Introduction to Algorithms
3rd Edition
ISBN: 9780262033848
Author: Thomas H. Cormen, Ronald L. Rivest, Charles E. Leiserson, Clifford Stein
Publisher: MIT Press
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Chapter 10.4, Problem 6E
Program Plan Intro
To show that the only two-pointers and one Boolean value in each node, so that the parent of a node or all of its children can be reached and identified in linear time.
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Create a binary linked tree, and traverse the tree by using the recursive function.
The structure of the tree is as follow: //PICTURE//
You should input the nodes in pre-order sequence. If a child of a node is NULL, input a space.
Write the function of create binary tree, pre-order to print the nodes, in-order to print the nodes and post-order to print the nodes. Also Count the height of the tree.
PLEASE USE C LANGUAGE
A binomial tree is a special type of rooted tree, defined recursively as follows:
Basis Step: The Oth binomial tree, Bo, is a single vertex.
Recursive Step: For k > 1, the kth binomial tree, B, consists of two copies of B-1 and an
additional edge making the root of one copy the child of the root of the other.
a) Draw B2
b) Find a (non-recursive) formula for the number of vertices in B, and prove your formula
correct.
c) Prove that there are C(k, l) vertices at level l of Be
(You may find Pascal's identity useful: C(n + 1, k) = C(n, k – 1) + C(n, k))
4. Write a recursive algorithm in pseudocode that finds the lowest common ancestor (LCA) of two
given nodes in a binary tree T. The lowest common ancestor is defined between two
nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node
to be a descendant of itself). If either p or q is null, the LCA is null. For this problem, Nodes have
left, right, and parent references as well as a field called level which stores the level of the
node in the tree. In the sample tree below, node 5 is on level 0, while nodes 4 and 6 are on level 1.
Write your solution here.
5 <- level 0
4 6 < level 1
function lowest_common_ancestor (Node P,
Node q)
Chapter 10 Solutions
Introduction to Algorithms
Ch. 10.1 - Prob. 1ECh. 10.1 - Prob. 2ECh. 10.1 - Prob. 3ECh. 10.1 - Prob. 4ECh. 10.1 - Prob. 5ECh. 10.1 - Prob. 6ECh. 10.1 - Prob. 7ECh. 10.2 - Prob. 1ECh. 10.2 - Prob. 2ECh. 10.2 - Prob. 3E
Ch. 10.2 - Prob. 4ECh. 10.2 - Prob. 5ECh. 10.2 - Prob. 6ECh. 10.2 - Prob. 7ECh. 10.2 - Prob. 8ECh. 10.3 - Prob. 1ECh. 10.3 - Prob. 2ECh. 10.3 - Prob. 3ECh. 10.3 - Prob. 4ECh. 10.3 - Prob. 5ECh. 10.4 - Prob. 1ECh. 10.4 - Prob. 2ECh. 10.4 - Prob. 3ECh. 10.4 - Prob. 4ECh. 10.4 - Prob. 5ECh. 10.4 - Prob. 6ECh. 10 - Prob. 1PCh. 10 - Prob. 2PCh. 10 - Prob. 3P
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Similar questions
- i. ii. Using an array-based representation of a binary tree, the following array represents a binary tree. 0 1 2 3 59 39 75 4 48 5 6 70 80 111. Draw the binary tree that is represented by the above array. 7 8 9 10 11 12 13 14 44 68 90 What is the postorder traversal of the binary tree that is created in (i). What is the parent of 72?arrow_forwardConsider a tree that has a relatively high (between 10 to 30) typical number of children of each node. Under what conditions would a static child pointer array implementation be usable, and under what conditions would it be a better choice? Under what conditions would a dynamic child pointer array implementation be usable, and under what conditions would it be a better choice?arrow_forwardA tree that is rooted and has a maximum of two children per node is commonly known as a binary tree. The purpose of this demonstration is to establish that the quantity of nodes present in a binary tree that are accountable for generating two children is precisely one unit less than the quantity of leaves in the same tree.arrow_forward
