Question

Answer the following questions so as to justify Proposition 7.10 a. What is the minimum number of external nodes for a binary tree with height \(h ?\) Justify your answer. b. What is the maximum number of external nodes for a binary tree with height \(h ?\) Justify your answer. c. Let \(T\) be a binary tree with height \(h\) and \(n\) nodes. Show that \\[ \log (n+1)-1 \leq h \leq(n-1) / 2 \\] d. For which values of \(n\) and \(h\) can the above lower and upper bounds on \(h\) be attained with equality?

Short answer

The min number of external nodes is 1; the max is \(2^h\). For \(\log(n+1) - 1 \leq h \leq \frac{(n-1)}{2}\), complete binary trees achieve the lower bound, and fully unbalanced trees (skew trees) achieve the upper bound.

Step-by-step solution

- Minimum Number of External Nodes for Height h

To find the minimum number of external nodes in a binary tree with height \(h\), consider that the tree is fully unbalanced. This means that each level of the tree has only one node. Thus, when the tree is at height \(h\), the tree has only one external node.

- Maximum Number of External Nodes for Height h

To find the maximum number of external nodes in a binary tree with height \(h\), consider a complete binary tree. In a complete binary tree, every level except possibly the last is fully filled. Therefore, the number of external nodes will be in the last level of the tree. For height \(h\), the number of external nodes at this level is \(2^h\).

- Lower Bound on Height

To prove the lower bound on the height \(h\) of a binary tree with \(n\) nodes, note that the minimum height occurs in the case of a complete binary tree, where the number of nodes \(n\) is given by \(2^h - 1\). Using the logarithm, we find: \[ h \geq \log_2(n+1) - 1. \]

- Upper Bound on Height

To prove the upper bound on the height \(h\) of a binary tree with \(n\) nodes, consider an unbalanced tree where each node has at most one child. This scenario gives the maximum height of a binary tree, making the height \(h\) equal to \(n - 1\). Thus, the upper bound on height is \(\frac{n-1}{2}\) since, in the worst-case scenario, every node could potentially be only one edge down in depth.

- Equality Conditions for Bounds on Height

For the lower bound \(\log(n+1) - 1\) to hold with equality, the tree must be a complete binary tree. For the upper bound \(\frac{n-1}{2}\) to hold with equality, the tree must be maximally unbalanced (i.e., a skew tree). Therefore, the values of \(n\) and \(h\) that can satisfy the conditions with equality are when n is the form of a complete binary tree or skew tree.

Key concepts

external nodes

External nodes, also known as leaves, are nodes in a binary tree that do not have children. They are at the ends of the branches. Understanding their count helps in analyzing the tree’s structure. In a binary tree of height \( h \), the minimal number of external nodes is \(1\), assuming a fully unbalanced tree. On the other hand, a complete binary tree (where every level except the last is full) at height \( h \) has a maximum of \(2^h\) external nodes. This insight is fundamental when trying to determine the depth and structure of a binary tree.

tree height

Tree height is the longest path from the root to an external node. It's a critical measure because it affects the efficiency of tree operations. In a worst-case scenario (a fully skewed tree), the height can be \(n-1\), where \( n \) is the number of nodes. In contrast, the minimal height (best-case scenario) is achieved in a complete binary tree and can be represented as \[ h \geq \log_2(n+1) - 1. \] Knowing how to calculate the height and understanding these bounds is crucial for analyzing algorithms that operate on binary trees.

complete binary tree

A complete binary tree is a highly efficient structure where every level, except possibly the last, is fully filled. The nodes are as far left as possible. This structure ensures the tree height is minimized. For a complete tree with \( n \) nodes, the height can be represented as \[\text{h \geq \log_2(n+1) - 1} \]. Understanding complete binary trees is important in optimizing various operations like search, insert, and delete, as they provide the lower bound for tree height.

unbalanced tree

An unbalanced tree is one where the height varies significantly. In the worst case, the tree becomes a linear structure (skew tree), where each node has only one child, making the height equal to \( n-1\). This structure is inefficient, as operations like search, insert, and delete can become linear in time complexity. Hence, balancing a tree is often necessary to ensure optimal performance. Knowing the impact of unbalanced trees is essential for designing algorithms and data structures.

node count

Node count refers to the total number of nodes in a binary tree. It’s a vital metric because it influences the tree's height and the efficiency of its operations. For a tree with \( n \) nodes, the height falls between \[ \log_2(n+1) -1 \leq h \leq (n-1)/2, \]. This range shows how the node count affects the tree’s structure and height. More nodes usually lead to greater height unless the tree is balanced. Thus, understanding the relationship between node count and tree height is key to effective tree management.