Children Sum Parent

Problem Statement - link #

Given a Binary Tree. Check whether all of its nodes have the value equal to the sum of their child nodes.

Your Task: You don’t need to read input or print anything. Your task is to complete the function isSumProperty() that takes the root Node of the Binary Tree as input and returns 1 if all the nodes in the tree satisfy the following properties. Else, it returns 0. For every node, data value must be equal to the sum of data values in left and right children. Consider data value as 0 for NULL child. Also, leaves are considered to follow the property.

Expected Time Complexity: O(N)
Expected Auxiliary Space: O(height of the tree)

Examples #

Example 1:

Input:
       1
     /   \
    4     3
   /  \
  5    N
Output: 0
Explanation: Here, 1 is the root node
and 4, 3 are its child nodes. 4 + 3 =
7 which is not equal to the value of
root node. Hence, this tree does not
satisfy the given conditions.

Example 2:

Input:
     10
    /
  10 
Output: 1
Explanation: Here, every node is sum of
its left and right child.

Constraints #

• 1 <= Number of nodes <= 10^5
• 1 <= Data of a node <= 10^5

Solutions #

class Solution
{
    public:
    //Function to check whether all nodes of a tree have the value 
    //equal to the sum of their child nodes.
    int isSumProperty(Node *root)
    {
     // Add your code here
     if(!root) return 1;
     if(!root->left && !root->right) return 1;
     int ls=0, rs=0;
     if(root->left) ls =root->left->data;
     if(root->right) rs =root->right->data;
     if(root->data != (ls+rs)) return 0;
     return isSumProperty(root->left)&&isSumProperty(root->right);
    }
};