- illustrates a recursive pseudocode description of theinsert operation on a k-d tree. Here, x is the key to be inserted into the k-d tree, T isthe pointer to the k-d tree and DISC is the discriminator.arrow_forwardGiven a binary tree, let an H-node be defined as a non-leaf node in the tree whose value is greater than or equal to its children nodes (1 or 2 children). Write a function called countHNodes(TreeNode* p), that returns the number of H-nodes in a binary tree (pointed by p). Use recursion in the function.arrow_forwardWrite a recursive function treeToList(Node root) that takesan ordered binary tree and rearranges the internal pointers to make a circular doubly linked list outof the tree nodes. The "previous" pointers should be stored in the "small" field and the "next"pointers should be stored in the "large" field. The list should be arranged so that the nodes are inincreasing order. Return the head pointer to the new list. The operation can be done in O(n) time -- essentially operating on each node once. Try the problem directly, or see the hints below.Hint #1 The recursion is key. Trust that the recursive call on each sub-tree works and concentrate onassembling the outputs of the recursive calls to build the result. It's too complex to delve into howeach recursive call is going to work -- trust that it did work and assemble the answer from there.Hint #2The recursion will go down the tree, recursively changing the small and large sub-trees into lists,and then append those lists together with the…arrow_forward
- Given a binary tree, let an H-node be defined as a non-leaf node in the tree whose value is greater than or equal to its children nodes (1 or 2 children). Write a function, countHNodes(), that returns the number of H-nodes in a binary tree (pointed by p) using recursion.arrow_forwardCreate a binary linked tree, and traverse the tree by using the recursive function. The structure of the tree is as follows: //check pic// You should input the nodes in pre-order sequence. If a child of a node is NULL, input a space. Write the function of create binary tree, pre-order to print the nodes, in-order to print the nodes and post-order to print the nodes. Count the height of the tree. Hints: Header file typedef char ElemType; typedef struct node//define the type of binary tree node { }BTnode; Source file #include <stdio.h> #include <stdlib.h> #include "tree.h" BTnode * createTree()//create the binary tree，return the root { BTnode *tnode;// tnode is the root char elem; ;//input the character //if the input is a space，set the pointer as NULL Else// if the input is not a space，generate the binary node and create its left…arrow_forwardGiven a Binary Tree (BT), convert it to a Doubly Linked List(DLL) in-place. The left and right pointers in nodes are to be used as previous and next pointers respectively in converted DLL. The order of nodes in DLL must be the same as in Inorder for the given Binary Tree. The first node of Inorder traversal (leftmost node in BT) must be the head node of the DLL.arrow_forward
- Suppose the nodes of a binary tree are unique and that you are given the order of elements as they are encountered in a preorder traversal and the order of the elements as they are encountered in a postorder traversal. Under what conditions can you accurately reconstruct the structure of the tree from these two traversal orders?arrow_forwardA binary tree is a kind of rooted tree that does not have more than two offspring per node at any point in its structure. Show that the number of nodes in a binary tree that are responsible for creating two offspring is precisely one fewer than the number of leaves in that tree. Demonstrate this by subtracting one from the total number of leaves.arrow_forwardThe BST remove algorithm traverses the tree from the root to find the node to remove. When the node being removed has 2 children, the node's successor is found and a recursive call is made. One node is visited per level, and in the worst-case scenario, the tree is traversed twice from the root to a leaf. A BST with N nodes has at least log2N levels and at most N levels. Therefore, the runtime complexity of removal is best case O(logN) and worst case O(N). Two pointers are used to traverse the tree during removal. When the node being removed has 2 children, a third pointer and a copy of one node's data are also used, and one recursive call is made. Thus, the space complexity of removal is always O(1)." I have to explain this clearly! and the advantages of the BST algorithimarrow_forward
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