Mix Order Traversal of a Binary Tree

Given a Binary Tree consisting of N nodes, the task is to print its Mix Order Traversal. 

Mix Order Traversal is a tree traversal technique, which involves any two of the existing traversal techniques like Inorder, Preorder and Postorder Traversal. Any two of them can be performed or alternate levels of given tree and a mix traversal can be obtained.

Examples:

Input: N = 6

Output: 7 4 5 1 3 6 
Explanation:
Inorder-Preorder Mix Traversal is applied to the given tree in the following order: 
Inorder Traversal is applied at level 0 
Preorder Traversal is applied at level 1 
Inorder Traversal at level 2.

Output: 4 5 7 1 6 3 
Explanation:
Inorder-Postorder Mix Traversal is applied to the given tree in the following order: 
Inorder Traversal is applied at level 0 
Postorder Traversal is applied at level 1 
Inorder Traversal at level 2.  

Approach:
The possible Mix Order Traversals are as follows:  

Inorder-Preorder Mix Traversal

Steps for inorder() will be:

• Perform Preorder Traversal on the left subtree.
• Print the current node.
• Perform Preorder Traversal on right subtree.

Steps for preorder() will be:

• Print the current node.
• Perform Inorder Traversal on left subtree(root->left).
• Perform Inorder Traversal on right subtree.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

void inOrder(struct node* root);
void preOrder(struct node* root);

// Node structure
struct node {
    char data;
    struct node *left, *right;
};

// Creates and initialize a new node
struct node* newNode(char ch)
{
    // Allocating memory to a new node
    struct node* n = (struct node*)
        malloc(sizeof(struct node));
    n->data = ch;
    n->left = NULL;
    n->right = NULL;
    return n;
}

// Perform Inorder Traversal
void inOrder(struct node* root)
{
    if (root) {
        preOrder(root->left);
        cout << root->data << " ";
        preOrder(root->right);
    }
}

// Perform Preorder Traversal
void preOrder(struct node* root)
{
    if (root) {
        cout << root->data << " ";
        inOrder(root->left);
        inOrder(root->right);
    }
}

// Driver Code
int main()
{
    // Given tree
    struct node* root = newNode('1');
    root->left = newNode('7');
    root->right = newNode('3');
    root->left->left = newNode('4');
    root->left->right = newNode('5');
    root->right->left = newNode('6');

    // Perform Mix order traversal
    inOrder(root);

    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;

class GFG{

// Node structure
static class node
{
    char data;
    node left, right;
};

// Creates and initialize a new node
static node newNode(char ch)
{

    // Allocating memory to a new node
    node n = new node();
    n.data = ch;
    n.left = null;
    n.right = null;
    return n;
}

// Perform Inorder Traversal
static void inOrder(node root)
{
    if (root != null)
    {
        preOrder(root.left);
        System.out.print(root.data + " ");
        preOrder(root.right);
    }
}

// Perform Preorder Traversal
static void preOrder(node root)
{
    if (root != null)
    {
        System.out.print(root.data + " ");
        inOrder(root.left);
        inOrder(root.right);
    }
}

// Driver Code
public static void main(String[] args)
{

    // Given tree
    node root = newNode('1');
    root.left = newNode('7');
    root.right = newNode('3');
    root.left.left = newNode('4');
    root.left.right = newNode('5');
    root.right.left = newNode('6');

    // Perform Mix order traversal
    inOrder(root);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to implement the above approach

# Node structure
class node:
    def __init__(self):
        self.data = 0
        self.left = None
        self.right = None

# Creates and initialize a new node
def newNode(ch):

    # Allocating memory to a new node
    n = node()
    n.data = ch
    n.left = None
    n.right = None
    return n

# Perform Inorder Traversal
def inOrder(root):
    if root != None:
        preOrder(root.left)
        print(root.data, end = " ")
        preOrder(root.right)

# Perform Preorder Traversal
def preOrder(root):
    if root != None:
        print(root.data, end = " ")
        inOrder(root.left)
        inOrder(root.right)

# Driver Code
# Given tree
root = newNode('1')
root.left = newNode('7')
root.right = newNode('3')
root.left.left = newNode('4')
root.left.right = newNode('5')
root.right.left = newNode('6')

# Perform Mix order traversal
inOrder(root)

# This code is contributed by divyeshrabadiya07.

C#

// C# program to implement
// the above approach
using System;
class GFG{

// Node structure
class node
{
    public char data;
    public node left, right;
};

// Creates and initialize a new node
static node newNode(char ch)
{

    // Allocating memory to a new node
    node n = new node();
    n.data = ch;
    n.left = null;
    n.right = null;
    return n;
}

