Tree: Inorder Traversal

Problem Statement :

In this challenge, you are required to implement inorder traversal of a tree.

Complete the inorder  function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values.

Input Format

Our hidden tester code passes the root node of a binary tree to your $inOrder* function.


1 <= Nodes in the Tree <= 500

Output Format

Print the tree's inorder traversal as a single line of space-separated values.

Sample Input

        /  \
       3    6
Sample Output

1 2 3 4 5 6

Solution :


                            Solution in C :

In Java :

void Inorder(Node root) {
    if (root == null) { return; }
    System.out.print( + " ");

In C++ :

/* you only have to complete the function given below.  
Node is defined as  

struct node
    int data;
    node* left;
    node* right;


void Inorder(node *root) {
    if (!root) return;
    printf("%d ", root->data);

In C :

/* you only have to complete the function given below.  
node is defined as  

struct node {
    int data;
    struct node *left;
    struct node *right;

void inOrder( struct node *root) {
      printf("%d ",root->data);

In python 3 :

Node is defined as
self.left (the left child of the node)
self.right (the right child of the node) (the value of the node)

def _preOrder(root, acc):
    if root:
        _preOrder(root.left, acc)
        _preOrder(root.right, acc)

def inOrder(root):
    acc = []
    _preOrder(root, acc)
    print(" ".join(map(str, acc)))
    #Write your code here

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