Problem 7.11 The set RBT of Red-Black Trees is defined recursively as follows: Base cases: (red) RBT, and (black) ERBT. Constructor cases: A, B are RBT's, thern if A. B start with black, then(red, A. B is an RBT. . if A, B start with red, then (black, A, B) is an RBT. For any RBT T, let be the number of red labels in T, o br be the number of black labels in T, and . nT : rT + bT be the total number of labels in T. Prove that nT 2nT 1 If T starts with a red label, then-STT (7.31)