# Postorder Successor

Posted: 2 Dec, 2020

Difficulty: Moderate

#### You are given a binary tree having ‘N’ distinct nodes and an integer ‘M’, you have to return the postorder successor of ‘M’.

#### Note:

```
The postorder successor of ‘M’ is defined as the next element to ‘M’ in the sequence of postorder traversal.
If the postorder traversal of a tree is 3 5 4 7 then the postorder successor of 5 is the next element to 5 i.e 4.
Return ‘-1’ if there is no postorder successor of ‘M’.
```

##### Input Format:

```
The first line contains an integer 'T' which denotes the number of test cases.
The first line of each test case contains elements of the tree in the level order form. The line consists of values of nodes separated by a single space. In case a node is NULL, we take -1 in its place.
The second line of each test case contains ‘M’, the node whose postorder successor is to be found out.
For example, the input for the tree depicted in the below image.
```

```
1
3 8
5 2 7 -1
-1 -1 -1 -1 -1 -1
```

##### Explanation :

```
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 3
Right child of 1 = 8
Level 3 :
Left child of 3 = 5
Right child of 3 = 2
Left child of 8 =7
Right child of 8 = null (-1)
Level 4 :
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 2 = null (-1)
Right child of 2 = null (-1)
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).
```

##### Note :

```
The above format was just to provide clarity on how the input is formed for a given tree.
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:
1 3 8 5 2 7 -1 -1 -1 -1 -1 -1 -1
```

#### Output Format:

```
Print an integer denoting the post-order successor of the node ‘M’ in the binary tree.
The output of each test case will be printed on a separate line.
```

#### Note:

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

#### Constraints:

```
1<= ‘T’ <= 5
1 <= ‘N’ <= 3000
1 <= ‘M’ <= 3000
1 <= nodeVal <= 10^9
Time Limit: 1 sec
```

Approach 1

8834The idea is to first store the Postorder traversal of the given tree in an array and then linearly search the given node and return the node next to it.

**Algorithm :**

- Let’s take an array ‘post’. To store postorder traversal.
- Perform postorder traversal on the given tree and store the elements in ‘post’.
- Now iterate through the ‘post’, one less than the size of ‘post’
- If the current element of ‘post’ is equal to the given node then return the next element (postorder successor) of the array ‘post’.

- If the given node is not found or it is the last element of the postorder traversal, then return -1.

Approach 2

We can optimize our space based on the following observation

- The root node is always the last element of postorder traversal. So, if the given node is a root node, then postorder traversal is -1.
- If the given node is the right child of the parent then the parent is the postorder successor.
- If the given node is the left child of the parent and the right child is NULL, then the parent is the postorder successor.
- If the given node is the left child of the parent and the right child of the parent is not NULL, then the postorder successor is the leftmost leaf node of the parent's right subtree.

**Algorithm :**

- Let ‘root’ be the root of the tree and ‘node’ be the given node to find the successor.
- If ‘root’ is null or ‘root’ is the given ‘node’ return -1
- Perform ‘DFS’ passing the parameter parent as NULL, child as ‘root’, and ‘node’, keeping the above observations in mind. It returns the postorder successor and stores it in variable ‘ans’.
- Finally, return ‘ans’.

**Description of DFS function:**

Let ‘postOrderTraversal’ be the name of the function which accepts three parameters: first, parent node, initially it is NULL, second, child node initially it is ‘root’ node, and third, ‘node’ whose successor is to be found.

postOrderTraversal( parent, child, node )

- If ‘child’ is ‘NULL’ return -1
- If ‘child’ -> data is equal to the given node.
- If parent right is equal to NULL or parent right is NULL then return parent data.
- Else, store the parent right child in the ‘temp’ variable.
- Run a loop While temp left is not NULL or temp right is not equal to NULL
- If temp left exist then move temp to temp left and continue
- Else move to temp right.

- Return temp data

- Call function on parameter current child as a parent, the current child left as child and node, and store the return value in variable ‘d’.
- If ‘d’ is equal to -1 ( we have not found a postorder successor) Call function on the parameter, current child as a parent, current child right as child and node, and store the return value in variable ‘d’.
- Return ‘d’.

SIMILAR PROBLEMS

# Postorder Traversal

Posted: 22 Jul, 2021

Difficulty: Easy

# Preorder Traversal

Posted: 22 Jul, 2021

Difficulty: Easy

# Root to Leaf Path

Posted: 23 Jul, 2021

Difficulty: Moderate

# Maximum Sum BST

Posted: 27 Jul, 2021

Difficulty: Hard

# Vertical Sum in BST

Posted: 27 Jul, 2021

Difficulty: Moderate