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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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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 ?

Here below we propose a recursive version and we leave to the interested reader the task of implementing an iterative version on his own. Insert() follows a similar approach. The candidate key for insertion is searched inside the tree. If it is found, the process returns. Otherwise if the search reaches a null node, the key is inserted in that position. The process can be either recursive or iterative. root) return NULL; return ((root->v__ < k) ? find(root->right, k) : (root->v__ > k ? root) { root = new binaryTree(k); return; } if (root->v__ < k) insert(root->right, k); else if (root->v__ > k) insert(root->left, k); } Complexity Average time complexity of “insert” and “find” is) , where expresses the number of nodes in the tree.

Otherwise the top of the stack is popped out and the visit continues towards the right children. root) return 0; left = diameter(root->left, dim); right = diameter(root->right, dim); if (left + right > dim) dim = left + right; return ((left > right) ? left : right) + 1; } Complexity Time complexity is and space complexity is . 3 Implementing an in-order visit for a Binary Tree Solution A non-recursive in-order visit can be implemented by using a stack. In a loop: First the left children are pushed until a leaf is reached Then if the stack is empty, we leave the loop.

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