Max heap is a complete binary tree. A complete binary tree is a binary tree in which all levels are completely filled and all the nodes in the last level are as left as possible. Max heap should also meets this criteria: the parent’s key is larger than both children’s keys. The largest value is at the root. Heap is mostly used to implement priority queue.
Since it is complete (no holes in the tree), it can be represented as an array. Each node’s position in the tree is corresponding to the index in the array.
The relationship between the nodes has the following formulas:
parentPos = (pos-1)/2
left = 2*pos+1
right=2*pos+2
So we can use array to implement max heap.
Table of Content
Insert
We declare three attributes: heap, length and maxSize. heap is an array. maxSize is the max length of the array. It is used to initialize the array. length is actual number of elements in the array. It starts from 0.
To insert a new element, we put the new value at the end of the array. Then we move it up based on its value compared to its parent. If it’s value larger than its parent, it should be switched the position with its parent, until it is below a larger node and above a smaller one. We call this process heap up (or trickle up).
Java
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import java.util.*; public class MaxHeap<T extends Comparable<T>> { T[] heap; int length; int maxSize; //Time O(1) Space O(1) @SuppressWarnings("unchecked") public MaxHeap(int capacity) { maxSize = capacity; heap = (T[]) new Comparable[capacity]; length = 0; // means queue is empty } //Insert a new value, Time O(h), Space O(1), n is number of items in heap //h is height of heap (complete binary tree) h = log(n) public void insert(T value) { if (length == maxSize) { System.out.println("heap is full, cannot insert " + value); return; } heap[length]= value; heapUp(length++); } //Time O(h), Space O(1) private void heapUp(int pos) { int parentPos = (pos-1)/2; T value = heap[pos]; while (pos > 0 && (value.compareTo(heap[parentPos]) > 0)) { heap[pos] = heap[parentPos]; pos = parentPos; parentPos= (parentPos-1)/2; } heap[pos] = value; } } |
Javascript
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class MinHeap { //Constructor, Time O(1) Space O(1) constructor(capacity) { this.maxSize = capacity; this.length = 0; this.heap = {}; } //Insert a new value, Time O(h), Space O(1), n is number of items in heap //h is height of heap (complete binary tree) h=log(n) insert(value) { if (this.length >= this.maxSize) { console.log("heap is full, cannot insert"); return; } this.heap[this.length] = value; this.heapUp(this.length++); } //Time O(h), Space O(1) heapUp(pos) { var parentPos = Math.floor((pos-1)/2); var value = this.heap[pos]; while (pos > 0 && value < this.heap[parentPos]) { this.heap[pos] = this.heap[parentPos]; pos = parentPos; parentPos= Math.floor((parentPos-1)/2); } this.heap[pos] = value; } } |
Python
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class MaxHeap : #Time O(1) Space O(1) def __init__(self, capacity) : self.maxSize = capacity self.length = 0 self.heap = [None]*capacity #Insert a new value, Time O(h), Space O(1), n is number of items in heap #h is height of heap (complete binary tree) h = log(n) def insert(self, value): if self.length >= self.maxSize : print("heap reach max size, return") return self.heap[self.length] = value # start from last self.heapUp(self.length) #Time O(h), Space O(1) def heapUp(self, pos) : parentPos = int((pos-1)/2) value = self.heap[pos] while pos > 0 and value > self.heap[parentPos] : self.heap[pos] = self.heap[parentPos] pos = parentPos parentPos = int((parentPos-1)/2) self.heap[pos] = value; self.length += 1 |
Doodle
Remove
The remove operation is to remove the element with the highest value, which is the first one in the array. When we remove the first element in the array, we move the last element to fill out the empty spot. But this element’s value might be smaller than its children’s. To correct this, we compare its value with its children’s values, and switch with the child with the larger value, until it is below a larger node and above a smaller one. This process is called heap down (or trickle down). Heap down is more complicated than heap up because it has two children to compare.
