You are given an arbitrary binary tree consisting of N nodes where each node is associated with a certain value. You need to check whether the given tree is a sum tree or not.
A binary tree is a sum tree if the value of each node is equal to the sum of nodes present in the left and the right subtree. An empty tree is a sum tree with 0 sums. A leaf node is also considered a sum tree with a sum equal to the value of the leaf node.
Input Format :
The first line of the input contains a single integer T, representing the number of test cases.
The first line of each test case will contain the values of the tree’s nodes in the level order form ( -1 for NULL node). Refer to the example for further clarification.
Example :
Consider the binary tree

The input of the tree depicted in the image above will be like:
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).
Output Format :
For each test case, print ‘true’ if the given binary tree is a sum tree else, print ‘false’. The output of each test case will be printed in 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
Time Limit: 1sec
2
3 1 2 -1 -1 -1 -1
3 -1 1 2 -1 -1 -1
true
false
For the first test case, the given tree is

For each level, we can see that the value of each node is equal to the sum of the left and the right subtree.
For the second test case, the given tree is

We can see that the sum of a subtree of a node with root 1 is not equal to the value of the node. Hence, the answer is ‘false’.
Can you use any kind of tree traversal to find the answer.
O(N ^ 2), where N is the number of nodes in the tree.
We will be calculating the sum of the same subtrees again and again in case of a skewed tree. Thus, the effective time complexity will be O(N ^ 2).
O(N), where N is the number of nodes in the tree.
There will be N nodes at max at any given point of time in the recursion stack. Thus, the effective space complexity will be O(N).