Question

When given a number of nodes we are able to calculate the min depth of the binary tree by doing log2(n)

Where n is the number of nodes.

If you draw the tree out for the maximum depth for example 12 nodes you work out that the maximum depth can only be 4 if the tree is to remained balanced.

                0
               /   \
              0     0
            /  \   / \
           0    0 0   0
          /\     \     \ 
        0   0      0    0

Sorry for the bad ascii art. Does anyone know of a forumla that is able to calculate the max depth of a binary tree when given the number of nodes? Or at least point me in the right direction?

Answer 1

By using the root element :

int maxHeight(BinaryTree *p) {
  if (!p) return 0;
  int left_height = maxHeight(p->left);
  int right_height = maxHeight(p->right);
  return (left_height > right_height) ? left_height + 1 : right_height + 1;
}

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By using the number of nodes and some math logic (which I definitely cannot express properly (I’m by no means a math guru); but here it is) :

Observation :

• 2-3 nodes => maxDepth = 1 (2 = 2^1, 3 = 2^1,.. < 2^2 )
• 4-7 nodes => maxDepth = 2 (4 = 2^2, 5 = 2^2,.., 6 = 2^2,.., 7 = 2^2,… < 2^3)
• 8-15 nodes => maxDepth = 3
• …

Analysis :

• m => max Depth (actual the INT part of the depth, discard any decimal places)

• n => number of nodes

• ln => natural logarithm (=log[e])

• 2^m = n

• ln(2^m) = ln(n)

• m*ln(2) = ln(n)
• m = ln(n)/ln(2)

Conclusion :

Now, if m = 2,… , then the maximum depth is 2. Just get the int part of it. 😉

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NOTE: I’m definitely re-inventing the wheel here; but that’s probably part of the fun of dealing with something you know nothing about; and doing it, solely following your instinct and observations… 🙂