# Download A collection of Tree Programming Interview Questions Solved by Dr Antonio Gulli PDF By Dr Antonio Gulli

Programming interviews in C++ approximately bushes

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Extra resources for A collection of Tree Programming Interview Questions Solved in C++

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Deleting the key would leave keys to the internal node but joining the neighbor would add keys plus one more key coming from the neighbor's parent. The total is still keys. The number of branches from a node is one more than the number of keys stored in that node. A B-tree is kept balanced by requiring that all leaf-nodes are at the same depth. More formally a B-tree of order is defined as a tree which satisfies the following properties: Every node has at most children. Every non-leaf node (except for the root) has at least children.

The search within a node can be linear or binary. Insertion To insert a new key, search the tree to find the leaf node, where the new key should be added: If the node contains fewer than, then there is room for the new element, which can be inserted by following the order. Otherwise if the node is full and it must be split into two nodes: The median element is selected among the leaf's elements and the new element. Values less than the median one are put in the new left node and values greater than the median one are put in the new right node The median acts as a separation value.

If the node is the root, we need to create a new root above the old one. If the splitting goes all the way up to the root, it creates a new root with a single separator key and two children. When a node is split, one key goes to the parent, but one additional key is added. Deletion The idea is to find and delete the item, then restructure the tree to regain its invariants. This operation is left as an exercise. l->leaf__) { for (unsigned j = 0; j < t__ ; j++) m->children__[j] = l->children__[j + t__]; } // one key less l->n__ = t__ - 1; // make room for (int j = n__ ; j >= i+1; j--) children__[j + 1] = children__[j]; // connect the new node children__[i + 1] = m; // make room for (int j = n__ - 1; j >= i; j--) keys__[j + 1] = keys__[j]; //copy the middle key keys__[i] = l->keys__[t__ - 1]; n__++; } friend class Btree; }; class Btree { BtreeNode * root; unsigned t__; public: Btree(unsigned t) : root(NULL), t__(t) {}; void traverse() { if(root) root->traverse(); } BtreeLink search(tKey k) { return(root ?