Root to leaf path with given sum in binary tree

Question

For a given binary tree and a sum, I have written the following function to check whether there is a root to leaf path in that tree with the given sum.

/* 
//A binary tree node
struct Node
{
    int data;
    struct Node* left, * right;
}; 
*/

bool hasPathSum(Node *node, int sum)
{
    if(!node)
        return sum==0;

    return ( hasPathSum(node->left,  sum-node->data) || 
             hasPathSum(node->right, sum-node->data) );
}

Are there any edge cases in which the code will break? Also, do comment on the code style.

Answer 1

• Give your operators some breathing space.

      if (!node) {
          return sum == 0;
      }
  
      return hasPathSum(node->left,  sum - node->data) || 
             hasPathSum(node->right, sum - node->data);
  

  Note that return expression needs no parenthesis.

• sum - node->data seems more natural to be expressed once:

      sum -= node->data;
      return hasPathSum(node->left,  sum) || 
             hasPathSum(node->right, sum);
  

• I see no edge cases except possible overflows.