# Handshaking Lemma and Interesting Tree Properties

**What is Handshaking Lemma?**

Handshaking lemma is about an undirected graph. In every finite undirected graph, an even number of vertices will always have an odd degree. 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 the Handshaking lemma.

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**1) In a k-ary tree where every node has either 0 or k children, the 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 into two cases.

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

* Case 2 *(Root is Internal Node): For trees with more than 1 node, the root is always an 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 a degree k. All leaves have degree 1. Applying the Handshaking lemma to such trees, we get the 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.

**Alternate Proof:** (Without using Handshaking Theorem)

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

There are I-1 internal nodes that 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.

**2) In a Binary tree, the****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 a number of nodes with 2 children are T. Proof can be divided into 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., the degree of the 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., the degree of the 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.

We have discussed proof of two important properties of Trees using Handshaking Lemma. Many GATE questions have been asked on these properties, the following are a few links.

GATE-CS-2015 (Set 3) | Question 35

GATE-CS-2015 (Set 2) | Question 20

GATE-CS-2005 | Question 36

GATE-CS-2002 | Question 34

GATE-CS-2007 | Question 43

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