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 structurestruct node { char data; struct node *left, *right;};// Creates and initilize a new nodestruct 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 Traversalvoid inOrder(struct node* root){ if (root) { preOrder(root->left); cout << root->data << " "; preOrder(root->right); }}// Perform Preorder Traversalvoid preOrder(struct node* root){ if (root) { cout << root->data << " "; inOrder(root->left); inOrder(root->right); }}// Driver Codeint 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;} |
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Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Node structurestatic class node{ char data; node left, right;};// Creates and initilize a new nodestatic 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 Traversalstatic void inOrder(node root){ if (root != null) { preOrder(root.left); System.out.print(root.data + " "); preOrder(root.right); }}// Perform Preorder Traversalstatic void preOrder(node root){ if (root != null) { System.out.print(root.data + " "); inOrder(root.left); inOrder(root.right); }}// Driver Codepublic 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 |
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C#
// C# program to implement// the above approachusing System;class GFG{// Node structureclass node{ public char data; public node left, right;};// Creates and initilize a new nodestatic 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 Traversalstatic void inOrder(node root){ if (root != null) { preOrder(root.left); Console.Write(root.data + " "); preOrder(root.right); }}// Perform Preorder Traversalstatic void preOrder(node root){ if (root != null) { Console.Write(root.data + " "); inOrder(root.left); inOrder(root.right); }}// Driver Codepublic 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 |
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Output:
7 4 5 1 3 6
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 structurestruct node { char data; struct node *left, *right;};// Creates and initilize a new nodestruct 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 Traversalvoid preOrder(struct node* root){ if (root) { cout << root->data << " "; postOrder(root->left); postOrder(root->right); }}// Perform Postorder Traversalvoid postOrder(struct node* root){ if (root) { preOrder(root->left); preOrder(root->right); cout << root->data << " "; }}// Driver Codeint 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;} |
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Java
// Java Program to implement// the above approachclass GFG{// Node structurestatic class node{ char data; node left, right;};// Creates and initilize a new nodestatic 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 Traversalstatic void preOrder(node root){ if (root != null) { System.out.print(root.data + " "); postOrder(root.left); postOrder(root.right); }}// Perform Postorder Traversalstatic void postOrder(node root){ if (root != null) { preOrder(root.left); preOrder(root.right); System.out.print(root.data + " "); }}// Driver Codepublic 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 |
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C#
// C# Program to implement// the above approachusing System;class GFG{// Node structureclass node{ public char data; public node left, right;};// Creates and initilize a new nodestatic 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 Traversalstatic void preOrder(node root){ if (root != null) { Console.Write(root.data + " "); postOrder(root.left); postOrder(root.right); }}// Perform Postorder Traversalstatic void postOrder(node root){ if (root != null) { preOrder(root.left); preOrder(root.right); Console.Write(root.data + " "); }}// Driver Codepublic 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 |
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Output:
A F D B E C
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 structurestruct node { char data; struct node *left, *right;};// Creates and initilize a new nodestruct 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 Traversalvoid inOrder(struct node* root){ if (root) { postOrder(root->left); cout << root->data << " "; postOrder(root->right); }}// Perform Postorder Traversalvoid postOrder(struct node* root){ if (root) { inOrder(root->left); inOrder(root->right); cout << root->data << " "; }}// Driver Codeint 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;} |
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Java
// Java Program to implement// the above approachimport java.util.*;class GFG{// Node structurestatic class node { char data; node left, right;};// Creates and initilize a new nodestatic 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 Traversalstatic void inOrder(node root){ if (root != null) { postOrder(root.left); System.out.print(root.data + " "); postOrder(root.right); }}// Perform Postorder Traversalstatic void postOrder(node root){ if (root != null) { inOrder(root.left); inOrder(root.right); System.out.print(root.data + " "); }}// Driver Codepublic 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 |
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C#
// C# Program to implement// the above approachusing System;class GFG{// Node structureclass node { public char data; public node left, right;};// Creates and initilize a new nodestatic 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 Traversalstatic void inOrder(node root){ if (root != null) { postOrder(root.left); Console.Write(root.data + " "); postOrder(root.right); }}// Perform Postorder Traversalstatic void postOrder(node root){ if (root != null) { inOrder(root.left); inOrder(root.right); Console.Write(root.data + " "); }}// Driver Codepublic 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 |
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Output:
F D B A E C
Time Complexity: O(N)
Auxiliary Space: O(N)
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