What is Handshaking Lemma?
Handshaking lemma is about undirected graph. In every finite undirected graph number of vertices with odd degree is always even. The handshaking lemma is a consequence of the degree sum formula (also sometimes called the handshaking lemma)

How is Handshaking Lemma useful in Tree Data structure?
Following are some interesting facts that can be proved using Handshaking lemma.

1) In a k-ary tree where every node has either 0 or k children, following property is always true.

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

Proof:
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.

2) In Binary tree, number of leaf nodes is always one more than nodes with two children.

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

Proof:
Let number of nodes with 2 children be T. Proof can be divided in three cases.
Case 1: There is only one node, the relationship holds
as 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 = (S + 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

Therefore, in all three cases, we get T = L-1.