Picture above is Figure 188a predecessorx k x points to a no

(Picture above is Figure 18.8(a))

predecessor(x, k): x points to a node containing a key k. The function returns the predecessor key of k; it returns Nil if the predecessor of k does not exist in the tree. For example, in the B-tree in Figure 18.8 (a), the predecessor of P is O and the predecessor of Q is P.

Note that the predecessor of k is interpreted as the key immediately preceding k in the implicit ordering in the tree. So the predecessor of k may be equal to k if the tree has multiple occurrences of k. In case the predecessor of k is interpreted as the largest key less than k, the above algorithm needs to be modified to skip the multiple occurrences of k, please resolve this

Solution

predecessor(x, k) { let k = x.keyi; if ( !x.leaf ){ // x is not a leaf s: let temp= largestKey(x.ci); // return the largest key in the subtree pointed to by ci if(temp<=x){ x=x.ci; goto s; } else return temp; else // x is a leaf { if ( i 2 ) return x.keyi1 else predecessorInParent(x, x.parent); // attempt to find the predecessor in the parent of x } } Here the modified code in the if block will take care if the predecessor is less than or equal then it will call back the function by moving to right...