My six line solution


  • 1
    F
    bool hasPathSum(TreeNode *root, int sum) 
    {
    	if (root == nullptr)
    		return false;
    
    	if (root->left == nullptr && root->right == nullptr)
    		return (sum - root->val == 0);
    	else
    		return (hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val));
    
    }

  • 0
    Z

    Great mind think alike!


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