inconsistency in concat behavior in pandas 0.21.0
#### Code Sample, a copy-pastable example if possible

```python
In [1]: from collections import OrderedDict
In [2]: import pandas as pd
In [3]: pd.__version__
Out[3]: u'0.21.0'

# the following works as expected:
In [4]: df1 = pd.DataFrame([[1, 2, 3]], columns=['a', 'b', 'c'])
In [5]: df2 = pd.DataFrame(columns=['a', 'b', 'c'])
In [6]: pd.concat([df1, df2])
Out[6]: 
   a  b  c
0  1  2  3

# however, the following seems like it does an identical thing but throws an error:
In [7]: od3 = OrderedDict([('a', [1]), ('b', [2]), ('c', [3])])
In [8]: od4 = OrderedDict([('a', []), ('b', []), ('c', [])])
In [9]: df3 = pd.DataFrame(od3)
In [10]: df4 = pd.DataFrame(od4)
In [11]: pd.concat([df3, df4])
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-11-ac4fae34c928> in <module>()
----> 1 pd.concat([df3, df4])
#...
ValueError: Shape of passed values is (3, 1), indices imply (3, 0)
```

Rewriting it in copy/pastable form...

This works:
```python
df1 = pd.DataFrame([[1, 2, 3]], columns=['a', 'b', 'c'])
df2 = pd.DataFrame(columns=['a', 'b', 'c'])
pd.concat([df1, df2])
```

And this does not:
```python
od3 = OrderedDict([('a', [1]), ('b', [2]), ('c', [3])])
od4 = OrderedDict([('a', []), ('b', []), ('c', [])])
df3 = pd.DataFrame(od3)
df4 = pd.DataFrame(od4)
pd.concat([df3, df4])
```

#### Problem description

This is also described [here](https://stackoverflow.com/questions/47194806/pandas-inconsistency-in-concat-behavior).

In the above code, `df1` equals `df3`, and `df2` and `df4` are both empty dataframes with the same column names (although, strangely, `df2` and `df4` aren't equal according to `df2.equals(df4)`) but `pd.concat([df1, df2])` works while `pd.concat([df3, df4])`results in `ValueError`.  This did not happen in previous versions of Pandas, but when I upgraded to 0.21.0, it started happening.

Oddly, as the stackoverflow link notes, using the `drop_duplicates()` method on either `df3` or `df4` (or both) results in the `concat()` working, even though neither of them contains any duplicates.

#### Expected Output

The expected output of `pd.concat([df3, df4])` is

```python
   a  b  c
0  1  2  3
```

#### Output of ``pd.show_versions()``

<details>

[paste the output of ``pd.show_versions()`` here below this line]

INSTALLED VERSIONS
------------------
commit: None
python: 2.7.13.final.0
python-bits: 64
OS: Darwin
OS-release: 16.7.0
machine: x86_64
processor: i386
byteorder: little
LC_ALL: None
LANG: en_US.UTF-8
LOCALE: None.None

pandas: 0.21.0
pytest: 3.2.1
pip: 9.0.1
setuptools: 32.1.0
Cython: 0.23.4
numpy: 1.13.3
scipy: 0.19.1
pyarrow: None
xarray: None
IPython: 5.3.0
sphinx: None
patsy: 0.4.1
dateutil: 2.6.1
pytz: 2017.3
blosc: None
bottleneck: None
tables: None
numexpr: None
feather: None
matplotlib: 2.0.2
openpyxl: None
xlrd: None
xlwt: None
xlsxwriter: None
lxml: None
bs4: 4.6.0
html5lib: 0.999999999
sqlalchemy: 1.1.14
pymysql: 0.7.11.None
psycopg2: 2.7.3.1 (dt dec pq3 ext lo64)
jinja2: 2.9.6
s3fs: None
fastparquet: None
pandas_gbq: None
pandas_datareader: 0.5.0

</details>
