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Last Updated: 22 Jul, 2021

Easy

```
For the given tree below,
Postorder traversal for the given tree will be [4, 5, 2, 3, 1]. Hence, the answer is [4, 5, 2, 3, 1].
```

```
The first line contains the elements of the tree in the level order form separated by a single space. If any node does not have a left or right child, take -1 in its place. Refer to the example for further clarification.
```

```
Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 in its place.
For example, the input for the tree depicted in the below image would be :
```

```
1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1
Explanation :
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 2
Right child of 1 = 3
Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6
Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)
Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)
The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level.
The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null (-1).
```

```
The output contains single line of 'n' space-separated integers denoting the post-order traversal of the given binary tree.
```

```
You do not need to print anything. It has already been taken care of. Just implement the function.
```

In this approach, we will create a recursive function **postorder(node, ans) **that will traverse all the nodes and store the postorder traversal in an array **ans**. We will first recursively call this function for its left and right subtree to traverse the nodes in the left and right subtree, and then insert the node’s value into** ans **array for each node.

- Defining the
**postorder(node, ans)**- If the
**node**is an empty node, we will return from the function. - Now we will traverse to the left subtree.
- Create a recursive call
**postorder**(left child of the**node**,**ans**). - Now we will traverse to the right subtree.
- Create a recursive call
**postorder**(right child of the**node**,**ans**). - Insert
**node**value into the**ans**array

- If the
- We will define an array
**ans**to store the postorder traversal of the given tree. - We will call
**postorder**(**root, ans**). - Return the array
**ans**.

In this approach, we will traverse each node iteratively and insert it into **ans** array. We will use a stack **nodeStack** to maintain the order of traversal. We will insert the **root** node into the **nodeStack**.We will iterate till **nodeStack** is not empty -:

- We will delete the top node from the stack, insert the
**node’s**value into the**ans**array. - We will insert its left and right child into
**nodeStack**to traverse the left and right subtree.

At last, we will reverse and return the **ans** array.

**Algorithm:**

- We will declare a stack
**nodeStack**to maintain the order of traversal. - We will declare an array
**ans**to store the postorder traversal. - Insert
**root**node into**nodeStack**. - While
**nodeStack**is not empty, do the following steps:- Set
**cur**as a top node of the**nodeStack**. - Delete
**cur**from**nodeStack**. - Insert
**cur**node’s value into**ans**array. - We will first insert the left child of
**cur**and then the right child because the right child will be removed first from the stack. - If the left child of
**cur**is not empty, do the following:- Insert left child of
**cur**into**nodeStack**.

- Insert left child of
- If the right child of
**cur**is not empty, do the following:- Insert right child of
**cur**into**nodeStack**.

- Insert right child of

- Set
- Reverse the
**ans**array - Return
**ans**array corresponding to a postorder traversal of the given tree.

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