Why recursion function doesn't need return - recursion

Question

I was trying to merge two binary trees and if I use "return root;" at last, the system told me "error: cannot find symbol: variable root". It works when I add "return root;" to every conditions. I don't understand why. Furthermore, in my past experience, it seems I don't have to write return statements after recursion function calls. Can someone explain this? Thank you!
public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
if (t1 == null && t2 == null) {
return null;
}
else if (t1 == null) {
TreeNode root = new TreeNode(t2.val);
root.left = mergeTrees(null, t2.left);
root.right = mergeTrees(null, t2.right);
}
else if (t2 == null) {
TreeNode root = new TreeNode(t1.val);
root.left = mergeTrees(t1.left, null);
root.right = mergeTrees(t1.right, null);
}
else {
TreeNode root = new TreeNode(t1.val + t2.val);
root.left = mergeTrees(t1.left, t2.left);
root.right = mergeTrees(t1.right, t2.right);
}
return root;
}

Answer 1

Your variable root is declared inside your if, else if, and else statements, therefore it loses scope once you get to the return outside your if statement block. If you want to return root, I would suggest this.
public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
TreeNode root; //Declare it here so it doesn't lose scope
if (t1 == null && t2 == null) {
return null;
}
else if (t1 == null) {
root = new TreeNode(t2.val);
root.left = mergeTrees(null, t2.left);
root.right = mergeTrees(null, t2.right);
}
else if (t2 == null) {
root = new TreeNode(t1.val);
root.left = mergeTrees(t1.left, null);
root.right = mergeTrees(t1.right, null);
}
else {
root = new TreeNode(t1.val + t2.val);
root.left = mergeTrees(t1.left, t2.left);
root.right = mergeTrees(t1.right, t2.right);
}
return root;
}

Answer 2

Unless you are programming in JavaScript and use var the variables in most algol languages has block scope.. Thus:
{ // new block scope started
TreeNode root = new TreeNode(t2.val); // New variable TreeNode created
} // Everything created in this scope gets destroyed
root; // Would no compile as root is an unbound variable
If you want to create a variable to be returned you need to create it in the same scope as you return. Thus:
{
TreeNode root = null;
// .. you logic goes here
return root;
} // root goes out of scope here
Trivia: I mentioned JavaScript and in that language using var your code would work:
function mergeTrees(t1, t2) {
if (t1 === null && t2 === null) {
return null;
}
else if (t1 === null) {
var root = new TreeNode(t2.val);
root.left = mergeTrees(null, t2.left);
root.right = mergeTrees(null, t2.right);
} else if (t2 === null) {
var root = new TreeNode(t1.val);
root.left = mergeTrees(t1.left, null);
root.right = mergeTrees(t1.right, null);
} else {
var root = new TreeNode(t1.val + t2.val);
root.left = mergeTrees(t1.left, t2.left);
root.right = mergeTrees(t1.right, t2.right);
}
return root;
}
The reason for this is that var associates the the closest function as scope and the blocks in the conditional doesn't introduce a new scope. ES6 has fixed this with let and const which works the same as other algol languages though and it's better style to use them.