Question

A binary tree node is called full if the node contains 2 children. Use a proof by induction to prove that in any binary tree, the number of leaves in the tree is equal to the number of full nodes plus one. (Hint: your inductive step should consider two cases: the k+1 node becomes the only child of a node that was previously a leaf; and the k+1 node becomes the second child of a node that previously only had one.)

Answer 1

$\text{Proof. We use proof by induction.}$

$$\forall k \in \mathbb{N}$ let $P(k)$ be\ the\ proposition\ that\ a \ binary \ tree \ with \ k \ nodes \ has \ n \ full \ nodes \and \ n+1 \ leaves$
Inductive Step: VKEN , we must show that P(K) > PIK+1) we assume P(K) is frue for the purpose of induction , and we must show that plk+1) is true. Now we have two cases: Case 1° - than both the number of full nodes in the binary free and the number of leaves in the binary free do not change Casezo Enlarging If we add a node to an existing leaf, =n42 a binary free by adding a node to an existing node that has 1 child, makes an initial binary free with in full nodes now have nto full nodes. Since, the barant node becomes a full rates, then the number of leaves in the binary free becames (mt) + This the number of full nodes we start with is exactly one less than the number of leaves, and adding a node to the binary free Eith changs neither number or increase both by exactly one, so the difference between the number of full nodes and the number of leaves will always be 1. So P(Kt1) is true we have P(K) => P(K+l) is Since true. Therefore, the number of full nodes plus one is equal to the number of leaves in a nonempty binary Proved P(k+1) is true free.