The number of leaf nodes in a rooted tree of n nodes, with each node having 0 or 3 children is

a) n/2 b) (n-1)/3 c) (n-1)/2 d) (2n+1)/3

The answer is: d) (2n+1) / 3

Let us derive the answer assuming the ternary tree a full ternary tree.

The maximum number of nodes in a ternary tree with height h is (3^(h + 1) - 1) / 2

(3^(h + 1) - 1) /2 = n

(3^(h + 1) - 1) = 2n

3^(h + 1) = 2n + 1

3^h * 3 = 2n+1

3^h = (2n+1) /3

At level h i.e. the number of leaf nodes in a ternary tree is 3^h.

So the number of leaf nodes in a ternary tree of n nodes is (2n+1)/3

The number of leaf nodes in a rooted tree of n nodes, with each node having 0 or m children is

The maximum number of nodes in an m'ary tree with height h is (m^(h+1) - 1) / (m - 1).

(m^(h + 1) - 1) / (m-1) = n

(m^(h + 1) - 1) = n(m-1)

m^(h + 1) = n(m-1) + 1

m^h *m = n(m-1) + 1

m^h = (n(m-1) + 1) / m

m^h = (nm - n + 1) / m

At level h i.e. the number of leaf nodes in an m'ary tree is m^h.

So the number of leaf nodes in an m'ary tree of n nodes is (nm - n + 1) / m.

Suppose m=2 i.e. a binary tree, the number of leaf nodes in a binary tree of n nodes is (2n - n + 1) / 2 = (n+1)/2.

Actual proof is not this straight forward. Let us consider a ternary tree which is not a full ternary tree but has either 0 or 3 children.

Here the number of nodes on level 0 is 1, on level 2 is 3, on level 3 is 6. The number of leaf nodes are 7 i.e. on level 1 there is one leaf node and on level 2 there are 6 leaf nodes.According to the formula the number of leaf nodes is (2n+1)/3.

Here n= 10

(2n+1)3 = (2*10+1)/3 = 21/ 3 = 7 which is the right answer, so how is it possible? A detailed proof is given here.

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