sas forks

#include #include using namespace STD;intf[ -][ -];inttree[ -][ -];intnode[ -][2];voidInit () {memset(Tree,0,sizeof(tree));memset(F,0,sizeof(f));intSolveintLintR) {if(F[l][r])returnF[L][R]; for(inti = l; I intx = Solve (l, I-1);inty = Solve (i +1, R);if(x = =0) x =1;if(Y = =0) y =1;if(x * y + f[i][i] > F[l][r]) {F[l][r] = x*y + f[i][i]; Tree[l][r] = i; } }returnF[L][R];}voidPrintintNumintLintR) {if(Tree[l][num-1]) {printf("%d", Tree[l][num-1]); Print (tree[l][num-1], L,

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.ExampleGiven 4 points:,,, (1,2) (3,6) (0,0) (1,3) .The maximum number is 3 .Leecode on the original topic, see my previous blog Invert binary tree flipping two fork trees.Solution One://recursionclassSolution { Public: /** * @param root:a TreeNode, the root of the binary tree * @return: Nothing*/ voidInvertbinarytree (TreeNode *root) { if(!root)return; TreeNode*tmp = root->Left ; Root-

to a single string. stringSerialize (treenode*root) {Ostringstream out; QueueQ; if(Root) q.push (root); while(!Q.empty ()) {TreeNode*t =Q.front (); Q.pop (); if(t) { out' '; Q.push (t-Left ); Q.push (t-Right ); } Else { out"# "; } } return out. STR (); } //decodes your encoded data to tree.Treenode* Deserialize (stringdata) { if(Data.empty ())returnnullptr; Istringstreaminch(data); QueueQ; stringVal; inch>>Val; TreeNode*res =NewTreeNode (Stoi (Val)),

Treenode:def __init__ (self, x): Self.val = x self.left = None self.right = Noneclass Soluti On: # @param preorder, a list of integers # @param inorder, a list of integers # @return a tree node def build Tree (self, Preorder, inorder): Lookup = {} for I, Num in enumerate (inorder): lookup[num] = i Return Self.buildtreerecu (lookup, preorder, inorder, 0, 0, Len (inorder)) def buildtreerecu (self, lookup, preorder, I Norder, Pre_start, In_start, in_end): if In_start = = In_end:return None

portion is divided into two parts, the first part is smaller than their own (left subtree part),The second part is bigger than yourself (right subtree part), so applying this relationship can be used to check whether a two-fork search tree is being iterated. ”The code is as follows 1#include 2#include 3#include 4#include 5#include 6 #defineMAX 100027 8 using namespacestd;9 intNum[max];Ten One BOOLIshou (intStartintend) { A if(End start) { - return true; - } the intState

1#include 2#include 3#include 4 using namespacestd;5 6 classSolution7 {8 Public:9vectorint> Preordertraversal (TreeNode *root)Ten { Onevectorint>result; A ConstTreeNode *p; -stackConstTreeNode *>s; - thep =Root; - if(P! =nullptr) S.push (p); - while(!s.empty ()) - { +p =s.top (); - S.pop (); +Result.push_back (p->val); A at if(P->right! = nullptr) S.push (p->Right ); - if(P->left! = nullptr) S.push (p->left);//Stack LIFO So first Ju

, then , the pointer moves to the left child node, whereas if the right child node is 1 larger than its parent node, if the left child node is present, the queue is queued, the pointer moves to the right child node, and so on until the queue is empty, see the code below:Solution Four:classSolution { Public: intLongestconsecutive (treenode*root) { if(!root)return 0; intres =0; QueueQ; Q.push (root); while(!Q.empty ()) { intLen =1; TreeNode*t =Q.front (); Q.pop (); wh

Title: Enter a binary search tree to convert the two-fork search tree into a sorted doubly linked list. The space complexity is O (1).The stupidest way to do this is to put the middle sequence traversal in an array or list and iterate through the concatenation again. But requires space complexity of O (1), there is no way to do so, this time can refer to this solution: http://blog.csdn.net/my_jobs/article/details/47666909 judge a tree is not a binary search tree, Can imitate the last method.The

Given A binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can SE E ordered from top to bottom.For example:Given The following binary tree, 1 You should return [1, 3, 4] .Credits:Special thanks to @amrsaqr for adding this problem and creating all test cases.This problem requires us to print out a two-fork tree the rightmost number of each line, in fact, is to find a binary tree sequence traversal of a deformation, we just need to save

, so the queue holds the elements of the current layer./*** Definition of TreeNode: * public class TreeNode {* public int val; * Public TreeNode left, right; * PU Blic TreeNode (int val) {* This.val = val; * This.left = This.right = null; *} *}*/ Public classSolution {/** * @paramroot:the root of binary tree. * @return: Level order A list of lists of integers*/ PublicArraylistLevelorder (TreeNode root) {//Write your code herequeueNewLinkedlist(); ArrayListNewArraylist(); if(Root = =NUL

