GATE | Gate IT 2005 | Question 50
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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Last Updated :
28 Jun, 2021
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