# New PDF release: A collection of Tree Programming Interview Questions Solved

By Dr Antonio Gulli

ISBN-10: 1499749007

ISBN-13: 9781499749007

Programming interviews in C++ approximately bushes

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Additional info for A collection of Tree Programming Interview Questions Solved in C++

Example text

7 Checking if two binary trees are structurally identical Solution A solution can be provided by recursion. We consider as base cases when both nodes are null, or if only one of them is null. Otherwise the algorithm checks if both nodes contain the same value and if both left children and right children are recursively structurally identical. r2) return false; return (r1->v__ == r2->v__ && structIdentical(r1->left, r2->left) && structIdentical(r1->right, r2->right)); } Complexity Time complexity is and space complexity is .

Root) return; else std::cerr << " v==" << root->v__ << " "; if (root->left) { std::cerr << " ->left "; dump(root->left); } if (root->right) { std::cerr << " ->right "; dump(root->right); } } } Complexity This method has complexity , where and are the numbers of nodes in the two BSTs. 25 Finding out if there is a triplet in a Balanced BST that adds to zero Solution The optimal solution is to convert the BST into a double linked list. Then it is possible to iterate through the list and for each negative element we try to find a pair in the list,which sum is equal to the key of the current node multiplied by -1.

Push(tmp->right); } } Complexity Time complexity is and space complexity is . 6 Counting the number of leaves in a tree Solution A solution can be provided by modifying the level order visit where we increment a counter every time we reach a leaf node. push(tmp->right); } } return count; } Complexity Time complexity is and space complexity is . 7 Checking if two binary trees are structurally identical Solution A solution can be provided by recursion. We consider as base cases when both nodes are null, or if only one of them is null.