;
public class Node {
public int data;
public Node left;
public Node right;
public Node( int data) {
this .data = data;
left = null ;
right = null ;
}
} public class BinaryTree {
public static void Flatten(Node parent) {
Stack<Node> c_lev = new Stack<Node>();
Stack<Node> n_lev = new Stack<Node>();
c_lev.Push(parent.left);
c_lev.Push(parent.right);
bool lev = true ;
Node prev = parent;
while (c_lev.Count > 0) {
while (c_lev.Count > 0) {
Node curr = c_lev.Pop();
if (curr == null )
continue ;
prev.right = curr;
prev.left = null ;
prev = curr;
if (!lev)
n_lev.Push(curr.left);
n_lev.Push(curr.right);
if (lev)
n_lev.Push(curr.left);
}
lev = !lev;
c_lev = n_lev;
n_lev = new Stack<Node>();
}
prev.left = null ;
prev.right = null ;
}
public static void Print(Node parent) {
Node curr = parent;
while (curr != null ) {
Console.Write(curr.data + " " );
curr = curr.right;
}
}
public static void Main( string [] args) {
Node root = new Node(1);
root.left = new Node(5);
root.right = new Node(2);
root.left.left = new Node(6);
root.left.right = new Node(4);
root.right.left = new Node(9);
root.right.right = new Node(3);
Flatten(root);
Print(root);
}
} |
Output:
1 2 5 6 4 9 3
Time Complexity: O(N)
Space Complexity: O(N) where N is the size of Binary Tree.