Flatten Binary Tree in order of Zig Zag traversal
Last Updated :
17 Feb, 2023
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
#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 flatten(node* parent)
{
stack<node *> c_lev, n_lev;
c_lev.push(parent->left);
c_lev.push(parent->right);
bool lev = 1;
node* prev = parent;
while (c_lev.size()) {
while (c_lev.size()) {
node* curr = c_lev.top();
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;
while (n_lev.size())
n_lev.pop();
}
prev->left = NULL;
prev->right = NULL;
}
void print(node* parent)
{
node* curr = parent;
while (curr != NULL)
cout << curr->data << " " , curr = curr->right;
}
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);
flatten(root);
print(root);
return 0;
}
|
Java
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 ;
}
}
static void flatten(node parent)
{
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;
while (c_lev.size() != 0 ) {
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 );
flatten(root);
print(root);
}
}
|
Python3
class node:
def __init__( self , key):
self .data = key
self .left = None
self .right = None
def flatten(parent):
c_lev = []
n_lev = []
c_lev.append(parent.left)
c_lev.append(parent.right)
lev = True
prev = parent
while ( len (c_lev) > 0 ):
while ( len (c_lev) > 0 ):
curr = c_lev.pop()
if (curr = = None ):
continue
prev.right = curr
prev.left = None
prev = curr
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
def printTree(parent):
curr = parent
while (curr is not None ):
print (curr.data, end = " " )
curr = curr.right
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 )
flatten(root)
printTree(root)
|
Javascript
<script>
class node {
constructor(data)
{
this .data = data;
this .left = null ;
this .right = null ;
}
};
function flatten()
{
var c_lev = [], n_lev = [];
c_lev.push(parent.left);
c_lev.push(parent.right);
var lev = 1;
var prev = parent;
while (c_lev.length!=0) {
while (c_lev.length!=0) {
var curr = c_lev[c_lev.length-1];
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==0?1:0;
c_lev = n_lev;
n_lev = [];
}
prev.left = null ;
prev.right = null ;
}
function print()
{
var curr = parent;
while (curr != null )
{
document.write(curr.data + " " );
curr = curr.right;
}
}
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);
flatten();
print();
</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);
}
}
|
Time Complexity: O(N)
Space Complexity: O(N) where N is the size of Binary Tree.
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