// Perform Inorder Traversal
static void inOrder(node root)
{
    if (root != null)
    {
        preOrder(root.left);
        Console.Write(root.data + " ");
        preOrder(root.right);
    }
}

// Perform Preorder Traversal
static void preOrder(node root)
{
    if (root != null)
    {
        Console.Write(root.data + " ");
        inOrder(root.left);
        inOrder(root.right);
    }
}

// Driver Code
public static void Main(String[] args)
{

    // Given tree
    node root = newNode('1');
    root.left = newNode('7');
    root.right = newNode('3');
    root.left.left = newNode('4');
    root.left.right = newNode('5');
    root.right.left = newNode('6');

    // Perform Mix order traversal
    inOrder(root);
}
}

// This code is contributed by sapnasingh4991

Javascript

<script>

// Javascript program to implement
// the above approach

// Node structure
class node
{
    constructor()
    {
        this.data = 0;
        this.left = null;
        this.right =  null;
    }
};

// Creates and initialize a new node
function newNode(ch)
{

    // Allocating memory to a new node
    var n = new node();
    n.data = ch;
    n.left = null;
    n.right = null;
    return n;
}

// Perform Inorder Traversal
function inOrder(root)
{
    if (root != null)
    {
        preOrder(root.left);
        document.write(root.data + " ");
        preOrder(root.right);
    }
}

// Perform Preorder Traversal
function preOrder(root)
{
    if (root != null)
    {
        document.write(root.data + " ");
        inOrder(root.left);
        inOrder(root.right);
    }
}

// Driver Code
// Given tree
var root = newNode('1');
root.left = newNode('7');
root.right = newNode('3');
root.left.left = newNode('4');
root.left.right = newNode('5');
root.right.left = newNode('6');
// Perform Mix order traversal
inOrder(root);


</script>

Output:

7 4 5 1 3 6

The time complexity for both inOrder and preOrder traversals is O(n), where n is the number of nodes in the tree. This is because each node is visited once.
Auxiliary space complexity:

The auxiliary space of newNode function is O(1), because it only allocates memory for a single node.
The auxiliary space complexity of inOrder and preOrder functions is O(h), where h is the height of the tree. 

Preorder-Postorder Mix Traversal

Steps for preorder() are as follows:

• Print the current node.
• Perform Postorder traversal on left subtree.
• Perform Postorder Traversal on the right subtree.

Steps for postorder() are as follows:

• Perform preorder traversal on the left subtree.
• Perform preorder traversal on right subtree.
• Print the current node.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

void preOrder(struct node* root);
void postOrder(struct node* root);

// Node structure
struct node {
    char data;
    struct node *left, *right;
};

// Creates and initialize a new node
struct node* newNode(char ch)
{
    // Allocating memory to a new node
    struct node* n = (struct node*)
        malloc(sizeof(struct node));
    n->data = ch;
    n->left = NULL;
    n->right = NULL;
    return n;
}

// Perform Preorder Traversal
void preOrder(struct node* root)
{
    if (root) {
        cout << root->data << " ";
        postOrder(root->left);
        postOrder(root->right);
    }
}

// Perform Postorder Traversal
void postOrder(struct node* root)
{
    if (root) {
        preOrder(root->left);
        preOrder(root->right);
        cout << root->data << " ";
    }
}

// Driver Code
int main()
{
    // Given tree
    struct node* root = newNode('A');
    root->left = newNode('B');
    root->right = newNode('C');
    root->left->left = newNode('F');
    root->left->right = newNode('D');
    root->right->right = newNode('E');

    // Starting Mix order traversal
    preOrder(root);

    return 0;
}

Java

// Java Program to implement
// the above approach
class GFG{

// Node structure
static class node
{
    char data;
    node left, right;
};

// Creates and initialize a new node
static node newNode(char ch)
{
    // Allocating memory to a new node
    node n = new node();

    n.data = ch;
    n.left = null;
    n.right = null;
    return n;
}

// Perform Preorder Traversal
static void preOrder(node root)
{
    if (root != null)
    {
        System.out.print(root.data + " ");
        postOrder(root.left);
        postOrder(root.right);
    }
}

// Perform Postorder Traversal
static void postOrder(node root)
{
    if (root != null)
    {
        preOrder(root.left);
        preOrder(root.right);
        System.out.print(root.data + " ");
    }
}

// Driver Code
public static void main(String[] args)
{
    // Given tree
    node root = newNode('A');
    root.left = newNode('B');
    root.right = newNode('C');
    root.left.left = newNode('F');
    root.left.right = newNode('D');
    root.right.right = newNode('E');

