## Complete Binary Tree and Perfect Binary Tree

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Sets and maps are important and useful abstractions. We've seen various ways to implement an abstract data type for sets and maps, since data structures that implement sets can be used to implement maps as well.

It's time to look at an implementation of sets that is asymptotically efficient and useful in practice: Binary trees have two advantages above the asymptotically more efficient hash table: Second, they store their values or keys, in the case of a map in order, which makes range queries and perfect binary tree and complete binary tree iteration possible.

For simplicity, we will implement sets of some type we will call value. We assume we have a comparison function compare: An important property of a search tree perfect binary tree and complete binary tree that it can be used to implement an ordered set or ordered map easily: Although the signature above doesn't show them, ordered sets generally provide operations for finding the minimum and maximum elements of the set, for iterating over all the elements between two elements, and for extracting or iterating over ordered subsets of the elements between a range:.

A binary search tree is a binary tree with the following representation invariant: For any node nevery node in left n has a value less than that of nand every node in right n has a value more than that of n.

And the entire left **perfect binary tree and complete binary tree** right subtrees satisfy the same invariant. Given such a tree, how do you perform a lookup operation? Start from the root, and at every node, if the value of the node is what you are looking for, you are done; otherwise, recursively look up in the left or right subtree depending on the value stored at the node.

Adding an element is similar: This is a nondestructive update, so as the recursion completes, a new tree is constructed that is just like the old one except that it has a new node if needed. What is the running time of those operations? Since add is just a lookup with an extra node creation, we focus on the lookup operation. Clearly, the run time of add is O hwhere h is the height of the tree. What's the worst-case height of a tree? Clearly, a tree of n nodes all in a single long branch imagine adding the numbers 1,2,3,4,5,6,7 in order into a binary search tree.

So the worst-case running time of lookup is still O n for n the number of nodes in the tree. What is a good shape for a tree that would allow for fast lookup? A perfect binary tree has the largest number of nodes n for a given height h: If a tree with n nodes is kept balanced, its height is Perfect binary tree and complete binary tree lg nwhich leads to a lookup operation running in time O lg n.

How can we keep a tree balanced? It can become unbalanced during element addition or deletion. Most approaches involve adding or deleting an element just like in a normal binary search tree, followed by some kind of tree surgery to rebalance the tree. Similarly, element deletion proceeds as in a binary search tree, followed by some corrective rebalancing action. Examples of balanced binary search tree data structures include. In each of these, we ensure asymptotic complexity of O lg n by enforcing a stronger invariant on the data structure than just the binary search tree invariant.

Red-black trees are a fairly simple and very efficient data structure for maintaining a balanced binary tree. The idea is to strengthen the rep invariant so a tree has height logarithmic in n. To help enforce the invariant, we color each node of the tree either red or black.

Where it matters, we consider the color of an empty tree to be black. If a tree satisfies these two conditions, it must also be the case that every subtree of the tree also satisfies the conditions. If a subtree violated either of the conditions, the whole tree would also.

With this invariant, the longest possible path from the root to an empty perfect binary tree and complete binary tree would alternately contain red and black nodes; therefore it is at most twice as long as the shortest possible path, which only contains black nodes. If n is the number perfect binary tree and complete binary tree nodes in the tree, the longest path cannot have a length greater than twice the length of the paths in a perfect binary tree and complete binary tree binary tree: Another way to see this **perfect binary tree and complete binary tree** to think about just the black nodes in the tree.

Suppose we snip all the red nodes out of the trees by connecting black nodes to their closest black descendants. Then we have a tree whose leaves are all at depth BH, and whose branching factor ranges between 2 and 4. How do we check for membership in red-black trees? Exactly the same way as for general binary trees. More interesting is the add operation.

We add by replacing the empty node that a standard add into a binary search tree would. We also color the new node red to ensure that invariant 2 is preserved. However, we may destroy invariant 1 in doing so, by producing two red nodes, one the parent of the other. In order to restore this invariant we will need to consider not only the two red nodes, but their parent. Otherwise, the red-red conflict cannot be fixed while preserving black depth.

The next figure shows all the possible cases that may arise:. Notice that in each of these trees, the values of the nodes in a,b,c,d must have the same relative ordering with respect to x, y, and z: Therefore, we can perform a local tree rotation to restore the invariant locally, while possibly breaking invariant 1 one level up in the tree:.

By performing a rebalance of the tree at that level, and all the levels above, we can locally and incrementally enforce invariant 1. In the end, we may end up with two red nodes, one of them the root and the other the child of the root; this we can easily correct by coloring the root black. The SML code which really shows the power of pattern matching! This code walks back up the tree from the point of insertion fixing the invariants at every level. At red nodes we don't try to fix the invariant; we let the recursive walk go back until a black node is found.

When the walk reaches the top the color of the root node is restored to black, which is needed if balance rotates the root. Removing an element from a red-black tree works analogously. We start with BST element removal and then do rebalancing. Here is code to remove elements from a binary tree. The key is that when an interior nonleaf node is removed, then we simply splice it out if it has zero or one children; if it has two children, we find the next value in the tree, which must be found inside its perfect binary tree and complete binary tree child.

Deleting a black element from the tree creates the possibility that some path in the tree has too few black nodes, breaking the black-height invariant 2 ; the solution is to consider that path to contain perfect binary tree and complete binary tree "doubly-black" node. A series of tree rotations can then eliminate the doubly-black node by propagating the "blackness" up until a red node can be converted to a black node, or until the root is reached and it can be changed from doubly-black to black without breaking the invariant.