Proving Binary Tree Properties: 2^(h-1) Leaf Nodes & 2^(h-1) - 1 Internal Nodes

[craig1928]

2 questions that came up in a past paper.

A binary tree which is both full and complete and has h levels contains a total of 2^(h-1)
leaf nodes. Prove that this is the case for all h > 0.

Show that with h levels the number of internal nodes is 2^(h−1) − 1

proof by induction apparently

anyone can help? I don't really need to understand it if it came up in an exam just to know it off by heart.

Sorry if I've posted in the wrong section.

thanks.

 

[Mark44]

Proof by induction is the way to go. A tree with 2 levels has 2 leaf nodes (= 2^{2 - 1} nodes). A tree with 3 levels has 4 leaf nodes (= 2^{3 - 1}).

Assume that a tree with k levels has 2^{k - 1} leaf nodes, and then use this to show that a tree with k + 1 levels has 2^k leaf nodes.

 

[Mene]

I would first like to clarify that proving binary tree properties is a mathematical concept and not a scientific one. However, as a scientist, I am trained in logical thinking and problem-solving, which can be applied to this situation.

To prove that a binary tree with h levels contains a total of 2^(h-1) leaf nodes for all h > 0, we can use mathematical induction. This method involves proving a statement for a base case (in this case, h = 1) and then showing that if the statement holds for a particular value of h, it also holds for the next value (h+1).

Base case (h = 1):
For a binary tree with 1 level, there is only one node, which is a leaf node. Therefore, the statement holds for h = 1.

Inductive step:
Assume that for a binary tree with h levels, the statement holds (i.e. it contains 2^(h-1) leaf nodes). Now, let's consider a binary tree with h+1 levels. We know that this tree can be divided into two smaller trees, each with h levels. By our assumption, each of these smaller trees contains 2^(h-1) leaf nodes. Therefore, the total number of leaf nodes in the binary tree with h+1 levels would be 2^(h-1) + 2^(h-1) = 2^(h+1-1). Thus, the statement holds for h+1 levels as well.

Therefore, by the principle of mathematical induction, we can conclude that a binary tree with h levels contains 2^(h-1) leaf nodes for all h > 0.

Now, to prove that the number of internal nodes in a binary tree with h levels is 2^(h-1) - 1, we can again use mathematical induction.

Base case (h = 1):
For a binary tree with 1 level, there are no internal nodes, and the statement holds (2^(1-1) - 1 = 0).

Inductive step:
Assume that for a binary tree with h levels, the statement holds (i.e. it contains 2^(h-1) - 1 internal nodes). Now, let's consider a binary tree with h+1 levels. We know that this tree can be divided into two smaller trees, each with h levels. By