    // Starting Mix order traversal
    preOrder(root);
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 Program to implement the above approach

# Node structure
class node:
    def __init__(self):
        self.data = '0'
        self.left = None
        self.right = None

# Creates and initialize a new node
def newNode(ch):
    # Allocating memory to a new node
    n = node()

    n.data = ch
    n.left = None
    n.right = None
    return n

# Perform Preorder Traversal
def preOrder(root):
    if root != None:
        print(root.data,  end = " ")
        postOrder(root.left)
        postOrder(root.right)

# Perform Postorder Traversal
def postOrder(root):
    if root != None:
        preOrder(root.left)
        preOrder(root.right)
        print(root.data, end = " ")

# Given tree
root = newNode('A')
root.left = newNode('B')
root.right = newNode('C')
root.left.left = newNode('F')
root.left.right = newNode('D')
root.right.right = newNode('E')

# Starting Mix order traversal
preOrder(root)

# This code is contributed by divyesh072019.

C#

// C# Program to implement
// the above approach
using System;
class GFG{

// Node structure
class node
{
    public char data;
    public node left, right;
};

// Creates and initialize a new node
static node newNode(char ch)
{
    // Allocating memory to a new node
    node n = new node();

    n.data = ch;
    n.left = null;
    n.right = null;
    return n;
}

// Perform Preorder Traversal
static void preOrder(node root)
{
    if (root != null)
    {
        Console.Write(root.data + " ");
        postOrder(root.left);
        postOrder(root.right);
    }
}

// Perform Postorder Traversal
static void postOrder(node root)
{
    if (root != null)
    {
        preOrder(root.left);
        preOrder(root.right);
        Console.Write(root.data + " ");
    }
}

// Driver Code
public static void Main(String[] args)
{
    // Given tree
    node root = newNode('A');
    root.left = newNode('B');
    root.right = newNode('C');
    root.left.left = newNode('F');
    root.left.right = newNode('D');
    root.right.right = newNode('E');

    // Starting Mix order traversal
    preOrder(root);
}
}

// This code is contributed by Rohit_ranjan

Javascript

<script>
    // Javascript Program to implement the above approach

    // Node structure
    class node
    {
        constructor() {
           this.left;
           this.right;
           this.data;
        }
    }

    // Creates and initialize a new node
    function newNode(ch)
    {
        // Allocating memory to a new node
        let n = new node();

        n.data = ch;
        n.left = null;
        n.right = null;
        return n;
    }

    // Perform Preorder Traversal
    function preOrder(root)
    {
        if (root != null)
        {
            document.write(root.data + " ");
            postOrder(root.left);
            postOrder(root.right);
        }
    }

    // Perform Postorder Traversal
    function postOrder(root)
    {
        if (root != null)
        {
            preOrder(root.left);
            preOrder(root.right);
            document.write(root.data + " ");
        }
    }

    // Given tree
    let root = newNode('A');
    root.left = newNode('B');
    root.right = newNode('C');
    root.left.left = newNode('F');
    root.left.right = newNode('D');
    root.right.right = newNode('E');

    // Starting Mix order traversal
    preOrder(root);

    // This code is contributed by rameshtravel07.
</script>

Output:

A F D B E C

Time Complexity: O(N) where N is the number of nodes in the tree.
Auxiliary Space: O(log(N))

Inorder-Postorder Mix Traversal 

Steps for inorder() are as follows:

• Perform Postorder Traversal on the left subtree.
• Print the current node.
• Perform Postorder Traversal on the right subtree.

Steps for postorder() will be: 

• Perform Inorder Traversal on left subtree.
• Perform Inorder Traversal on right subtree.
• Print the current node.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

void inOrder(struct node* root);
void postOrder(struct node* root);

// Node structure
struct node {
    char data;
    struct node *left, *right;
};

// Creates and initialize a new node
struct node* newNode(char ch)
{

    // Allocating memory to a new node
    struct node* n = (struct node*)
        malloc(sizeof(struct node));
    n->data = ch;
    n->left = NULL;
    n->right = NULL;
    return n;
}

// Perform Inorder Traversal
void inOrder(struct node* root)
{
    if (root) {
        postOrder(root->left);
        cout << root->data << " ";
        postOrder(root->right);
    }
}

// Perform Postorder Traversal
void postOrder(struct node* root)
{
    if (root) {
        inOrder(root->left);
        inOrder(root->right);
        cout << root->data << " ";
    }
}

