Final answer:
A complete binary tree with maximum nodes has all levels filled and a maximum node count derived from the sum of a geometric series, Mmax(h) = 2^(h+1) - 1. One with the minimum nodes still has all but the last level filled and has min(h) = 2^h nodes. The height h in terms of nodes n is found by solving n = 2^(h+1) - 1, yielding h = log2(n + 1) - 1.
Step-by-step explanation:
Complete Binary Tree Structure and Node Calculations
A complete binary tree of height h with the maximum number of nodes is a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible. For a tree of height h, the maximum number of nodes it can have, Mmax(h), is given by the sum of a geometric series:
Mmax(h) = 1 + 2 + 2^2 + ... + 2^h = 2^(h+1) - 1
This is because at each level i (0 ≤ i ≤ h), there are 2^i nodes, summing over all levels gives us the total number of nodes.
A complete binary tree of height h with the minimum number of nodes has all levels completely filled except for the last, which has just one node. The minimum number of nodes that this tree can have, min(h), is given by:
min(h) = 2^h
This considers only one node at the last level which is at the leftmost position.
To derive the height of a binary tree, h, in terms of the number of nodes, n, we can use the formula for the sum of a geometric series and solve for h:
n = 2^(h+1) - 1
h = log2(n + 1) - 1
This uses the fact that the height of the tree is the exponent that gives us the nearest number of nodes in a perfectly filled tree.