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Height of Spiral Tree

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Problem statement

You are given a special kind of binary tree where the leaf nodes are connected to each other in such a fashion that they form a circular doubly linked list. You need to find the height of the Binary Tree.

The left pointer of leaf node will act as a previous pointer of circular doubly linked list and its right pointer will act as next pointer of circular doubly linked list.

A binary tree (BT) is a data structure in which each node has at most two children.

For the given binary tree:

In the above binary tree, 4, 6 and 7 are leaf nodes and they form a circular doubly linked list.

Here, the height of the Binary Tree is 4.

Detailed explanation ( Input/output format, Notes, Images )

Input Format:

The first line contains an integer 'T' which denotes the number of test cases. Then the test cases follow.

The first and the only line of each test case 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 below.

Example:

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).

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 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Output Format:

For each test case, you need to return the height of the given binary tree.

Constraints:

1 <= T <= 100
0 <= N <= 1000
-10^6 <= data <= 10^6 and data != -1
where N is the number of nodes in the tree, T represents the number of test cases and data denotes data contained in the node of the binary tree.

Duplicate elements can be in the right subtree or left subtree.

Time Limit : 1 sec

Note:

You are not required to print the expected output, it has already been taken care of. Just implement the function.

Sample Input 1:

2
1 2 3 -1 4 -1 -1 -1 -1
1 2 3 4 8 5 6 -1 -1 -1 -1 -1 -1 7 -1 -1 -1

Sample Output 1:

3
4

Explanation of Input 1:

Here we have 2 test cases, hence there are 2 binary trees

Test Case 1:

We can see that the height of the given tree is 3.

Test Case 2:

We can see that the height of the given tree is 4.

Sample Input 2:

2
1 3 2 4 5 -1 7 -1 -1 -1 6 -1 -1 -1 -1
1 2 -1 3 -1 -1 -1

Sample Output 2

4
3

Hint 1

How to check if the node is a leaf node?

Approaches (1)

Approach 1

Recursive Solution

We would be doing the standard way of finding the height of a Binary Tree.
We will recursively calculate the height of the Binary Tree by taking the maximum of height of the left subtree and height of the right subtree and adding 1 to it.
The left and right node are NULL for a simple Binary Tree. For this binary tree, it is not the case.
For a leaf node, we would check if the node’s left’s right is the node itself or not. If it is the same, then it means that the node is a leaf node.

Time Complexity

O(N), where N is the number of nodes in the binary tree. We are recursively traversing through all the nodes of the tree.

Space Complexity

O(H), Stack space for recursive calls of N nodes of the binary tree and we are also using a hashmap which can have at most H elements in it. Here, H is the height of the binary tree which can become N for a skewed tree.

(100% EXP penalty)

Height of Spiral Tree

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