// Driver Code
int main()
{
    // Given tree
    struct node* root = newNode('A');
    root->left = newNode('B');
    root->right = newNode('C');
    root->left->left = newNode('F');
    root->left->right = newNode('D');
    root->right->right = newNode('E');

    // Starting Mix order traversal
    inOrder(root);

    return 0;
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG{

// Node structure
static class node
{
    char data;
    node left, right;
};

// Creates and initialize a new node
static node newNode(char ch)
{

    // Allocating memory to a new node
    node n = new node();
    n.data = ch;
    n.left = null;
    n.right = null;
    return n;
}

// Perform Inorder Traversal
static void inOrder(node root)
{
    if (root != null)
    {
        postOrder(root.left);
        System.out.print(root.data + " ");
        postOrder(root.right);
    }
}

// Perform Postorder Traversal
static void postOrder(node root)
{
    if (root != null)
    {
        inOrder(root.left);
        inOrder(root.right);
        System.out.print(root.data + " ");
    }
}

// Driver Code
public static void main(String[] args)
{
    // Given tree
    node root = newNode('A');
    root.left = newNode('B');
    root.right = newNode('C');
    root.left.left = newNode('F');
    root.left.right = newNode('D');
    root.right.right = newNode('E');

    // Starting Mix order traversal
    inOrder(root);
}
}

// This code is contributed by sapnasingh4991

Python3

# Python3 Program to implement the above approach

# Node structure
class node:
    def __init__(self):
        self.data = '0'
        self.left = None
        self.right = None

# Creates and initialize a new node
def newNode(ch):
    # Allocating memory to a new node
    n = node()
    n.data = ch
    n.left = None
    n.right = None
    return n

# Perform Inorder Traversal
def inOrder(root):
    if root != None:
        postOrder(root.left)
        print(root.data, end = " ")
        postOrder(root.right)

# Perform Postorder Traversal
def postOrder(root):
    if root != None:
        inOrder(root.left)
        inOrder(root.right)
        print(root.data, end = " ")

# Given tree
root = newNode('A')
root.left = newNode('B')
root.right = newNode('C')
root.left.left = newNode('F')
root.left.right = newNode('D')
root.right.right = newNode('E')

# Starting Mix order traversal
inOrder(root)

# This code is contributed by decode2207.

C#

// C# Program to implement
// the above approach
using System;
class GFG{

// Node structure
class node
{
    public char data;
    public node left, right;
};

// Creates and initialize a new node
static node newNode(char ch)
{

    // Allocating memory to a new node
    node n = new node();
    n.data = ch;
    n.left = null;
    n.right = null;
    return n;
}

// Perform Inorder Traversal
static void inOrder(node root)
{
    if (root != null)
    {
        postOrder(root.left);
        Console.Write(root.data + " ");
        postOrder(root.right);
    }
}

// Perform Postorder Traversal
static void postOrder(node root)
{
    if (root != null)
    {
        inOrder(root.left);
        inOrder(root.right);
        Console.Write(root.data + " ");
    }
}

// Driver Code
public static void Main(String[] args)
{
    // Given tree
    node root = newNode('A');
    root.left = newNode('B');
    root.right = newNode('C');
    root.left.left = newNode('F');
    root.left.right = newNode('D');
    root.right.right = newNode('E');

    // Starting Mix order traversal
    inOrder(root);
}
}

// This code is contributed by sapnasingh4991

Javascript

<script>
    // Javascript Program to implement the above approach

    // Node structure
    class node
    {
        constructor() {
           this.left;
           this.right;
           this.data;
        }
    }

    // Creates and initialize a new node
    function newNode(ch)
    {

        // Allocating memory to a new node
        let n = new node();
        n.data = ch;
        n.left = null;
        n.right = null;
        return n;
    }

    // Perform Inorder Traversal
    function inOrder(root)
    {
        if (root != null)
        {
            postOrder(root.left);
            document.write(root.data + " ");
            postOrder(root.right);
        }
    }

    // Perform Postorder Traversal
    function postOrder(root)
    {
        if (root != null)
        {
            inOrder(root.left);
            inOrder(root.right);
            document.write(root.data + " ");
        }
    }

    // Given tree
    let root = newNode('A');
    root.left = newNode('B');
    root.right = newNode('C');
    root.left.left = newNode('F');
    root.left.right = newNode('D');
    root.right.right = newNode('E');

    // Starting Mix order traversal
    inOrder(root);

  // This code is contributed by mukesh07.
</script>

Output:

F D B A E C

Time Complexity: O(N) 
Auxiliary Space: O(N)