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GATE | Gate IT 2005 | Question 50
  • Difficulty Level : Medium
  • Last Updated : 24 Dec, 2015

In a binary tree, for every node the difference between the number of nodes in the left and right subtrees is at most 2. If the height of the tree is h > 0, then the minimum number of nodes in the tree is:
(A) 2h – 1
(B) 2h – 1 + 1
(C) 2h – 1
(D) 2h


Answer: (B)

Explanation:

Let there be n(h) nodes at height h.

In a perfect tree where every node has exactly 
two children, except leaves, following recurrence holds.

n(h) = 2*n(h-1) + 1

In given case, the numbers of nodes are two less, therefore
n(h) = 2*n(h-1) + 1 - 2
     = 2*n(h-1) - 1

Now if try all options, only option (b) satisfies above recurrence.

Let us see option (B)
n(h) = 2h - 1 + 1

So if we substitute 
n(h-1) = 2h-2 + 1, we should get n(h) = 2h-1 + 1

n(h) =  2*n(h-1) - 1
     =  2*(2h-2 + 1) -1
     =  2h-1 + 1.


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