a. To write the preorder(),postorder() and inorder() traversal of the trees when the definition of visit() function is given. Given: visit() function is given as: if (p->left != 0 && p->el - p->left->el < 2) p->left->el += 2; In-order traversal: • From root node “10”, traverse the left subtree which is node “5”. • From node “5”, traverse the left subtree, which is node “4”. • From node “4” traverse the left subtree, which is node “0”. • Node “0” has no left child, so visit node “0”. • For node “0”, there is no left child, so no modification. • The next node to visit is node “4”. • Node “4” has left child, node “0”, but the difference between the data value of the node and left child is “4” which is greater than “2”, therefore no modification. • The next node to visit is node “5”. • Node “5” has left child, node “4” and the difference between the data value of the node and left child is “1”, which is less than “2”. • Therefore increment the value of the node “4” by “2”, and node “4” becomes node “6”. • From node “5”, traverse the right subtree, which is node “6”. • Since node “6” has no left child, visit node “6”. • Node “6” has no left child, so there is no modification. • From node “6”, traverse the right subtree, which is node “7”. • Since node “7” has no left child, visit node “7”. • Node “7” has no left child, so there is no modification. • From node “7”, traverse the right subtree, which is node “8”. • Since node “8” has no left child, visit node “8”. • Node “8” has no left child, so there is no modification. • The next node to visit is node “10”. • Node “10” has left child, node “5”, but the difference between the data value of the node and left child is “5” which is greater than “2”, therefore no modification. • From node “10”, traverse the right subtree, which is node “20”. • From node “20”, traverse the left subtree, which is node “15”. • From node “15”, traverse the left subtree, which is node “13”. • Node “13” has no left child, so visit node “13”. • For node “13”, there is no left child, so no modification. • The next node to visit is node “15”. • Node “15” has left child, node “13”, but the difference between the data value of the node and left child is “2” which is not less than “2”, therefore no modification. • The next node to visit is node “20”. • Node “20” has left child, node “15”, but the difference between the data value of the node and left child is “5” which is greater than “2”, therefore no modification. • From node “20”, traverse the right subtree, which is node “30”. • Node “30” has no left child, so visit node “30”. • Node “30” has no left child, so there is no modification. Pre-order traversal: • First visit the root node “10”. • Node “10” has left child, node “5”, but the difference between the data value of the node and left child is “5” which is greater than “2”, therefore no modification. • From node “10”, traverse the left subtree which is node “5”. • Visit node “5”. • Node “5” has left child, node “4” and the difference between the data value of the node and left child is “1”, which is less than “2”. • Therefore increment the value of the node “4” by “2”, and node “4” becomes node “6”. • From node “5”, traverse the left subtree which is node “4” which is incremented to “6” in the last step. Now call this node “4” as node “6”. • Visit node “6”. Node “6” has left child, node “0”, but the difference between the data value of the node and left child is “6” which is greater than “2”, therefore no modification. • From node “6”, traverse the left subtree which is node “0”. • Visit node “0”. • For node “0”, there is no left child, so no modification. • From node “5”, traverse the right subtree, which is node “6”. • Visit node “6”. • Node “6” has no left child, so there is no modification. • From node “6”, traverse the right subtree, which is node “7”. • Visit node “7”. • Node “7” has no left child, so there is no modification. • From node “7”, traverse the right subtree, which is node “8”. • Node “8” has no left child, so there is no modification. • From node “10”, traverse the right subtree, which is node “20”. • Visit node “20”. • Node “20” has left child, node “15”, but the difference between the data value of the node and left child is “5” which is greater than “2”, therefore no modification. • From node “20”, traverse the left subtree which is node “15”. • Visit node “15”. • Node “15” has left child, node “13”, but the difference between the data value of the node and left child is “2” which is not then “2”, therefore, no modification. • From node “15”, traverse the left subtree which is node “13”. • Visit node “13”. • Node “13” has no left child, so there is no modification. • From node “20”, traverse the right subtree, which is node “30”. • Visit node “30”. • Node “30” has no left