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Last Updated: 11 Feb, 2021

Moderate

```
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.
```

```
Consider the binary tree
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```
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).
```

```
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.
```

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

```
1 <= T <= 5
1 <= N <= 3000
Time Limit: 1sec
```

- The simple approach is to use DFS for finding the sum for each subtree.
- Let getSubtreeSum() be an integer function that takes the tree’s root as input and returns the sum of the given subtree.
- Let isSumTree() a boolean function which returns true if the given tree is sum tree or not.
- The base case will be when the root does not exist, or the current node is a leaf node, then the answer will be true.
- Let 'leftSum' = 0, denote the sum of the left subtree of the given node and 'rightSum' = 0, denotes the sum of the right subtree of the given node,
- Recursively call for left and right subtrees, i.e. ‘leftSum’ = getSubtreeSum(root->left) and 'rightSum' = getSubtreeSum(root->right)
- Now if ‘leftSum’ + ‘rightSum’ = root->data, root->left and root->right is also a sumTree then return true.
- Else, return false.

- Since we were repeatedly calculating the sums of the left and the right subtrees in the first method, we can use some other ways to find the sum to improve the efficiency.
- We can use the property:
- If the node is a leaf, it is already the sum tree.
- Else, if we pass the value of the subtree to the root node, considering the fact that the given subtree is a sum tree, the overall value of the root should be twice the value of the given node.

- Hence, we can get the sum of the given subtree directly as follows:
- If the current node is a leaf node, then the sum is equal to the node’s value.
- Else, the sum of the subtree rooted with the given node is twice the value of the given node.