Here's a simple left-rotate pseudo-rotate that performs a left-rotate operation on a binary search tree. What this does is update the max property of the node so that it completes in o(1) time.
left-rotate(t, x): t is the tree, x is the node y := xright x.right := y.left y.left = x t.Pseudo-** explanation of root = yleft-rotate:
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In go, we can define a struct to represent the nodes of the interval tree. Each node has a max property that represents the maximum value in that node and all of its children.
Here's a pseudo ** for the left-rotate operation
type node struct func leftrotate(node *node) func max(a, b int) int return b}
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The following is a pseudo-** of the left-rotate operation of the interval tree node implemented by the go language
type intervaltreenode struct func leftrotate(node *intervaltreenode) func max(a, b int) int return b}
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When designing a data structure for an interval tree, each node usually contains interval information and some auxiliary information, such as the maximum value (max). In order to ensure the efficiency of operations, interval trees usually need to support fast insertion, deletion, and query operations, and the time complexity of each operation is o(log n).
In your question, you're asking for a process called left-rotate, which updates the max property of the node. This process is often used to balance binary search trees (BSTs), such as **l trees or red-black trees, but it can also be used in interval trees, especially when nodes violate some of the properties of interval trees.
Here's a pseudo-Left-Rotate process that assumes you have a node struct that contains information about the interval and the maximum value (max) property
Type Node struct Left-Rotate function func left-rotate(x *node) If the right child of y is greater than the maxvalue of x, if y also needs to be updatedright.maxvalue > x.maxvalue returns a new root node return y}interval represents the type of interval that the node contains. maxvalue is the maximum value property on the node. Left and right are pointers to the left and right subtrees of the node. The left-rotate function swaps the x node and its right child y, and updates the pointer. After the swap, x becomes the right child of y, and the original right child of y becomes the left child of x.
Updating the maxvalue property of x is a critical step to ensure that the max property is still correct after the rotation. This requires comparing the maxvalue of x with the maxvalue of y and the maxvalue of the right child of y, taking the largest of the three as the new maxvalue of x.
Note that this pseudo-section is generic and doesn't take into account specific implementation details, such as the definition of an interval type, or how nodes specifically store and manage intervals. In practice, you may need to adapt this paragraph to your specific needs.
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Unlike an interval tree, a binary search tree (BST) typically contains a single interval rather than a single value in its nodes, and is designed to maintain some additional properties to support efficient interval queries. However, there is no direct left-rotate operation for interval trees, because the rotation operation is mainly used to adjust the balance of the binary search tree, such as in the **l tree or the red-black tree.
If your interval tree implements a BST-based structure, and does have a max attribute that represents the maximum value of the interval range represented by each node, and assumes that the maximum value attribute of the relevant node and its ancestor needs to be updated each time the interval is inserted or deleted, then you must ensure that the max attribute is correct when performing the rotation operation.
Here's a simplified left-rotate pseudo-operation that works with any type of BST node, including nodes that may store compartments. During the rotation process, we make sure to recalculate the maximum property of the parent node and the newly rotated child node
Defines the interval tree node structure, which contains the maximum value attribute node lefthand operation, with x as the rotation axis node func leftrotate(x *node) *node updates the parent node y to the parent node yparent = x.parent if x is the root node, then update the root node to y if xparent == nil else if x == x.parent.left else sets x to y's left child node yleft = x x.parent = y update the maximum value property of the node after rotation xmaxvalue = max(x.interval.max(),getmaxofsubtree(x.left), getmaxofsubtree(x.right)) y.maxvalue = max(y.interval.max(),x.maxvalue, getmaxofsubtree(y.right)) return y returns a new subtree root node} helper function to get the maximum value of the subtree func getmaxofsubtree(node *node) valuetype return nodemaxvalue}
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In go, we can use the map type to represent the interval tree, where the key of the map is the left endpoint of the interval and the value is the corresponding node. In the left-rotate operation, we need to update the max property of the node, which can be done in o(1) time.
Here's a simple pseudo-implementation:
type node struct func leftrotate(node *node, left int, right int) to update the information of the original node nodeleft = newnode.mid node.mid = right Update the max property newnodemax = updatemax(node, newnode) to add the new node to the map tree[newnode.left] = &newnode tree[newnode.mid] = &newnode}func updatemax(node *node, newnode *node) int if newnode.right < node.right return max}
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The following is a pseudo-**:type interval struct type node struct func updatemax(node *node) max := node.interval.high if node.left != nil &&node.left.max > max if node.right != nil &&node.right.max > max node.max = max}func leftrotate(root, x *node) *node return y}Insert a description here.
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The following is the interval tree node structure written in the Go language and the pseudo-**::, which updates the max attribute and performs the left-rotate operation in O(1) time
package maintype interval struct type node struct update the maximum value property of the nodefunc updatemax(node *node) if noderight != nil &&node.right.interval.max > max node.interval.max = max} left rotation operation (left-rotate) func leftrotate(root, x *node) *node yparent = x.parent sets the parent node of y to the parent node of x if xparent == nil else if x == x.parent.left else y.left = x sets x to the left child of y xparent = y set y to the parent node of x updatemax(x) update the maximum value property of the related node after rotation updatemax(y) return root}func main().Insert a description here.
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In an interval tree, nodes typically contain an interval (usually the left and right endpoints) and an associated value. When doing a left-rotate operation, we usually need to update the max property of the node to make sure it reflects the maximum value in the subtree. Here's a pseudo-lefthand operation that acts on an interval tree node that updates the max property in o(1) time:
Let's say we have an interval tree node struct type intervalnode struct left-rotate function, func leftrotate(x *intervalnode) *intervalnode, update y's parent pointer yparent = x.parent if x is the root node, update the root node if xparent == nil else } sets the left child of y to x yleft = x updates x., the parent pointer of xparent = y updates the max property x. for x and ymax = max(x.value, x.right.max) y.max = max(y.value, y.left.max) returns a new root node return y} max helper function func max(a, b int) int return b}Note that this pseudo-assumes that each node has a pointer to its parent node, which is necessary in the actual interval tree implementation. In the actual go, you may need to adjust this pseudo based on the specific interval tree implementation.
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In this pseudo, we will implement a left-rotate operation for the interval tree, which is used to update the nodes and max properties of the interval tree in o(1) time.
function left-rotate(t, x): Gets the right child of x y y = xright If y's left child is not empty, set x to y's parent if yleft != null: y.left.parent = x sets the parent of y to yparent = x.parent If the parent node of x is empty, x is the root node, and set y to the root node if xparent == null: t.root = y If x is the left child of its parent, set y to the left child of the x parent else if x == xparent.left: x.parent.left = y If x is the right child of its parent, set y to the right child of x's parent else: xparent.right = y sets x to y's left child node yleft = x sets the parent node of x to y xparent = y update the max property xmax = max(x.left.max, x.right.max) y.max = max(y.left.max, y.right.max)Insert a description here.
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