# 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.

Constraints

1 <= Nodes in the Tree <= 500

Output Format

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

Sample Input

1
\
2
\
5
/  \
3    6
\
4
Sample Output

1 2 3 4 5 6```

### Solution :

```                            ```Solution in C :

In Java :

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

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;
Inorder(root->left);
printf("%d ", root->data);
Inorder(root->right);
}

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) {
if(root==NULL){
return;
}
else{
inOrder(root->left);
printf("%d ",root->data);
inOrder(root->right);
}
}

In python 3 :

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

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

def inOrder(root):
acc = []
_preOrder(root, acc)
print(" ".join(map(str, acc)))
```

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