child, so there is no modification. Post-order traversal: • From node “10”, traverse the left subtree, which is node “5”. • From node “5”, traverse the left subtree, which is node “4”. • From node “4”, traverse the left subtree, which is node “0”. • Node “0” has no left and right subtree, so visit node “0”. • For node “0”, there is no left child, so no modification. • Next node to visit is node “4”. • Node “4” has left child, node “0”, but the difference between the data value of the node and left child is “4” which is greater than “2”, therefore no modification. • From node “5”, traverse the right subtree, which is node “6”. • From node “6”, traverse the right subtree, which is node “7”. • From node “7”, traverse the right subtree, which is node “8”. • Node “8” has no left and right subtree, so visit node “8”. • For node “8”, there is no left child, so no modification. • Next node to visit is node “7”. • For node “7” there is no left child, so no modification. • Next node to visit is node “6”. • For node “6”, there is no left child, so no modification. • Next node to visit is node “5”. • Node “5” has left child, node “4” and difference between the data value of the node and left child is “1”, which is less than “2”. • Therefore increment the value of the node “4” by “2”, and node “4” becomes node “6”. • From node “10”, traverse the right subtree, which is node “20”. • From node “20”, traverse the left subtree, which is node “15”. • From node “15”, traverse the left subtree, which is node “13”. • Node “13” has no left and right subtree, so visit node “13”. • For node “13”, there is no left child, so no modification. • Next node to visit is node “15”. • Node “15” has left child, node “13”, but the difference between the data value of the node and left child is “2” which is not less than “2”, therefore no modification. • From node “20”, traverse the right subtree, which is node “30”. • Node “30” has no left and right subtree, so visit node “30”. • For node “30”, there is no left child, so no modification. • Next node to visit is node “20”. • Node “20” has left child, node “15”, but the difference between the data value of the node and left child is “5” which is greater than “2”, therefore no modification. Therefore for all the three traversal, value of the node “4” becomes . b. To write the preorder(),postorder() and inorder() traversal of the trees when the definition of visit() function is given. Given: visit() function is given as: if (p->left == 0) p->right = 0; In-order traversal: • From root node “10”, traverse the left subtree which is node “5”. • From node “5”, traverse the left subtree, which is node “4”. • From node “4” traverse the left subtree, which is node “0”. • Node “0” has no left child, so visit node “0”. • For node “0”, there is no left child and right child, so no modification. • The next node to visit is node “4”. • Node “4” has left child but no right child, so no modification. • Next node to visit is node “5”. • Node “5” has left child and right child, so no modification. From node “5”, traverse the right subtree, which is node “6”. • Since node “6” has no left child, visit node “6”. • Node “6” has no left child, but have right child. • So remove right subtree of node “6”. • Node “7” and node “8” is removed from the tree. • The next node to visit is node “10”. • Node “10” left child and right child, so no modification. • From node “10”, traverse the right subtree, which is node “20”. • From node “20”, traverse the left subtree, which is node “15”. • From node “15”, traverse the left subtree, which is node “13”. • Node “13” has no left child, so visit node “13”. • For node “13”, there is no left child and right child, so no modification. • The next node to visit is node “15”. • Node “15” has left child and no right child, so no modification. • The next node to visit is node “20”. • Node “20” has left child and right child, so no modification. • From node “20”, traverse the right subtree, which is node “30”. • Node “30” has no left child, so visit node “30”. • Node “30” has no left child and right child, so there is no modification. Pre-order traversal: • First visit the root node “10”. • Node “10” has left child and right child, so no modification. • From node “10”, traverse the left subtree which is node “5”. • Visit node “5”. • Node “5” has left child and right child, so no modification. • From node “5”, traverse the left subtree which is node “4”. • Visit node “4”. • Node “4” has left child and no right child, so no modification. • From node “4”, traverse the left subtree which is node “0”. • Visit node “0”. • For node “0”, there is no left child and right child, so no modification. • From node “5”, traverse the right subtree, which is node “6”. • Visit node “6”. • Node “6” has no left