Difference Between Two Recursive Methods - recursion

Question

I wanted to reflect a binary tree, such that all nodes on the left ended up on the right, and vice versa.
Something like :
A
/ \
B C
/ / \
D E F
would become
A
/ \
C B
/ \ \
F E D
I noticed that, while writing my solution, this code worked:
static Tree getReflection(Tree root) {
if(root == null) {
return null;
}
Tree reflect = root;
Tree subRight = getReflection(root.right);
Tree subLeft = getReflection(root.left);
reflect.left = subRight;
reflect.right = subLeft;
return reflect;
}
And yet, this one doesn't:
static Tree getReflection(Tree root) {
if(root == null) {
return null;
}
Tree reflect = root;
reflect.left = getReflection(root.right);
reflect.right = getReflection(root.left);
return reflect;
}
Can someone explain to me why? To me, they seem like identical methods, except one uses temporary tree variables.

Answer 1

Look at your first statement in each: when you assign
reflect = root
, the two variables now point to the same memory location. Now, let's look at the operation of the second routine:
Tree reflect = root;
// reflect and root now refer to exactly the same tree.
reflect.left = getReflection(root.right);
// reflect the right subtree; make that the new left subtree.
reflect.right = getReflection(root.left);
// Grab that new subtree, re-reflect it, and put it back on the right.
The original left subtree is lost, replaced by a reflection of the right.
In the first routine, you saved them in local variables until you'd done both reflections.

Answer 2

It is because in the second function (the one that doesn't work), you are assigning the reflected result to your left node and then using that as input to the reflection that you assign to your right node.
Tree reflect = root;
reflect.left = getReflection(root.right);
reflect.right = getReflection(root.left);