
Answer:


Answer:

Questions:

/* ======================================================== Delete( x, r_{x} ) from a node N in B^{+}tree ======================================================== */ Delete( x, r_{x}, N ) { Delete x, r_{x} from node N; /* ================================================== Take care of the special case: N = root ================================================== */ if ( N == root ) { /* ================================================================= The root node does not need be have ≥ ⌊(n+1)/2⌋ pointers !!! ================================================================= */ if ( N has ≥ 2 pointers ) { return; } else { // Root node is empty ===> make left most child of root the new root root = root.leftMostPointer; return; } } // ALternative solution (shorter  but less clear) // // if ( N == root ) // { // if ( N is empty (or has 0 keys) ) // { // root = root.leftMostPointer; // make leftMost child the new root // // return; // } // } // Continue here to handle nonroot internal node deletion..... /* ==================================== Check for underflow condition... ==================================== */ if ( N has ≥ ⌊(n+1)/2⌋ pointers /* At least half full*/ ) { return; // Done } else { /*  N underflowed: fix the size of N with transfer or merge  */ /* ======================================================== Always try transfer first !!! (Merge is only possible if 2 nodes are half filled) ======================================================== */ if ( leftSibling(N) has ≥ ⌊(n+1)/2⌋ + 1 pointers ) { 1. transfer last key from leftSibling(N) through parent into N as the first key; 2. transfer right subtree link into N as the first link } else if ( rightSibling(N) has ≥ ⌊(n+1)/2⌋ + 1 pointers ) { 1. transfer first key from rightSibling(N) through parent into N as the last key; 2. transfer left subtree link into N as the last link } /* ========================================================== Here: can't solve underflow with a transfer Because: BOTH sibling nodes have minimum # keys/pointers (= half filled !!!) Solution: merge the 2 half filled nodes into 1 node ========================================================== */ else if ( leftSibling(N) exists ) { /* =============================================== merge N with left sibling node =============================================== */ 1. Merge (1) leftSibling(N) + (2) key in parent node + (3) N into the leftSibling(N) node; 2. Delete ( transfered key, right subtree ptr, parent(N) ); // Recurse !! } else // Node N must have a right sibling node !!! { /* =============================================== merge N with right sibling node =============================================== */ 1. Merge (1) N + (2) key in parent node + (3) rightSibling(N) into the node N; 2. Delete ( transfered key, right subtree ptr, parent(N) ); // Recurse !! } } 

Questions:
