A Binary tree has n leaf nodes. The number of nodes of degree 2 in T is:

If we count the number of edges in a ternary tree, we see that every node except the root has an incoming edge into it. If E is the number of edges, then n = E + 1, i.e. number of edges in a ternary tree is one less than the total number of nodes.

a) log

_{2}*n*b) n-1 c) n d) 2^h
The answer is b) n-1

Let n1 be the number of nodes of degree one

n0 be the number of nodes of degree 0, leaf nodes are the ones which have degree 0

n2 be the number of nodes of degree 2, root node is one example of a degree 2 node in the figure below.

n be the total number of nodes in the tree.

we have, n = n0 + n1 + n2 ------------- 1

If we count the number of edges in a binary tree, we see that every node except the root has an incoming edge into it. If E is the number of edges, then n = E + 1, i.e. number of edges in a binary tree is one less than the total number of nodes.

Each n0 node contributes no edge,

each n1 node contributes one edge,

and each n2 node contributes 2 edges to the total number of edges, E.

E = 0*n0 + 1 * n1 + 2* n2

E = n1 + 2n2

We know that n = E + 1

n = n1 + 2n2 + 1 -------------- 2

Solving equations 1 and 2, we get

n0 + n1 + n2 = n1 + 2n2 + 1

n0 = n2 + 1

The number of leaf nodes in a binary tree is one more than the number of degree 2 nodes.

In the above binary there are 6 nodes in total, n = 6

We know that n = n0 + n1 + n2

n = 3 + 1 + 2

Also we know that n = E + 1

E is the total number of edges, E = 5, every node has one incoming edge except the root node.

E = n1 + 2n2

n1 = {2} , which contributes edge 'c'

n2 = {1, 3}, node 1 contributes edges 'a' and 'b', node 3 contributes edges 'd' and 'e'.

n0 = {4,5,6} doesn't contribute any edge.

E = 1 + 2 * 2 = 5

n = E+ 1 = 6

n0 = n2 + 1 = 2 + 1 = 3

In a ternary tree we have n = n0 + n1 + n2 + n3 ------------- 3

n3 is the number of nodes of degree 3.
Each n0 node contributes no edge,

each n1 node contributes one edge,

each n2 node contributes 2 edges,

each n3 node contributes 3 edges to the total number of edges, E.

each n3 node contributes 3 edges to the total number of edges, E.

E = 0*n0 + 1 * n1 + 2* n2 + 3* n3

E = n1 + 2n2 + 3n3

We know that n = E + 1

n = n1 + 2n2 + 3n3 + 1 -------------- 4

Solving equations 3 and 4, we get

n0 + n1 + n2 +n3= n1 + 2n2 + 3n3 + 1

n0 = n2 + 2n3 + 1 -------------- 5

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

n0 = n2 + 2n3 + 1

n2 = 0, n1 = 0 as there are only nodes having degree 0 or degree 3

then n0 = 2n3 + 1

we know that n = n0 + n1 + n2 + n3

n = (2n3 + 1) + 0 + 0 + n3

n3 = (n - 1) / 3

then n0 = 2n3 + 1 = 2 ((n-1)/3) + 1 = (2n + 1) / 3

In an m'ary tree we have n = n0 + n1 + n2 + n3 + ........... +n

_{m}
Equation 5 can be generalised to

n0 = n2 + 2n3 + 3n4 + ......... + (m-1)n

_{m}+ 1
The number of leaf nodes in a m'ary tree of n nodes, with each node having 0 or m children is

n1= 0, n2 = 0, n3 = 0,.........................n

_{m-1}= 0 as there are only nodes having degree 0 or m
n0 = 0 + 0 + 0 +.............. (m-1)n

_{m}+ 1
n0 = (m-1)n

_{m}+ 1
we know that n = n0 + n1 + n2 + n3 + ........... +n

_{m}
n = ((m-1)n

_{m}+ 1) + 0 + 0 +..................................+ n_{m}
n = m*n

_{m}+ 1
n

_{m}= (n - 1) / m
then n0 = (m-1)n

_{m}+ 1 = ((m-1) (n-1) / m )+ 1 = (m*n -m -n + m + 1) / m
n0 = (m*n - n + 1) / m

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