Created
June 21, 2020 18:25
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| // Assume heap is represented as an array list: heap = [...] | |
| // Assume min-heap | |
| function siftDown(idx) { | |
| let leftChildIdx = (2 * idx) + 1; | |
| let rightChildIdx = (2 * idx) + 2; | |
| // base case | |
| // make sure the heap has children | |
| // it has to at least have a left child | |
| if(leftChildIdx >= heap.length) return; | |
| // start at the current node and find the new min-value among the three nodes (parent, left child, right child) | |
| let node = heap[idx]; | |
| let minValueIdx = idx; | |
| // check the left child | |
| if(heap[leftChildIdx] !== null && node > heap[leftChildIdx]) { | |
| node = heap[leftChildIdx]; | |
| minValueIdx = leftChildIdx; | |
| } | |
| // now check the right child | |
| if(heap[rightChildIdx] !== null && node > heap[rightChildIdx]) { | |
| node = heap[rightChildIdx]; | |
| minValueIdx = rightChildIdx; | |
| } | |
| // if the min-value is not the initial node, do the swap and then recursion | |
| if(minValueIdx !== idx) { | |
| let temp = heap[idx]; | |
| heap[idx] = heap[minValueIdx]; | |
| heap[minValueIdx] = temp; | |
| siftDown(minValueIdx); | |
| } | |
| } |
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