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

By Dr Antonio Gulli

ISBN-10: 1499749007

ISBN-13: 9781499749007

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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.

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. However if the tree is not balanced, complexity is). 15 Print a BST in order Solution A BST is a binary tree, so the in-order visit starts with the left subtree, and then it continues with the current node and finally the right subtree. node) return; printInorder(node->left); std::cout << " n=" << node->v__; printInorder(node->right); } Complexity Time complexity is), where is the number of nodes in the tree.

Tmp) p = p->next; // break first half p->next = NULL; // start next half q = tmp->next; tmp->next = NULL; // recurisve tmp->prex = doubleListToBST(head); tmp->next = doubleListToBST(q); return tmp; } Complexity Time complexity is) , where is the number of nodes in the list. 21 Converting a BST into a Double Linked List Solution Converting a BST into a double linked list can be easily achieved recursively. The right child of a node can be used as forward pointer of the double linked list and the left child as backward pointer.

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A collection of Tree Programming Interview Questions Solved in C++ by Dr Antonio Gulli

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