A reducing sugar is easily oxidized. What makes it easy to oxidize? The presence of an "oxo" group, either an aldehyde or a ketone. Right now you are probably looking at a picture of maltose or glucose and saying to yourself that you do not see any double-bonded oxygens in the structures. You see a ring with a lot of hydroxyls hanging off it but no carbonyl groups. Many sugars exist in a ring structure--it is the most energetically favorable structure. But these rings can open to the straight-chain structures where you will see the carbonyl structure. Glucose and maltose will have an aldehyde group and fructose will have a ketone group. The open-chain form of the sugar is what can be oxidized and is, therefore a reducing sugar (that is, it reduces something else, often silver  or copper cations to silver or copper metal). 

How can you tell if a ring will open to expose a carbonyl group? Look closely at the closed ring structure. You will see that one of the members of the five- or six-membered rings is an oxygen atom. Now look at the two carbons adjacent to the oxygen. Do either (or both) have a hydroxyl group (-OH) attached directly to the carbon? If so, that ring structure is a hemi-acetal (will open to an aldehyde) or a hemi-ketal (will open to a ketone). Look at the structure of glucose. The #1 carbon in the ring is next to an oxygen in the ring. It also has a hydroxyl connected directly to it. It will open up to the chain formation and can be oxidized. Look at the maltose structure and you will see that one of the two rings has the same structure but the other ring, instead of having a hydroxyl group attached to the #1 carbon, has an entire sugar ring attached there. That portion of the maltose molecule will NOT open up. That means that both glucose and maltose have the same reducing power even though there are two rings in maltose.

Now let's look at sucrose. You will see a five-membered ring (the fructose ring) and a six-membered ring (the glucose). Look for the oxygen in each ring structure. Once you find them, look at the carbons adjacent to the oxygens. Do any of them have a free hydroxyl? No!

The glucose is attached to the fructose through that position and the fructose is attached to the glucose at that position. Sucrose will form an open chain structure.

The carbon that can change from a hemi-ketal or hemi-acetal to a ketone or aldehyde is called the "anomeric" position. If the anomeric carbon is blocked (i.e., instead of -OH it is -OCH3 or -O-sugar or -O-almost-anything) the sugar cannot open up and the sugar is non-reducing. If the hydroxyl is there, then it is a reducing sugar.

Let's try some examples. Look at mannose (tri Wikipedia for structures). Is it a reducing sugar? Does it have an -OH  on a carbon in the ring where the carbon is next to an oxygen in the ring? Yes! It is a reducing sugar. Now look at glyceraldehyde.  !?! Glyceraldehyde does not have a ring-structure. But it is a sugar in the open-chain form. Does it have a C=O carbonyl structure? Yes! It is also a reducing sugar. 

Now look up inositol. It is a six-membered ring. Does it have an oxygen in the ring? No! It is a non-reducing sugar alcohol. Now let's try a hard one. Look up the sugar raffinose. It has 3 rings. First look for the oxygens in the rings (or any exposed carbonyl groups). You should find 3. Do any of those have a carbon holding a hydroxyl group right next to them? No! Raffinose is a non-reducing sugar.

Two more. Look up methylglucoside in Wikipedia. There are two types of methylglucoside, the alpha- and the beta-form. Look closely. There is an oxygen in the ring but the carbon that has a hydroxyl attached when the molecule is glucose now has a -OCH3 (methoxy group) attached. The methyl group blocks the sugar from opening into the open-chain form. Methylglucoside is non-reducing.

Simple rules: 

1. Is the sugar in a ring structure or is it in an open-chain structure? If open-chain structure, does it have a ketone or an aldehyde group? If yes, it is a reducing sugar. If no, then it isn't.

2. Is the sugar in a ring structure? Is there an oxygen in the ring? If no, then it is not a reducing sugar (but see below.)

3. If there is an oxygen in the ring structure, does either carbon adjacent to that oxygen have an -OH group attached directly to it? If yes, it is a reducing sugar. If no, it is not.

Caveats. Everything here is true for 99% of the sugar-like molecules you will ever see. BUT there can be some trickery. Do not bother with this strange structures unless you really want to be a sugar scientist.

A. The -OH at the reducing (or anomeric) position can be replaced with these groups and it is still reducing: -SH, -NH2, but not with -SCH3, -NHOAc, or other derivatized structures. That is because these molecules will still open up and will hydrolyze to either ketones or aldehydes. 

B. The -O- in the ring can be replaced with -S- or -NH-. That is because these can also open up to form ketones or aldehydes. 

A reducing sugar is easily oxidized. What makes it easy to oxidize? The presence of an "oxo" group, either an aldehyde or a ketone. Right now you are probably looking at a picture of maltose or glucose and saying to yourself that you do not see any double-bonded oxygens in the structures. You see a ring with a lot of hydroxyls hanging off it but no carbonyl groups. Many sugars exist in a ring structure--it is the most energetically favorable structure. But these rings can open to the straight-chain structures where you will see the carbonyl structure. Glucose and maltose will have an aldehyde group and fructose will have a ketone group. The open-chain form of the sugar is what can be oxidized and is, therefore a reducing sugar (that is, it reduces something else, often silver  or copper cations to silver or copper metal). 

