Data Structure (Tree Continued)
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The document provides an overview of tree traversal methods including preorder, inorder, and postorder with corresponding examples and illustrations. It outlines the rules and processes for converting general trees to binary trees and traversing them using stack-based methods. Additionally, it includes exercises to practice binary tree creation and traversal techniques.
Data Structure (Tree Continued)
- 3. Types of Tree Traversal • Preorder : Node – Left - Right • Inorder : Left – Node - Right • Postorder : Left – Right - Node
- 4. Ilustration (1) H A L K B C J Preorder (NLR) : HA KCB JL Inorder (LNR) : HA KCB J L Postorder (LRN) : HA KCB JL
- 5. Ilustration (2) Preorder (NLR) : ABDE C GH Inorder (LNR) : Postorder (LRN) : A B H CD E F GI I F ABD E CGHI F ABDE CGHI F
- 6. Case Prefix : / + A- B DH Infix : Postfix: / + ^ - A B F * D * ^ E = A + BDH – F G - K A + B * D^ H- F )( / G( - )E = K H - KG GK-F /+A -B D H* ^ G K-F /+A -BDH *^ GK -F Subtree Kiri Subtree Kanan Root
- 7. Make binary tree from this general tree and do the tree traversal process: Exercise A B H C D E F GI R P Q
- 8. TREE TRAVERSAL INTO BINARY TREE
- 9. Example 1 Preorder : Inorder : J B H CD E F GI BDEI F GJC H DIEB F CJH G
- 10. Example 2 Inorder : Postorder : J B H CD E F GI IEDH F CBG J DIEB F CJH G
- 11. Make binary tree from these result of tree traversal: • Inorder : BCDFGKMPSUWY Preorder : MFDBCKGSPWUY • Postorder : EGHCIMFBNPJLKDA Inorder : EBGCHFMIANJPDLK Exercise
- 13. Node have 2 Using Preorder Rules : • If scanned node has 2 children then PUSH its right son (RS). • If scanned node hasn’t child then POP element of stack.
- 14. Node have 2 Using Preorder Notes : • Arrow down (↓) means PUSH. • Arrow up (↑) means POP. • LS : Left Son • RS : Right Son
- 15. Node have 2 Using Preorder A B H CD E F GI Preorder (NLR) : Head B D E I C F G H A BDEICFGHA LS RS LS RS B RS B LS RS RS RS C RS C
- 16. Node have 2 Using Inorder Rules and Note : • If scanned node has left child then PUSH itself (Scanned Node Address). • If scanned node hasn’t child then POP element of stack. • Add = Address.
- 17. Node have 2 Using Inorder A B H CD E F GI Inorder (LNR) : Head B D E I E B C F G DIEBFGHCA LS RS LS Add. B Add. E RS LS Add. C Add. BAdd. E Add. C RS RS H RSC A
- 18. Node have 2 Using Postorder Rules : • If scanned node has 2 children then PUSH itself (Scanned Node Address) and its right child. • If scanned node has only 1 child then PUSH itself (Scanned Node Address). • If scanned node hasn’t child then POP element of stack.
- 19. Node have 2 Using Postorder A B H CD E F GI Postorder (LRN) : Head B D E I E D C F G IEDHGFACB LS RS LS RS B Add. E LS RS B Add. D G F A C H Add B RS Add C Add D Add E RS C Add C Add F RS Add G Add G Add F RS C B Add B
- 20. Traverse this tree using stack (3 ways): Exercise 1 A B H C D E FI
- 21. Make binary tree from this statement then traverse binary tree using stack (3 ways): Exercise 2 E = A + (B-D)H – F GK
- 22. Make binary in linked list form from these result of tree traversal and traverse the result using stack in postorder way: Preorder : RFCAEDLHPVTZXWY Inorder : ACDEFHLPRTVWXYZ Exercise 3
- 23. Make binary tree from this general tree and traverse the result using preorder, inorder, and postorder: Exercise 4 S B L D E F J K MC G H I N P X A PY
- 24. Contact Person: Adam Mukharil Bachtiar Informatics Engineering UNIKOM Jalan Dipati Ukur Nomor. 112-114 Bandung 40132 Email: adfbipotter@gmail.com Blog: http://adfbipotter.wordpress.com Copyright © Adam Mukharil Bachtiar 2012