In every finite undirected graph, an even number of vertices will always have odd degree :

The handshaking lemma is a consequence of the degree sum formula (also sometimes called the handshaking lemma):

sum(deg(V)) = 2(E)

# k-ary tree where every node has either 0 or k children¶

```
L = (k - 1)*I + 1
Where L = Number of leaf nodes
I = Number of internal nodes
```

## Proof¶

### With handshaking lemma¶

Proof can be divided in two cases.

Case 1 (Root is Leaf):There is only one node in tree. The above formula is true for single node as L = 1, I = 0.

Case 2 (Root is Internal Node): For trees with more than 1 nodes, root is always internal node. The above formula can be proved using Handshaking Lemma for this case. A tree is an undirected acyclic graph.

Total number of edges in Tree is number of nodes minus 1, i.e., |E| = L + I – 1.

All internal nodes except root in the given type of tree have degree k + 1. Root has degree k. All leaves have degree 1. Applying the Handshaking lemma to such trees, we get following relation.

```
Sum of all degrees = 2 * (Sum of Edges)
Sum of degrees of leaves +
Sum of degrees for Internal Node except root +
Root's degree = 2 * (No. of nodes - 1)
Putting values of above terms,
L + (I-1)*(k+1) + k = 2 * (L + I - 1)
L + k*I - k + I -1 + k = 2*L + 2I - 2
L + K*I + I - 1 = 2*L + 2*I - 2
K*I + 1 - I = L
(K-1)*I + 1 = L
```

So the above property is proved using Handshaking Lemma, let us discuss one more interesting property.

### Without it¶

Since there are I internal nodes, each having K children, therefore total children in the tree = K * I.

There are I-1 internal nodes which are children of some other node (root has been excluded hence one less than the total number of internal nodes)

That is, out of these K* I children, I-1 are internal nodes and therefore the rest (K*I – (I-1)) are leaves.

Hence L = (K-1)*I + 1.

# In Binary tree,¶

```
L = T + 1
Where L = Number of leaf nodes
T = Number of internal nodes with two children
```

## Proof¶

### Case 1¶

```
T = 0
L = 1
```

### Case 2¶

Root has two children, i.e., degree of root is 2.

```
Sum of degrees of nodes with two children except root +
Sum of degrees of nodes with one child +
Sum of degrees of leaves + Root's degree = 2 * (No. of Nodes - 1)
Putting values of above terms,
(T-1)*3 + S*2 + L + 2 = (S + T + L - 1)*2
Cancelling 2S from both sides.
(T-1)*3 + L + 2 = (T + L - 1)*2
T - 1 = L - 2
T = L - 1
```

### Case 3:¶

Root has one child, i.e., degree of root is 1.

```
Sum of degrees of nodes with two children +
Sum of degrees of nodes with one child except root +
Sum of degrees of leaves + Root's degree = 2 * (No. of Nodes - 1)
Putting values of above terms,
T*3 + (S-1)*2 + L + 1 = (S + T + L - 1)*2
Cancelling 2S from both sides.
3*T + L -1 = 2*T + 2*L - 2
T - 1 = L - 2
T = L - 1
```