By Dr Antonio Gulli
Programming interviews in C++ approximately timber
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In a loop: First the left children are pushed until a leaf is reached Then if the stack is empty, we leave the loop. Otherwise the top of the stack pops out and the current node is visited. pop(); std::cout << " v=" << root->v__; //right root = root->right; } } Complexity Time complexity is and space complexity is 4 Implementing a post-order visit for a Binary Tree Solution During post-ordering every single node is visited twice: the first time when moving towards the left children and then again, when moving towards the right children.
The right child of a node can be used as forward pointer of the double linked list and the left child as backward pointer. A special case happens at the beginning, when the tail pointer of the double linked list is not defined as detailed in the code. root) return; if (root->left) bstToDoubleList(root->left, head, tail); root->left = *tail; if (*tail) (*tail)->right = root; else *head = root; *tail = root; if (root->right) bstToDoubleList(root->right, head, tail); }; Complexity Time complexity is) , where is the number of nodes in the list.
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.
A collection of Tree Programming Interview Questions Solved in C++ by Dr Antonio Gulli