How can you tell if a ring will open to expose a carbonyl group? Look closely at the closed ring structure. You will see that one of the members of the five- or six-membered rings is an oxygen atom. Now look at the two carbons adjacent to the oxygen. Do either (or both) have a hydroxyl group (-OH) attached directly to the carbon? If so, that ring structure is a hemi-acetal (will open to an aldehyde) or a hemi-ketal (will open to a ketone). Look at the structure of glucose. The #1 carbon in the ring is next to an oxygen in the ring. It also has a hydroxyl connected directly to it. It will open up to the chain formation and can be oxidized. Look at the maltose structure and you will see that one of the two rings has the same structure but the other ring, instead of having a hydroxyl group attached to the #1 carbon, has an entire sugar ring attached there. That portion of the maltose molecule will NOT open up. That means that both glucose and maltose have the same reducing power even though there are two rings in maltose.

Now let's look at sucrose. You will see a five-membered ring (the fructose ring) and a six-membered ring (the glucose). Look for the oxygen in each ring structure. Once you find them, look at the carbons adjacent to the oxygens. Do any of them have a free hydroxyl? No!

The glucose is attached to the fructose through that position and the fructose is attached to the glucose at that position. Sucrose will form an open chain structure.

The carbon that can change from a hemi-ketal or hemi-acetal to a ketone or aldehyde is called the "anomeric" position. If the anomeric carbon is blocked (i.e., instead of -OH it is -OCH3 or -O-sugar or -O-almost-anything) the sugar cannot open up and the sugar is non-reducing. If the hydroxyl is there, then it is a reducing sugar.

Let's try some examples. Look at mannose (tri Wikipedia for structures). Is it a reducing sugar? Does it have an -OH  on a carbon in the ring where the carbon is next to an oxygen in the ring? Yes! It is a reducing sugar. Now look at glyceraldehyde.  !?! Glyceraldehyde does not have a ring-structure. But it is a sugar in the open-chain form. Does it have a C=O carbonyl structure? Yes! It is also a reducing sugar. 

Now look up inositol. It is a six-membered ring. Does it have an oxygen in the ring? No! It is a non-reducing sugar alcohol. Now let's try a hard one. Look up the sugar raffinose. It has 3 rings. First look for the oxygens in the rings (or any exposed carbonyl groups). You should find 3. Do any of those have a carbon holding a hydroxyl group right next to them? No! Raffinose is a non-reducing sugar.

Two more. Look up methylglucoside in Wikipedia. There are two types of methylglucoside, the alpha- and the beta-form. Look closely. There is an oxygen in the ring but the carbon that has a hydroxyl attached when the molecule is glucose now has a -OCH3 (methoxy group) attached. The methyl group blocks the sugar from opening into the open-chain form. Methylglucoside is non-reducing.

Simple rules: 

1. Is the sugar in a ring structure or is it in an open-chain structure? If open-chain structure, does it have a ketone or an aldehyde group? If yes, it is a reducing sugar. If no, then it isn't.

2. Is the sugar in a ring structure? Is there an oxygen in the ring? If no, then it is not a reducing sugar (but see below.)

3. If there is an oxygen in the ring structure, does either carbon adjacent to that oxygen have an -OH group attached directly to it? If yes, it is a reducing sugar. If no, it is not.

Caveats. Everything here is true for 99% of the sugar-like molecules you will ever see. BUT there can be some trickery. Do not bother with this strange structures unless you really want to be a sugar scientist.

A. The -OH at the reducing (or anomeric) position can be replaced with these groups and it is still reducing: -SH, -NH2, but not with -SCH3, -NHOAc, or other derivatized structures. That is because these molecules will still open up and will hydrolyze to either ketones or aldehydes. 

B. The -O- in the ring can be replaced with -S- or -NH-. That is because these can also open up to form ketones or aldehydes. 

To Saadi Gharib-Bibalan:

You are absolutely right that sugars that have free ketones or aldehydes are reducing sugars. But based on the question, that seemed to be a tautological answer. That meant that I had to explain how to tell if a carbohydrate had a potential free ketone or aldehyde. 

I must remind you that all monosaccharides are NOT reducing sugars. The most obvious being methylglucoside. Depending on the level of definition, sorbitol is another non-reducing monosaccharide. 

It does little good to tell someone that a free anomeric position is required to make a reducing sugar if you do not explain what the anomeric position is and how you identify it in the ring form. You also need to explain what "free" means in this case, which I tried to do. Based on the question from Jaruwan Kongprasuk I would guess that Jaruwan is a beginning student who needed a deeper explanation.

All previous answers are correct. The fact that sucrose is not a reducing sugar (the same holds strue for trehalose) provides an explanation as regard to the ease of crystallization (which is so difficult for non-reducing sugars).

Because of its non-reducing nature the structure of sucrose can be written in two ways:

alpha-D-Glcp 1-2 beta-D-Fruf or beta-D-Fruf 2-1 alpah-D-Glcp