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Last Updated: 11 Feb, 2021

Check If Binary Tree Is Sum Tree Or Not

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Problem statement

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: 

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


01 Approach

  • 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.