Lecture 23
... are efficient because half the tree is eliminated at each comparison (cf binary search with arrays). In searching for an item in a binary search tree only involves going left or right for each node as we descend the tree. This is similar to choosing first or second half during a binary search for an ...
... are efficient because half the tree is eliminated at each comparison (cf binary search with arrays). In searching for an item in a binary search tree only involves going left or right for each node as we descend the tree. This is similar to choosing first or second half during a binary search for an ...
lecture 9
... • Lemma: A red-black tree with n internal nodes has height at most 2 lg(n +1) • Definition: Black-height, bh(x), is the number of black nodes on any path from x to a leaf (not counting x itself). • Proof: We first prove a claim: The sub-tree rooted at any node x contains at least 2bh(x) –1 internal ...
... • Lemma: A red-black tree with n internal nodes has height at most 2 lg(n +1) • Definition: Black-height, bh(x), is the number of black nodes on any path from x to a leaf (not counting x itself). • Proof: We first prove a claim: The sub-tree rooted at any node x contains at least 2bh(x) –1 internal ...
Lecture 14
... hierarchical data (eg: organizational structure) • Trees allows us to find things efficiently – Navigation is O(log n) for a “balanced” tree with n nodes ...
... hierarchical data (eg: organizational structure) • Trees allows us to find things efficiently – Navigation is O(log n) for a “balanced” tree with n nodes ...
INF 431 PC – 2 Balanced Binary trees
... • One way to implement a priority queue is by using a binary heap: binary heap is a binary tree with two additional properties: – The tree is complete, i.e. every level of the tree is full, except perhaps the bottom level, and it is filled left-to-right. – For every node in the tree, the value of th ...
... • One way to implement a priority queue is by using a binary heap: binary heap is a binary tree with two additional properties: – The tree is complete, i.e. every level of the tree is full, except perhaps the bottom level, and it is filled left-to-right. – For every node in the tree, the value of th ...
if - Read
... The code is broken into two cases If the right subtree of node x is nonempty, then the successor of x is just the left-most node in the right subtree, which found by calling Tree-Minimum(right) If the right subtree of x is empty and x has a successor y, then y is the lowest ancestor of x whose left ...
... The code is broken into two cases If the right subtree of node x is nonempty, then the successor of x is just the left-most node in the right subtree, which found by calling Tree-Minimum(right) If the right subtree of x is empty and x has a successor y, then y is the lowest ancestor of x whose left ...
Applications of Trees
... { while(t != null) { if(s == label(t)) return t; if(s < label(t) t = leftSubTree(t); else t = rightSubTree(t); ...
... { while(t != null) { if(s == label(t)) return t; if(s < label(t) t = leftSubTree(t); else t = rightSubTree(t); ...
Lists and Trees (continued)
... has more than one potential successor • Defines a partial order CS-2301 D-term 2009 ...
... has more than one potential successor • Defines a partial order CS-2301 D-term 2009 ...
Midterm (with solution)
... terms of H, what is the complexity in time (big-Oh notation) of the following operations. (a) Printing BST using the in order traversal. (b) Printing BST using the post order traversal. (c) Printing BST using the pre order traversal. 3. Redo the previous question if we assume that BST is a degenerat ...
... terms of H, what is the complexity in time (big-Oh notation) of the following operations. (a) Printing BST using the in order traversal. (b) Printing BST using the post order traversal. (c) Printing BST using the pre order traversal. 3. Redo the previous question if we assume that BST is a degenerat ...
document
... This pathlength is determined by counting the number of links that must be followed to get from the root to the node The root is considered to be level 0, the children of the root are at level 1, the grandchildren of the root are at level 2, and so on ...
... This pathlength is determined by counting the number of links that must be followed to get from the root to the node The root is considered to be level 0, the children of the root are at level 1, the grandchildren of the root are at level 2, and so on ...
1a) Describe the characrteristics of a complete binary tree
... Q. Let A be a collection of Objects. Describe an efficient method for converting A to a set. That is remove all duplicates from the collection. A. Simply define an additional storage location such as an array and copy all the objects to it checking to see that they do not already exist before copyin ...
... Q. Let A be a collection of Objects. Describe an efficient method for converting A to a set. That is remove all duplicates from the collection. A. Simply define an additional storage location such as an array and copy all the objects to it checking to see that they do not already exist before copyin ...
Slides for Exam 3 review
... – Disjoint sets and up-tree representation • representative of each set • direction of pointers ...
... – Disjoint sets and up-tree representation • representative of each set • direction of pointers ...
Trees - GearBox
... ► Go to parent or children from a given node ► Add a root to an empty tree ► Add a child to a node ► Remove a node (can impose that the node be a leaf, for simplicity) ► Get the element associated to a node ► Replace the element associated to a node ...
... ► Go to parent or children from a given node ► Add a root to an empty tree ► Add a child to a node ► Remove a node (can impose that the node be a leaf, for simplicity) ► Get the element associated to a node ► Replace the element associated to a node ...
