Symmetric Tree Solution

Create a helper that checks if two trees are mirrors:

function isSymmetric(root):
    if root == null:
        return true
    return isMirror(root.left, root.right)

function isMirror(t1, t2):
    if t1 == null and t2 == null:
        return true
    if t1 == null or t2 == null:
        return false
    return (t1.val == t2.val and
            isMirror(t1.left, t2.right) and
            isMirror(t1.right, t2.left))

The insight: in a mirror, $t1$'s left corresponds to $t2$'s right, and $t1$'s right corresponds to $t2$'s left.

Time: $O(n)$. Space: $O(h)$.