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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 timber

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If the node is external and there is one chil (either on the left or on right side) then the child replaces the father. In any case the external node is freed. root) return NULL; else if (key < root->v__) root->left = deleteNode(root->left, key); else if (key > root->v__) root->right = deleteNode(root->right, key); else { // internal node if (root->left && root->right) { tmp = findMax(root->left); root->v__ = tmp->v__; root->left = deleteNode(root->left, root->v__); } else { tmp = root; if (root->left == NULL) root = root->right; if (root->right == NULL) root = root->left; delete tmp; } } return root; } Complexity Average time complexity is) , where is the number of nodes in the tree.

Root) return 0; int lLPSum = maxPathSum(root->left, res); int rLPSum = maxPathSum(root->right, res); int curr_sum = max((lLPSum + rLPSum + root->v__), max(lLPSum, rLPSum)); if (res < curr_sum) res = curr_sum; return max(lLPSum, rLPSum) + root->v__; } Complexity Time complexity is and space complexity is . 11 Reconstructing a tree given its pre-order and in-order traversals Solution In a preorder visit the root is always the leftmost element of each subtree. So we start to put the leftmost element of the preorder array of visits as the root of the whole tree.

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.

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