112. Path Sum

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

bool hasPathSum(TreeNode* root, int sum) {
    if (!root) return false;
    if (!root->left && !root->right && root->val == sum) return true; 
    return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
}