Topic 1367: Two fork search tree post-order traversal sequence time limit: 1 seconds memory limit: 32 trillion special sentence: No: 1359 resolution: 684 Title Description: Enter an array of integers to determine if the array is the result of a sequential traversal of a binary search tree. If yes, output yes, otherwise output No. Assume that any two digits of the input array are different. Input: Each test case consists of 2 rows: The first behavior is 1 integers n (1#include OJ AddressThe sword

The middle sequence traversal of a binary tree, that is, the left subtree, the root, the right sub-tree1 /**2 * Definition for binary tree3 * struct TreeNode {4 * int val;5 * TreeNode *left;6 * TreeNode *right;7 * TreeNode (int x): Val (x), left (null), right (null) {}8 * };9 */Ten classSolution { One Public: A voidDFS (vectorint> ans,treenode *root) { - if(!root) { - return; the } -DFS (ans,root->Left ); -Ans.push_back (root->val); -DFS (ans,root->Right ); +

- - ]; - } - } in for(i=0; i) - { to if(f[i]->pr==NULL) + returnF[i]; - } the returnNULL; * } $ Panax Notoginseng BOOLCMP (Tree *r1,tree *R2) - { the if(r1==nullr2==NULL) + return true; A Else if(r1==null| | r2==NULL) the return false; + if(r1->c!=r2->c) - return false; $ if(CMP (R1-GT;LT,R2-GT;LT) cmp (r1->rt,r2->rt)) | | (CMP (R1-GT;LT,R2-GT;RT) cmp (r1->rt,r2->LT)) ) $ return true; - return fa

#include "iostream"#include "Memory.h"using namespace Std;/*Find the Root method:Number of inputs per node: The root node is entered once when it enters itself,A non-root node is entered once when the parent node is entered, and then entered when you enterOnce, so I entered it two times. So you can calculate each node through the counterThe number of times it was entered to find the root node.*/const int MAX = 1002;struct node{int id;Char Val;int left, right;Node (char c, int i = 0) {id = i;val

), left (null), right (null) {}}; Bstnode *lltobstcore (ListNode **headaddr,intN) { if(N 0)returnNULL; Bstnode*left = Lltobstcore (headaddr, n/2); Bstnode*root =NewBstnode ((*HEADADDR)val); Root-left =Left ; *headaddr = (*HEADADDR)Next; Root-right = Lltobstcore (headaddr, n-n/2-1); returnRoot;} Bstnode*lltobst (ListNode *head) { if(!head)returnNULL; intn =0; ListNode *p =Head; for(; p; ++n, p = Pnext); returnLltobstcore (head, n);}intMain () {ListNode*head =NewListNode (1); ListNode*end =

Data structure Experiment two forks tree two: traverse the binary tree time limit:1000ms Memory limit:65536kbsubmit Statisticproblem DescriptionA sequence of characters of a binary tree that is known to iterate through the input, such as a abc,,de,g,,f,,, (which represents an empty node). Build a two-fork tree and traverse the two-tree in both the middle and post order ways.InputEnter multiple sets of data consecutively, with each set of data entering

When eating with a knife and fork, fix the left side with the fork, cut a small size with a knife, dipped in the sauce into the mouth. The American way of eating is to cut all the food into small pieces, and then take the right hand and take the fork slowly. However, if you eat juicy meat dishes, the gravy will flow into your dishes, not elegant. Therefore, it is best to take a bite. The basic principle of how to use a knife and a cross is to hold a knife or spoon in the right hand and take a

#include "stdio.h"#include "malloc.h"int count;typedef struct NODE{char data;struct node *lchild;struct node *rchild;}bitnode,*bitree;void Creatbitree (Bitree * BT)//pre-order calendar sequence Create two-tree linked list{Char Ch=getchar ();if (ch== ' # '){*bt=null;}Else{*bt= (bitree) malloc (sizeof (Bitnode));(*BT)->data=ch;Creatbitree ( (*BT)->lchild);Creatbitree ( (*BT)->rchild);}}int Proorder (Bitree root)//pre-order Calendar output binary tree node{if (root!=null){printf ("%c", root->data);

Why do I show cross forks by referencing images? Solution lt; html gt; lt; head gt; lt; title gt; image insertion example lt; title gt; lt; head gt; lt; body gt; lt; img nbsp; srcE: phpwwwrootstaticsimagesv9_logo.jpg nbsp; alt this is the PHP icon. why do I use the picture to show cross forks? Solution Image insertion example ------ Solution -------------------- The image cannot be

binary search tree converts an ordered array to a binary-Sorted, and converts the list to binary searches tree to convert the ordered list to a binary search and Kth smallest elemen T in a BST two forks the small element of the K in the search tree. This problem can be solved by recursion, we first look at the example given in the topic, because the binary search tree is characterized by left Solution One:classSolution { Public: TreeNode* Lowestcommo