

You are given the root node of a binary tree consisting of 'N' nodes. Your task is to return the maximum height difference of the tree.
The height of a Binary Tree is defined as the number of nodes present in the longest path from the root node to any leaf node of the tree. The height difference of a node is equal to the absolute difference of height of the left and right subtree.
For example:
For the given tree,
The maximum height difference is 1. The height difference of node 1 is the absolute difference of the height of subtree with root node 2, and the height of subtree with root node 3, which is 1. Hence the answer is 1.
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
The first line of the input contains an integer, 'T,’ denoting the number of test cases.
The first 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 for further clarification.
Output Format:
For each test case, print a single integer that denotes the maximum height difference for the given tree.
Print the output of each test case in a separate line.
Note:
You do not need to print anything. It has already been taken care of. Just implement the given function.
1 <= T <= 10
1 <= N <= 10^6
1 <= nodeVal <=10^9
Time limit: 1 sec
2
1 2 3 4 5 -1 -1 -1 -1 -1 -1
1 -1 2 -1 3 -1 4 -1 -1
1
3
For the first test case,
The maximum height difference is 1. The height difference of node 1 is the absolute difference of the height of subtree with root node 2, and the height of subtree with root node 3, which is 1. Hence the answer is 1.

For the second test case,
The maximum height difference is 3. The height difference of node 1 is the absolute difference of the height of subtree with root node 2, and the height of subtree with an empty node, which is 3. Hence the answer is 3.
2
1 2 3 4 -1 7 8 5 -1 9 -1 -1 -1 6 -1 -1 -1 -1 -1
10 20 11 -1 -1 -1 -1
3
0
Can we create a function to calculate the height of a node?
This approach will create a function height(node) that will calculate the height of the node.
We will travel through all nodes, call the height function, and calculate the height difference between the left and right subtree and update the maximum height difference of the tree. We will traverse all nodes using traverse(node) function.
Algorithm:
O(N^2), where N is the number of nodes in the tree.
In at worst case(For skew trees), the time complexity of height function will be O(N). We have to travel all the N nodes to calculate the height of the root node. We call the height function to find the height of the left and right subtree for each node. So the time complexity will be N*N. Hence, the overall time complexity is O(N^2).
O(N), where N is the number of nodes in the tree.
We use O(N) space complexity for recursive stack during recursion calls of height and traverse function in the worst case. Hence the overall space complexity is O(N).