child, but has right child. • So remove right subtree of node “6”. • Node “7” and node “8” is removed from the tree. • From node “10”, traverse the right subtree, which is node “20”. • Visit node “20”. • Node “20” has left child and right child, so no modification. • From node “20”, traverse the left subtree which is node “15”. • Visit node “15”. • Node “15” has left child and no right child, so no modification. • From node “15”, traverse the left subtree which is node “13”. • Visit node “13”. • Node “13” has no left child and right child, so no modification. • From node “20”, traverse the right subtree, which is node “30”. • Visit node “30”. • Node “30” has no left child and right child, so no modification. Post-order traversal: • From node “10”, traverse the left subtree, which is node “5”. • From node “5”, traverse the left subtree, which is node “4”. • From node “4”, traverse the left subtree, which is node “0”. • Node “0” has no left and right subtree, so visit node “0”. • For node “0”, there is no left child and right child, so no modification.. • Next node to visit is node “4”. • Node “4” has left child and no right child, so no modification. • From node “5”, traverse the right subtree, which is node “6”. • From node “6”, traverse the right subtree, which is node “7”. • From node “7”, traverse the right subtree, which is node “8”. • Node “8” has no left and right subtree, so visit node “8”. • For node “8”, there is no left child and right child, so no modification. • Next node to visit is node “7”. • For node “7” there is no left child, but has right child. • So remove right subtree of node “7”. • Node “8” is removed from the tree. • Next node to visit is node “6”. • For node “6”, there is no left child, but has right child. • So remove right subtree of node “6”. • Node “7” is removed from the tree. • Next node to visit is node “5”. • Node “5” has left child and right child, so no modification.. • From node “10”, traverse the right subtree, which is node “20”. • From node “20”, traverse the left subtree, which is node “15”. • From node “15”, traverse the left subtree, which is node “13”. • Node “13” has no left and right subtree, so visit node “13”. • For node “13”, there is no left child and right child, so no modification. • Next node to visit is node “15”. • Node “15” has left child and no right child, so no modification. • From node “20”, traverse the right subtree, which is node “30”. • Node “30” has no left and right subtree, so visit node “13”. • For node “30”, there is no left child and right child, so no modification. • Next node to visit is node “20”. • Node “20” has left child and right child, so no modification. Therefore for all the three traversal node “7” and node “8” will get removed. . c. To write the preorder(),postorder() and inorder() traversal of the trees when the definition of visit() function is given. Given: visit() function is given as: if (p->left == 0) p->left = new IntBSTNode(p->el - 1) In-order traversal: • From root node “10”, traverse the left subtree which is node “5”. • From node “5”, traverse the left subtree, which is node “4”. • From node “4” traverse the left subtree, which is node “0”. • Node “0” has no left child, so visit node “0”. • For node “0”, there is no left child, so create a node with value “-1”. • Assign created node as left child of node “0”. • The next node to visit is node “4”. • Node “4” has left child, therefore no modification. • The next node to visit is node “5”. • Node “5” has left child, therefore no modification. • From node “5”, traverse the right subtree, which is node “6”. • Since node “6” has no left child, visit node “6”. • Node “6” has no left child, so create a node with value “5”. • Assign created node as left child of node “6”. • From node “6”, traverse the right subtree, which is node “7”. • Since node “7” has no left child, visit node “7”. • Node “7” has no left child, so create a node with value “6”. • Assign created node as left child of node “7”. • From node “7”, traverse the right subtree, which is node “8”. • Since node “8” has no left child, visit node “8”. • Node “8” has no left child, so create a node with value “7”. • Assign created node as left child of node “8”. • The next node to visit is node “10”. • Node “10” has left child, therefore no modification. • From node “10”, traverse the right subtree, which is node “20”. • From node “20”, traverse the left subtree, which is node “15”. • From node “15”, traverse the left subtree, which is node “13”. • Node “13” has no left child, so visit node “13”. • For node “13”, there is no left child, so create a node with value “12”. • Assign created node as left child of node “13”. • The next node to visit is node “15”. • Node “15” has left child, therefore no modification. • The next node to