binary-tree

I have the implemented a binary search tree but I also want to make it generic. The code is the following: typedef struct treeNode { int data; struct treeNode *left; struct treeNode *right; } treeNode; and the functions: treeNode* FindMin(treeNode *node) { if(node==NULL) { /* There is no element in the tree */ return NULL; } if(node->left) /* Go to the left sub tree to find the min element */ return FindMin(node->left); else return node; } treeNode * Insert(treeNode *node,int data) { if(node==NULL) { treeNode *temp; temp = (treeNode *)malloc(sizeof(treeNode)); temp -> data = data; temp -> left

While i was studying for midterm about binary trees, i found a statement that any arbitrary n-node binary tree can be transformed into any other n-node binary tree with at most 2*n-2 rotations. Is there any proof for that? I found some kind of proof with asymptotic notations but it was not that clear. I mean could someone explain/show why it is true? And if it says that n-node binary tree, does it include the root? This answer is from CLRS 3rd Edition textbook question 13.2-4. Let LEFT = an entire left linked list binary tree RIGHT = an entire right linked list binary tree. You can easily

Given k sorted arrays of integers, each containing an unknown positive number of elements (not necessarily the same number of elements in each array), where the total number of elements in all k arrays is n, give an algorithm for merging the k arrays into a single sorted array, containing all n elements. The algorithm's worst-case time complexity should be O(n∙log k). Name the k-sorted lists 1, ..., k. Let A be the name of the combined sorted array. For each list, i, pop v off of i and push (i, v) into a min-heap. Now the heap will contain pairs of value and list id for the smallest entries in

Given a BST, I am required to find the number of left nodes of the tree. Example: ` +---+ | 3 | +---+ / \ +---+ +---+ | 5 | | 2 | +---+ +---+ / / \ +---+ +---+ +---+ | 1 | | 4 | | 6 | +---+ +---+ +---+ / +---+ | 7 | +---+` The answer should be 4, as (5, 1 , 4, 7) are all left nodes of the tree. What I am thinking of doing is: public int countLeftNodes() { return countLeftNodes(overallRoot, 0); } private int countLeftNodes(IntTreeNode overallRoot, int count) { if (overallRoot != null) { count += countLeftNodes(overallRoot.left, count++); count = countLeftNodes(overallRoot.right, count); }

This is a homework, I have difficulties in thinking of it. Please give me some ideas on recursions and DP solutions. Thanks a lot generate and print all structurally distinct full binary trees with n leaves in dotted parentheses form， "full" means all internal (non-leaf) nodes have exactly two children. For example, there are 5 distinct full binary trees with 4 leaves each. U can use recursion, on i-th step u consider i-th level of tree and u chose which nodes will be present on this level according to constraints: - there is parent on previous level - no single children present (by your