How traversals and merging in binary trees are done?

Question

1. Binary tree traversals

• Inorder traversal
• Postorder Traversal
• Preorder Traversal 

2. Formation of binary tree from its traversals. 
3. Merging two binary trees together

Answer 1

1. Traversals in Binary tree

• The traversal operation is frquently used operation on a binary tree.
•  This operation is used to visit each node in the tree exactly once. 
• There are 6(six) possible ways a tree can be traversed and they are as follows:

1. R   T_L  T_Y 
2. _​​​​T_L   R   T_R
3. T_L  T_R  R_​​​​​​​​​
4. T_R  T_L  R
5. T_R  R    T_L  
6.  R  T_R  T_L    

 where, T_L is left sub-treesof the node R. 

           T_R is right sub-trees of the node R. 

• Out of these six possible traversals, only three are fundamental , they are categorized as given below:

1. R   T_L  T_R  (preorder) 
2. T_L  R    T_R (inorder) 
3. T_L  T_R   R (post order) 

Example:

Inorder

1) The given binary tree is 

 

Find out the inorder, preorder and postorder traversals.

Solution:

Inorder:Left  root right

Step:1 LC A RC

Step:2 _B_A RC

Step:3 DBHEARC

Step:4 DBHEA_C_

Step:5 DBHEAIFJCG

Step:6 DBHEAIFJCG

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Algorithm

1. ptr=ROOT
2. if(ptr≠NULL) then
3. inorder(ptr→LC) 
4. visit(ROOT) 
5. inorder(ptr→RC) 
6. end if
7. stop

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Preoder:Root  LC  RC

Step:1 A LC RC

Step:2 A B_   C  _

Step:3 ABDEH CFIJG

Step:4 ABDEHCFIJG

Step:5 ABDEHCFIJG

Algorithm

1. ptr=ROOT
2. if(ptr≠NULL) then
3. visit(ROOT) 
4. preorder(ptr→LC) 
5. preorder(ptr→RC) 
6. end if
7. stop

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Postorder : LC  RC Root 

Step:1 LC RC A

Step:2 DHEB IJEGC A

Step:3 DHEBIJEGCA

Step:4 DHEBIJEGCA 

Algorithm

1. ptr=ROOT
2. if (ptr≠NULL) then
3. postorder(ptr→LC) 
4. postorder(ptr→RC) 
5. visit(ROOT) 
6. end if
7. stop

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2) To construct inorder, preoder and post order traversals for below binary tree:

Solution: 

Inorder: left root right 

Step:1 LC n₆ RC 

Step:2 n₁ n₂ n₃ n₄ n₅ n₆ n₇ n₈ n₉ 

Step:3 n₁ n₂ n₃ n₄ n₅  n₆ n₇ n₈ n₉ 

Preoder: Root LC RC 

Step:1 n₆ LC RC 

Step:2 n₆ n₂ n₁ n₄ n₃ n₅ n₉ n₇ n₈  

Step:3 n₆ n₂ n₁ n₄ n₃ n₅ n₉ n₇ n₈​​​​​​_{​​​​​}  

Postorder:LC  RC root 

Step1:  LC RC n₆ 

Step:2 n₁ n₃ n₅ n₄ n₂  n₈ n₇ n₉ n₆ 

Step:3 n₁ n₃ n₅ n₄ n₂ n₈ n₇ n₉ n₆​

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2. Formation of Binary tree from its traversals​​

• From a single traversal, it is not possible to construct a unqiue binary tree. 
• But, if two traversals are known then, we can construct a unique binary tree. 
• Out of two traversals, one should be inorder traversal and another preorder or postorder. 
• If the  two traversals are preorder and postorder then, a  binary tree cannot be constructed uniquely. 

Example 1:

Construct the binary tree from the inorder and preorder traversals of a binary tree as given below:

Inorder:D B H E A I F J C G 

Preoder:A B D E H C F I J G 

Solution:

The following steps need to be followed:

1. From the preoder traversal, it is evident that A​​​​​​ is the root node. 
2. In inorder traversal, all the nodes  which are on the left side of A belong to the left sub tree and those which are on right side of A belong to the right subtree. 
3. Now the problem reduces to form subtreees and the same procedure can be applied repeatedly. 

Step:1 

Step:2

(i) In left subtree , B will be the root(B D E H) 

(ii) In right subtree, C will be the root(C F I J G) 

Step:3

(i) In right subtree of B, E is the root from the preoder E H. 

From inorder, H will be left subtree of E. 

(ii) From left subtree of C, F is the root from the preorder F I J

From inorder, I is left of F and J is right of F. So, 

So, the final binary tree from inorder and preorder is 

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Example:2

Construct a binary tree from the inorder and postorder traversals given below:

Inorder: n₁  n₂  n₃ n₄  n₅ n₆ n₇  n₈  n₉ 

Postorder: n₁ n₃ n₅  n₄  n₂ n₈  n₇ n₉ n₆

 Solution:

Step:1

n₆ is the root from postorder.

Left subtree:​​​​ ^{​​​​​​} n₁  n₂ n₃ n₄ n₅ 

Right subtree: n₇  n₈  n₉ 

Step:2

From postorder : n₂ is the root of left subtree. 

From postorder: n₉ is the root for right subtree. 

Step:3 

n₄ is the root of n₂ right subtree. 

n₇ is the root of n₉ left subtree. 

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3. Merging two binary trees together

• Merge operation is applicable to trees which are reprsented by using a linked structure . 
• There are two ways that this operation can be carried out. 

Approach 1:

Suppose T₁ and T₂ are two binary trees. T₂ can be merged with T₁ if all the nodes from T₂, one by one , are inserted into the binary tree T₁  

Approach 2:

When the entire tree T₂ can be include as a subtree of T₁ To achieve this, we need that in either tree . There must be atleast one null subtree. 

(i) Before performing merging, we have to test for compatibility. If in both trees the root node has both the left and  right subtrees, then the merge  will be fail. 

(ii) If T₁ has left subtree(right  subtree) empty. Then T₂ will be added to the left subtree(the right subtree) of the T₁ and vice-versa. 

(iii) If T₁ is a tree with n₁ nodes and T₂ is a tree with n₂ nodes. Then the resultant tree T after merging is

T(n₁+n₂) =T₁(n₁) +T₂(n₂) 

If the binary tree contains an arithmetic expression, 

Then ,its

• Inorder will give infix expression
• Preorder will give prefix expression
• Postorder will give postfix expression. 

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