### Leftist Heap - OpenGenus IQ: Learn Computer Science

Leftist Heap - OpenGenus

### Leftist Tree / Leftist Heap - GeeksforGeeks

CMSC 341 Lecture 15 Leftist Heaps

### Leftist Heap - OpenGenus IQ: Learn Computer Science

Leftist tree - Wikipedia

### Leftist Tree / Leftist Heap - Tutorialspoint.dev

Leftist Tree / Leftist Heap - GeeksforGeeks

### Data Structures - Leftist Heap - Determining Time ...

Aug 02, 2017 · A leftist tree or leftist heap is a priority queue implemented with a variant of a binary heap. Every node has an s-value (or rank or distance) which is the distance to the nearest leaf. In contrast to a binary heap (Which is always a complete binary tree), a leftist tree may be very unbalanced.. Below are time complexities of Leftist Tree / Heap. ...

### Leftist Heaps - Cs.cmu.edu

Reading time: 40 minutes. A leftist heap is a modification priority queue implemented with variant of binary heap. Regarding binary heap, it is always a complete binary tree. But a leftist heap may be unbalanced sometimes. A node of a leftist tree contains following elements:

### Leftist Heap - Determining Time Complexity - Genera Codice

A leftist tree or leftist heap is a priority queue implemented with a variant of a binary heap. Every node has an s-value (or rank or distance) which is the distance to the nearest leaf. In contrast to a binary heap (Which is always a complete binary tree ), a leftist tree may be very unbalanced. Below are time complexities of Leftist Tree / Heap.

### 1 Recap: Leftist Heaps

The time complexity of merge (union) operation is said to be O ( lg. . ( n 1 + n 2)), where n 1 and n 2 are the numbers of elements in the merged heaps, respectively. I do not understand this - the algorithm has to go through all the elements of both rightmost paths of the original heaps - lengths of these paths are bound by O ( lg. .

### CMSC 341 Lecture 15 Leftist Heaps

Make Heap – Time Bound There are at most n/2h items at height h. Siftdown for all height h nodes is O(h•n/2h) time Total time = O(∑ h h•n/2h) [sum of time for each height] = O(n ∑ h (h / 2h)) [factor out the n] = O(n) [sum bounded by const]