A binary tree is balanced if the left and right branches do not differ in depth for more than one level.

Here is some code to test this (

*isTreeBalanced*) you can check on my

Gist for some test cases as well:

```
/*
is tree balanced
A tree is balanced if the left and right branches do not differ for more than 1 level in length
*/
private static Entry isTreeBalancedAux(BST root){
//null root is always balanced
if(root == null) return new Entry(true, -1);
Entry left, right;
//check subtrees
left = isTreeBalancedAux(root.left);
right = isTreeBalancedAux(root.right);
//if they are not balanced, we have our answer
if(!left.isBalanced || !right.isBalanced) return new Entry(false, 0);
//otherwise if they are balanced, but they differ for more than 1 level in depth, this node is not balanced
if(Math.abs(left.height - right.height) > 1) return new Entry(false, 0);
//finally, if all is good, return the max depth from this node incuding itself and mark it as balanced
return new Entry(true, Math.max(left.height, right.height) + 1);
}
public static boolean isTreeBalanced(BST root){
return isTreeBalancedAux(root).isBalanced;
}
```

I use an additional data structure (

*Entry*) to pass around two pieces of information:

- is the tree balanced so far

- what is the current depth of the path being analyzed