Week4_DAT305.docx DAT/305: Data Structures For Problem Solving April 20th, 2020 1.18 <

week4_DAT305.docx  DAT/305: Data Structures For Problem Solving  April 20th, 2020  1.18 < 20, so current node becomes 12. 18 > 12, and 12 has no right child, so 12's right child is assigned with the new node 18.  2.11 < 20, so current node becomes 12. 11 < 12, so current node becomes the existing node 11. 11 < 11 is false, so the algorithm's else branch executes, causing insertion as the right child of the existing node 11.  3.The tree has 3 levels. Iteration 1 visits the first level. Iteration 2 visits the second level. Iteration 3 visits the third level, whose node has no children, and inserts the new node as that node's child.  4.The tree has 8 levels. Each iteration visits a level, descending left or right depending on the new and current nodes' keys. A main benefit of a BST is that inserts require only O(logN) iterations to find the proper insert location in a nearly-full N-node tree.  1.The current node is updated to node 13. The insertion algorithm continues to search for an insertion location.   2.The new node is inserted as the right child because 18 > 12 and cur⇢right is null.  3.The new node becomes the tree's root.  4.The new node is inserted as the left child because 53 < 76 and cur⇢left is null.   5.The current node is updated to node 500. The insertion algorithm continues to search for an insertion location.   6  1.The tree is not empty so the algorithm first initializes cur = tree⇢root, or 7.  2.cur = cur⇢right, or 64, because the key is greater than the current node's k