Java
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//Remove from max (first one), Time O(h), Space O(1) public T remove() { if (length == 0) { System.out.println("heap is empty"); return null; } T max = heap[0]; heap[0] = heap[length-1]; //put last at first length--; heapDown(0); return max; } //Time O(h), space O(1) private void heapDown(int pos) { T item = heap[pos]; int index; while (pos < length/2) { int left = 2*pos + 1; int right = 2*pos + 2; if (right < length && heap[left].compareTo(heap[right]) < 0) index = right; else index = left; if (item.compareTo(heap[index]) >= 0) break; heap[pos] = heap[index]; pos = index; } heap[pos] = item; } |
Javascript
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//Remove from min (first one), Time O(h), Space O(1) remove() { if (this.length == 0) { console.log("heap is empty"); return null; } var min = this.heap[0]; this.heap[0] = this.heap[this.length-1]; //put last at first this.length--; this.heapDown(0); return min; } //Time O(h), Space O(1) heapDown(pos) { var item = this.heap[pos]; var index; while (pos < Math.floor(this.length/2)) { let left = 2*pos + 1; let right = 2*pos + 2; if (right < this.length && this.heap[left] > this.heap[right]) index = right; else index = left; if (item <= this.heap[index]) break; this.heap[pos] = this.heap[index]; pos = index; } this.heap[pos] = item; } |
Python
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#Remove from max (first one), Time O(h), Space O(1) def remove(self) : if self.length == 0: print("heap is empty") return None max = self.heap[0] self.heap[0] = self.heap[self.length-1] # put last at first self.length -= 1 self.heapDown(0) return max # Time O(h), Space O(1) def heapDown(self, pos) : item = self.heap[pos] index = 0 while pos < int(self.length/2) : left = 2*pos + 1 right = 2*pos + 2 if right < self.length and self.heap[left] < self.heap[right]: index = right else: index = left if item >= self.heap[index]: break self.heap[pos] = self.heap[index] pos = index self.heap[pos] = item; |
Doodle
Search
Since the underlying data structure is an array, we iterate through all elements to find the one matched with the key.
Java
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//Time O(n), Space O(1), n is number of items in heap public boolean search(T key) { for (int i = 0; i < length; i++) { if (key.compareTo(heap[i]) == 0) return true; } return false; } |
Javascript
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//Time O(n), Space O(1), n is number of items in heap search(key) { for (let i = 0; i < this.length; i++) { if (key == this.heap[i]) return true; } return false; } |
Python
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#Time O(n), Space O(1), n is number of items in heap def search(self, key) : for i in range(0, self.length) : if key == self.heap[i]: return True return False |
Traversal
Traversal is to visit all nodes in the tree in some specified order. Please note the heap is a tree in conceptual context, the underlying implementation is array. Therefore there are no actual nodes as we see in the binary tree implementation (there is no root node). But we still can traverse all elements in the same way as any tree traversal – by level order, preorder, inorder and postorder.
Level order is to visit node level by level from top to bottom. The order is the same as visiting array from index 0 to its length.
Java
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//Traverse as array, it is also level order of tree, Time O(n), Space O(n) public List<T> levelOrder() { List<T> list = new ArrayList<>(); for(int i = 0; i < length; i++) { list.add(heap[i]); } return list; } |
Javascript
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//Traverse as array, it is also level order of tree, Time O(n), Space O(n) levelOrder() { var str = []; for(let i = 0; i < this.length; i++) { str.push(this.heap[i]); } return str; } |
Python
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# Traverse as array, it is also level order of tree, Time O(n), Space O(n) def levelOrder(self) : s = [] for i in range(0, self.length): s.append(self.heap[i]) return s |
Preorder is to visit the root, traverse the left subtree, traverse the right subtree. The implementation is using dfs recursion. To get the children “node”, we use the formula 2*pos+1 for left child, and 2*pos+2 for right child.
