112. Path Sum

1.有點回溯的感覺，試完左樹試右樹。
2.注意葉子節點的判斷條件
3.root=NULL，sum=0時也要返回false。

class HasPathSum {
public:
    bool hasPathSum(TreeNode* root, int sum)
    {
        if (root == NULL)
        {
            return false;
        }
        pathSum = 0;
        return hasPathSumCore(root, sum);
    }
    bool hasPathSumCore(TreeNode*
 
 root, int sum) {

        if (root->left == NULL && root->right == NULL)
        {
            pathSum += root->val;
            if (sum == pathSum)
            {
                return true;
            }else
            {
                pathSum -= root->val;
                return
 
 false;
            }
        }
        pathSum += root->val;
        bool left = false;
        bool right = false;
        if (root->left != NULL)
        {
            left = hasPathSumCore(root->left, sum);
        }
        if (!left && root->right != NULL)
        {
            right =
 
 hasPathSumCore(root->right, sum);
        }
        if (left || right)
        {
            return true;
        }
        pathSum -=root->val;
        return false;
    }
    int pathSum;
};