Binary Heap: Height of subtree rooted at position i

Question

Given n values I created a Binary Max-Heap. I want to know the formula to calculate height of subtree rooted at any position i.

Answer 1

Each level in complete binary heap has 2^i nodes , so any binary heap must have 2^i nodes in level (i) except last level.

..............................(100)...................  2^0 nodes
............................./.....\..................  
...........................(19)...(36)................  2^1
........................../...\.../...\...............
.......................(17)..(3).(25).(1).............  2^2
......................./..\...........................
.....................(2).(7)..........................  last level.

because last level may be not complete so n < 2^h

Answer 2

The level of any node at position k is floor(log2(k + 1)).

So the height of the subtree rooted at position i is the difference of levels (assuming that a single leaf has height 0): height = floor(log2(n + 1)) - floor(log2(i + 1)).

We need to check if the subtree reaches the bottom of the heap or if it is one level shorter. We do this by comparing the left-most leaf to the last element:

if(pow(2, height) * (i + 1) - 1 >= n)
    height--;

Answer 3

simple algorithm:- go left till you hit null. this can be done in binary heap by left = 2*parent. The time complexity of algorithm is O(logn). Moreover you can directly evaluate it using following inequality :- 2^k*i < N , k < log2(N/i) , k = log2(floor(N/i)) . The above calculation can be done efficiently in O(log(log(n))) where you need to do binary search on the highest bit set in floor(N/i).

Finding highest bit set in integer :-

high = 63
low = 0
mid = 0
while(low<high) {

  mid = (low+high)>>1
  val = 1<<mid
  if(val==mid)
    return mid
  if(val>mid) {
     high = mid-1
  }
  else low = mid+1
}

return mid