Java
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//Traverse as tree preorder, Time O(n), Space (n) public List<T> preorder() { List<T> list = new ArrayList<>(); preorderRec(0, list); return list; } //Recursion helper, Time O(n), Space (h) private void preorderRec(int nodeIndex, List<T> list ) { if (nodeIndex > length-1) return; list.add(heap[nodeIndex]); preorderRec(2*nodeIndex + 1, list); preorderRec(2*nodeIndex + 2, list); } |
JavaScript
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//Traverse as tree preorder, Time O(n), Space (n) preorder() { var output = []; this.preorderRec(0, output); return output; } //Recursion helper, Time O(n), Space (h) preorderRec(nodeIndex, output) { if (nodeIndex > this.length-1) return; output.push(this.heap[nodeIndex]); this.preorderRec( 2*nodeIndex + 1, output); this.preorderRec( 2*nodeIndex + 2, output ) ; } |
Python
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# Traverse as tree preorder, Time O(n), Space (n) def preorder(self) : output = [] self.preorderRec(0, output) return output # Recursion helper, Time O(n), Space (h) def preorderRec(self, nodeIndex, output) : if nodeIndex > self.length-1 : return output.append(self.heap[nodeIndex]) self.preorderRec( 2*nodeIndex + 1, output) self.preorderRec( 2*nodeIndex + 2, output) |
Inorder is to traverse the left subtree, visit the root, traverse the right subtree. The implementation is using dfs recursion. To get the children “node”, we use the formula 2*pos+1 for left child, and 2*pos+2 for right child.
Java
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//Traverse as tree inorder, Time O(n), Space (n) public List<T> inorder() { List<T> list = new ArrayList<>(); inorderRec(0, list); return list; } //Recursion helper, Time O(n), Space (h) private void inorderRec(int nodeIndex, List<T> list) { if (nodeIndex > length-1) return; inorderRec(2*nodeIndex + 1, list); list.add(heap[nodeIndex]) ; inorderRec(2*nodeIndex + 2, list); } |
Javascript
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//Traverse as tree inorder, Time O(n), Space (n) inorder() { var output = []; this.inorderRec(0, output); return output; } //Recursion helper, Time O(n), Space (h) inorderRec(nodeIndex, output) { if (nodeIndex > this.length-1) return; this.inorderRec(2*nodeIndex + 1, output); output.push(this.heap[nodeIndex]) ; this.inorderRec(2*nodeIndex + 2, output ) ; } |
Python
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# Traverse as tree inorder, Time O(n), Space (n) def inorder(self) : output = [] self.inorderRec(0, output) return output # Recursion helper, Time O(n), Space (h) def inorderRec(self, nodeIndex, output) : if nodeIndex > self.length-1 : return self.inorderRec(2* nodeIndex +1, output) output.append(self.heap[nodeIndex]) self.inorderRec(2*nodeIndex + 2, output) |
Preorder is to visit the root, traverse the left subtree, traverse the right subtree. The implementation is using dfs recursion. To get the children “node”, we use the formula 2*pos+1 for left child, and 2*pos+2 for right child.
Java
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//Traverse as tree postorder, Time O(n), Space O(n) public List<T> postorder() { List<T> list = new ArrayList<>(); postorderRec(0, list); return list; } //Recursion helper, Time O(n), Space (h) private void postorderRec(int nodeIndex, List<T> list) { if (nodeIndex > length-1) return; postorderRec(2*nodeIndex + 1, list); postorderRec(2*nodeIndex + 2, list); list.add(heap[nodeIndex]) ; } |
JavaScript
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//Traverse as tree postorder, Time O(n), Space O(n) postorder() { var output = []; this.postorderRec(0, output); return output; } //recursion helper, Time O(n), Space (h) postorderRec(nodeIndex, output) { if (nodeIndex > this.length-1) return; this.postorderRec(2*nodeIndex + 1, output); this.postorderRec(2*nodeIndex + 2, output ) ; output.push(this.heap[nodeIndex]); } |
Python
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# Traverse as tree postorder, Time O(n), Space O(n) def postorder(self) : output = [] self.postorderRec(0, output) return output # Recursion helper, Time O(n), Space (h) def postorderRec(self, nodeIndex, output) : if (nodeIndex > self.length-1): return self.postorderRec(2*nodeIndex + 1, output) self.postorderRec(2*nodeIndex + 2, output) output.append(self.heap[nodeIndex]) |
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