Data Structures and Algorithm Analysis - Computer Science at ...

Sec. 5.2 Binary Tree Traversals 153205040 7520 to 40Figure 5.6 To be a binary search tree, the left child of the node with value 40must have a value between 20 and 40.Another problem that occurs when recursively processing data collections iscontrolling which members of the collection will be visited. For example, sometree “traversals” might in fact visit only some tree nodes, while avoiding processingof others. Exercise 5.20 must solve exactly this problem in the context of a binarysearch tree. It must visit only those children of a given node that might possiblyfall within a given range of values. Fortunately, it requires only a simple localcalculation to determine which child(ren) to visit.A more difficult situation is illustrated by the following problem. Given anarbitrary binary tree we wish to determine if, for every node A, are all nodes in A’sleft subtree less than the value of A, and are all nodes in A’s right subtree greaterthan the value of A? (This happens to be the definition for a binary search tree,described in Section 5.4.) Unfortunately, to make this decision we need to knowsome context that is not available just by looking at the node’s parent or children.As shown by Figure 5.6, it is not enough to verify that A’s left child has a valueless than that of A, and that A’s right child has a greater value. Nor is it enough toverify that A has a value consistent with that of its parent. In fact, we need to knowinformation about what range of values is legal for a given node. That informationmight come from any of the node’s ancestors. Thus, relevant range informationmust be passed down the tree. We can implement this function as follows.boolean checkBST(BinNode rt,int low, int high) {if (rt == null) return true; // Empty subtreeint rootkey = rt.element();if ((rootkey < low) || (rootkey > high))return false; // Out of rangeif (!checkBST(rt.left(), low, rootkey))return false; // Left side failedreturn checkBST(rt.right(), rootkey, high);}