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Tree Traversal

Tree Traversal. Pre Order Tree Traversal. Visit the root Traverse the left sub-tree, Traverse the right sub-tree. In Order Tree Traversal. Traverse the left sub-tree, Visit the root Traverse the right sub-tree. Post Order Tree Traversal. Traverse the left sub-tree,

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Tree Traversal

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  1. Tree Traversal

  2. Pre Order Tree Traversal • Visit the root • Traverse the left sub-tree, • Traverse the right sub-tree

  3. In Order Tree Traversal • Traverse the left sub-tree, • Visit the root • Traverse the right sub-tree

  4. Post Order Tree Traversal • Traverse the left sub-tree, • Traverse the right sub-tree • Visit the root

  5. Post Order Tree Traversal • A*(((B+C)*(D*E))+F) • A B C + D E * * F + *

  6. Tree Traversal

  7. Example • Print the pre, in and post order form of the following tree DAICBHEGF – in HBDIACEFG – pre ACIDBGFEH – post H E B F D I G A C

  8. Example class Node { private Object value; private Node left, right; public Node (Object value) { this.value = value; } ... }

  9. Example class Tree { private Node root; public Tree () { root = null; } ... }

  10. In Order traversal publicvoid inOrder () { inOrder (root); } privatevoid inOrder (Node current) { if (current != null) { inOrder(current.getLeft()); visit(current); inOrder(current.getRight()); } }

  11. Pre Order traversal publicvoid preOrder () { preOrder (root); } privatevoid preOrder (Node current) { if (current != null) { visit(current); preOrder(current.getLeft()); preOrder(current.getRight()); } }

  12. Post Order traversal publicvoid postOrder () { postOrder (root); } privatevoid postOrder (Node current) { if (current != null) { postOrder(current.getLeft()); postOrder(current.getRight()); visit(current); } }

  13. Example • Can we reconstruct a tree from it’s InOrder traversal ? • Are there two different trees with the same InOrder traversal • What about pre and post order ? D A

  14. Example • Can we reconstruct a tree given both it’s in order and pre order traversals ? in order - DAICBHEGF pre order - HBDIACEFG (find the root) Left sub tree DAICB BDIAC Right sub tree EGF EFG

  15. Reconstructing a binary tree ReconstructTree (inOrder in[] preOrder pre[]) if (n==0) return null root (T)  pre[1] find index i s.t in[i] = pre[1] left(T)  Reconstruct (in[1…i-1], pre [2…i]) right(T)  Reconstruct (in[i+1..n], pre [i+1…n]) return T

  16. Reconstructing a binary tree • Time complexity • This is the same recurrence relation of quick sort

  17. Binary Search Trees • In a BST the in order output is a sorted list of the tree nodes. • If we obtain a pre order path of a BST, we can sort the nodes and use the output to reconstruct the tree using the previous algorithm

  18. Reconstructing a binary tree ReconstructTree (inOrder in[] postOrder post[]) if (n==0) return null root (T)  post[n] find index i s.t in[i] = post[n] left(T)  Reconstruct (in[1…i-1], post [1…i-1]) right(T)  Reconstruct (in[i+1..n], post[i…n-1]) return T

  19. Level traversal (BFS) publicvoid BFS () { Queue q = new Queue (); if {root != null) q.enqueue (root); while (! q.isEmpty()) { Node current = (Node)queue.dequeue(); visit (current); if (current.getLeft() != null) q.enqueue(current.getLeft()); if (current.getRight() != null) q.enqueue(current.getRight()); } }

  20. Depth privateint depth (Node node) { if (current == null) { return0; } else { return Math.max(depth(node.getLeft()), depth(node.getRight())) + 1; } }

  21. privatevoid printLevel (Node current, int level) { if (current != null) { if (level == 0) { visit (current); } } else { printLevel (current.getLeft(), level - 1); printLevel (current.getRight(), level - 1); } } publicvoid printAllLevels () { int depth = depth (root); for (int i = 0; i < depth; i++); printLevel(root, i); } }

  22. Questions • 1. Draw a single Binary Tree such that each node contains a single character and: • a.  The pre-order traversal results in EXAMFUN • b.  The in-order traversal results in MAFXUEN

  23. Questions • 2. An in-order traversal of a Binary search tree yields a sorted list of elements in O(n) time. Does this fact contradict our comparison-based lower bound of nlogn time? Can something similar be done using a binary heap?

  24. Questions • 3. Is there a heap T storing seven distinct elements such that a preorder traversal of T yields the elements of T in sorted order? • How about an in-order traversal? • How about a post-order traversal? • How about a post-order traversal to yield the elements in reverse sorted order?

  25. Questions • 4. Let T be a binary search tree with more than a single node. Is it possible that the preorder traversal of T visits the nodes in the same order as the post-order traversal of T? • If so, give an example; otherwise, argue why this cannot occur. Likewise, is it possible that the preorder traversal of T visits the nodes in the reverse order of the post-order traversal of T? • If so, give an example; otherwise, argue why this cannot occur.

  26. Questions • 5. Given the following binary tree, print out the order of the nodes for an in-order traversal, preorder traversal, and post-order traversal. / (division operator) / \ / \ - * / \ / \ / \ 2 a ^ * / \ / \ b 2 * c / \ 4 a

  27. Questions • 6. Write a method that given a TreeNode N , returned the number of elements contained at the tree rooted by N.

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