visit is node “20”. • Node “20” has left child, therefore no modification. • From node “20”, traverse the right subtree, which is node “30”. • Node “30” has no left child, so visit node “30”. • Node “30” has no left child, so create a node with value “29”. • Assign created node as left child of node “30”. Pre-order traversal: • First visit the root node “10”. • Node “10” has left child, therefore no modification. • From node “10”, traverse the left subtree which is node “5”. • Visit node “5”. • Node “5” has left child, therefore no modification. • From node “5”, traverse the left subtree which is node “4”. • Visit node “4”. • Node “4” has left child, therefore no modification. • From node “4”, traverse the left subtree which is node “0”. • Visit node “0”. • For node “0”, there is no left child, so create a node with value “-1”. • Assign created node as left child of node “0”. • From node “0”, traverse the left subtree which is node “-1”. • For node “-1”, there is no left child, so again create a node and it will continue infinitely. Therefore, Pre-order traversal gives an infinite tree. Post-order traversal: • From node “10”, traverse the left subtree, which is node “5”. • From node “5”, traverse the left subtree, which is node “4”. • From node “4”, traverse the left subtree, which is node “0”. • Node “0” has no left and right subtree, so visit node “0”. • For node “0”, there is no left child, so create a node with value “-1”. • Assign created node as left child of node “0”. • Next node to visit is node “4”. • Node “4” has left child, therefore no modification. • From node “5”, traverse the right subtree, which is node “6”. • From node “6”, traverse the right subtree, which is node “7”. • From node “7”, traverse the right subtree, which is node “8”. •Node “8” has no left and right subtree, so visit node “8”. • Node “8” has no left child, so create a node with value “7”. • Assign created node as left child of node “8”. • Next node to visit is node “7”. • Node “7” has no left child, so create a node with value “6”. • Assign created node as left child of node “7”. • Next node to visit is node “6”. • Node “6” has no left child, so create a node with value “5”. • Assign created node as left child of node “6”. • Next node to visit is node “5”. • Node “5” has left child, therefore no modification. • From node “10”, traverse the right subtree, which is node “20”. • From node “20”, traverse the left subtree, which is node “15”. • From node “15”, traverse the left subtree, which is node “13”. • Node “13” has no left and right subtree, so visit node “13”. • For node “13”, there is no left child, so create a node with value “12”. • Assign created node as left child of node “13”. • Next node to visit is node “15”. • Node “15” has left child, therefore no modification. • From node “20”, traverse the right subtree, which is node “30”. • Node “30” has no left and right subtree, so visit node “30”. • For node “30”, there is no left child so create a node with value “29”. • Assign created node as left child of node “30”. • Next node to visit is node “20”. • Node “20” has left child, therefore no modification. Conclusion: Therefore, In the In-order and Post-order traversal operation, each node that is not have a left child then left node will be created. Data value of newly created node is “parent node value -1”, whereas, in case of preorder traversal, an infinite tree is generated. d. To write the preorder(),postorder() and inorder() traversal of the trees when the definition of visit() function is given. Given: visit() function is given as: { tmp = p->right; p->right = p->left; p->left = tmp; } In-order traversal: • From root node “10”, traverse the left subtree which is node “5”. • From node “5”, traverse the left subtree, which is node “4”. • From node “4” traverse the left subtree, which is node “0”. • Node “0” has no left child, so visit node “0”. • For node “0”, there is no left child and right child, so no modification. • The next node to visit is node “4”. • Node “4” has left child, node “0” and right child null. • On interchange left child of node “4” becomes null and right child becomes node “0”. • The next node to visit is node “5”. • Interchange the left and right subtree of node “5”. • Now the left subtree of node “5” is node “6” and right subtree is node “4”. • From node “5”, traverse the right subtree, which is node “4” since the subtree is interchanged. • Interchange the left and right subtree of node “4”. • Now the left subtree of node “4” is node “0” and right subtree is null. • The next node to visit is node “10”. • Interchange the left and right subtree of node “10”. • Now the left subtree of node “10” is node “20” and right subtree is node “5”. • From node “5”, traverse the left subtree, which is node “6”. • Interchange the left and right subtree of node “6”. • Now the left subtree of node “6” is node “7” and right subtree is null. • The next node to visit is node “5”. • Interchange the left and right subtree of node “5”. • Now the left subtree of node “5” is node “4” and right subtree is node “6”. • From node “5”, traverse the right subtree, which is node “6”. • The next node to visit is node “6”. • Interchange the left and right subtree of node “6”. • Now the left subtree of node “6” is null and right subtree is node “7”. • From node “6”, traverse the right subtree, which is node “7”. • Since node “7” has no left child, visit node “7”. • Interchange the left and right subtree of node “7”. • Now the left subtree of node “7” is null and right subtree is node “8”. Therefore, after In-order traversal, both subtree of root is swapped. Preorder traversal: • First visit the root node “10”. • Interchange the left and right subtree of node “10”. • Now the left subtree of node “10” is “20” and right subtree is node “5”. • The next node to visit is node “20”. • Interchange the left and right subtree of node “20”. • Now the left subtree of node “20” is “30” and right subtree is node “15”. • The next node to visit is node “30”. • For node “30”, there is no left child and right child, so no modification. • From node “20”, traverse the right subtree which is node “15”. • Interchange the left and right subtree of node “15”. •Now the left subtree of node “15” is null and right subtree is node “13”. • From node “20”, traverse the right subtree which is node “13”. • For node “13”, there is no left child and right child, so no modification. • The next node to visit is node “5”. • Interchange the left and right subtree of node “5”. • Now the left subtree of node “5” is “6” and right subtree is node “4”. • The next node to visit is node “6”. • Interchange the left and right subtree of node “6”. • Now the left subtree of node “6” is “7” and right subtree is null. • The next node to visit is node “7”. • Interchange the left and right subtree of node “7”. • Now the left subtree of node “7” is “8” and right subtree is null. • The next node to visit is node “8”. • For node “8”, there is no left child and right child, so no modification. • The next node to visit is node “4”. • Interchange the left and right subtree of node “4”. • Now the left subtree of node “4” is null and right subtree is “0”. Therefore, pre-order traversal results in mirror image of the tree. Post-order traversal: • From node “10”, traverse the left subtree, which is node “5”. • From node “5”, traverse the left subtree, which is node “4”. • From node “4”, traverse the left subtree, which is node “0”. • Node “0” has no left and right subtree, so visit node “0”. • For node “0”, there is no left and right child, so no modification. • From node “5”, traverse the right subtree, which is node “6”. • From node “6”, traverse the right subtree, which is node “7”. • From node “7”, traverse the right subtree, which is node “8”. • Node “8” has no left and right subtree, so visit node “0”. • For node “8”, there is no left and right child, so no modification. • The next node to visit is node “7”. • Interchange the left and right subtree of node “7”. • Now the left subtree of node “7” is “8” and right subtree is null. • The next node to visit is node “6”. • Interchange the left and right subtree of node “6”. • Now the left subtree of node “6” is “7” and right subtree is null. • The next node to visit is node “5”. • Interchange the left and right subtree of node “5”. • Now the left subtree of node “5” is “6” and right subtree is “4”. • From node “10”, traverse the right subtree, which is node “20”. • From node “20”, traverse the left subtree, which is node “15”. • From node “15”, traverse the left subtree, which is node “13”. • For node “13”, there is no left and right child, so no modification. • The next node to visit is node “15”. • Interchange the left and right subtree of node “15”. • Now the left subtree of node “15” is null and right subtree is “13”. • The next node to visit is node “30”. • For node “30”, there is no left and right child, so no modification. • The next node to visit is node “20”. • Interchange the left and right subtree of node “20”. • Now the left subtree of node “20” is “30” and right subtree is “15”. • The next node to visit is node “10”. • Interchange the left and right subtree of node “10”. • Now the left subtree of node “10” is “5” and right subtree is “20”. Therefore, post-order traversal results in mirror image of the tree.

EBK DATA STRUCTURES AND ALGORITHMS IN C

4th Edition

ISBN: 9781285415017

Author: DROZDEK

Publisher: YUZU

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Chapter 6, Problem 4E

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