2019春季考試第4題 7-4 Structure of a Binary Tree （30 分）
阿新 • • 發佈：2019-03-04
img node freopen root 遇到 題目 not 單詞 n) 
題外話：考試的時候花了一個小時做了27分，由於Siblings這個單詞不知道意思，所以剩下的3分就沒去糾結了，後來發現單詞是兄弟的意思，氣哭~~
這道題的麻煩之處在於如何從一個字符串中去找數字。先首先我們需要整行讀取（包括空格），這個我自己寫另一個函數，挺好用的：
string mygets() { char str[100]; fgets(str,sizeof(str),stdin); int len = strlen(str); if(str[len-1]==‘\n‘) str[len-1] = ‘\0‘; return str; }
用的時候只要 string str = mygets();就可以了，裏面已經去掉換行符了。
下一步是再寫一個從istart位置讀取數字的函數，先從istart位置讀取字符串（遇到空格，讀取結束），然後用atoi把字符串轉換成數字。
//字符串中istart位置開始取某個單詞（以空格結束） int getNum(string str,int istart) { int len = str.length(); string strA=""; for(int i=istart;i<len;++i) if(str[i]==‘ ‘) break; else strA += str[i]; //printf("%s\n",strA.c_str()); int a = atoi(strA.c_str()); return a; }
題目中的字符串格式太多了是吧，眼花繚亂：
15 is the root 8 and 2 are siblings 32 is the parent of 11 23 is the left child of 16 28 is the right child of 2 7 and 11 are on the same level It is a full tree拿字符串siblings舉例吧，想用find函數siblings字符串，如果返回不等於-1，表示就是此種類型，然後用以下代碼獲取數字8和2。註意find函數的用法，返回的是查找字符串的首地址，所以find "of"返回的是o出現的首地址，沒有找到of則返回-1，否則返回o的首地址。然後我們可以想象一下istart+1是f，istart+2是空格，istart+3就是我們所要的數字出現的首地址。
if(str.find("siblings")!=-1) { int a = getNum(str,0); int b = getNum(str,str.find("of")+3); }
以下是我的代碼：
PS：如果我的構樹代碼看不懂，可以看我寫的這篇文章讓你懂得已知先序（後序）與中序遍歷如何快速構樹：https://www.cnblogs.com/jlyg/p/10402919.html
樹的結構體中我加了level變量，記錄當前節點的水平深度。其他看代碼吧~~
#include<iostream> #include<cstdio> #include<cstdlib> #include<vector> #include<queue> #include<map> #include<set> #include<algorithm> #include<cstring> using namespace std; struct Node{ int val; int level; Node* left; Node* right; Node(){ left = right = NULL; level = 0; } }; map<int,int> valtoid; //通過值去找id map<int,Node*> valtonode; //通過值去找節點 //構造數 Node* Insert(Node* node,int val) { if(node==NULL) { node = new Node(); node->val= val; valtonode[val] = node; } else if(valtoid[val] < valtoid[node->val]) { node->left = Insert(node->left,val); node->left->level = node->level+1; } else { node->right= Insert(node->right,val); node->right->level = node->level+1; } return node; } Node* root = NULL; int n; //是不是滿二叉樹 bool isFullTree(Node* root) { queue<Node*> q; q.push(root); Node* node; while(!q.empty()) { node = q.front(); q.pop(); if(node->left) q.push(node->left); if(node->right) q.push(node->right); if(node->left&&!node->right || !node->left&&node->right) return false; } return true; } //字符串中istart位置開始取某個單詞（以空格結束） int getNum(string str,int istart) { int len = str.length(); string strA=""; for(int i=istart;i<len;++i) if(str[i]==‘ ‘) break; else strA += str[i]; //printf("%s\n",strA.c_str()); int a = atoi(strA.c_str()); return a; } //是不是兄弟節點 bool isSiblings(Node* node,Node* child1,Node* child2) { if(node==NULL) return false; if(node->left==child1&&node->right!=child2|| node->left==child2&&node->right!=child1) return false; else if(node->left==child1&&node->right==child2|| node->left==child2&&node->right==child1) return true; return isSiblings(node->left,child1,child2)||isSiblings(node->right,child1,child2); } //判斷這一個這句話是不是對的 bool isRight(string str) { if(str.find("root")!=-1) { int a = getNum(str,0); return a == root->val; } else if(str.find("siblings")!=-1) { int a = getNum(str,0); int b = getNum(str,str.find("of")+3); //兄弟節點，必須在同一層 if(valtonode[b]&&valtonode[a]&&valtonode[b]->level==valtonode[a]->level) { if(isSiblings(root,valtonode[a],valtonode[b])) return true; } return false; } else if(str.find("parent")!=-1) { int a = getNum(str,0); int b = getNum(str,str.find("of")+3); if(valtonode[b]&&valtonode[a]) { if(valtonode[a]->left == valtonode[b] || valtonode[a]->right == valtonode[b]) return true; } return false; } else if(str.find("left")!=-1) { int a = getNum(str,0); int b = getNum(str,str.find("of")+3); if(valtonode[b]&&valtonode[a]) { if(valtonode[b]->left==valtonode[a]) return true; } return false; } else if(str.find("right")!=-1) { int a = getNum(str,0); int b = getNum(str,str.find("of")+3); if(valtonode[b]&&valtonode[a]) { if(valtonode[b]->right==valtonode[a]) return true; } } else if(str.find("same")!=-1) { int a = getNum(str,0); int b = getNum(str,str.find("and")+4); if(valtonode[b]&&valtonode[a]) { if(valtonode[b]->level==valtonode[a]->level) return true; } return false; } else if(str.find("full")!=-1) { return isFullTree(root); } return false; } string mygets() { char str[100]; fgets(str,sizeof(str),stdin); int len = strlen(str); if(str[len-1]==‘\n‘) str[len-1] = ‘\0‘; return str; } int main() { #ifndef ONLINE_JUDGE freopen("1.txt","r",stdin); #endif scanf("%d",&n); int post[n]; for(int i=0;i<n;++i) scanf("%d",&post[i]); for(int i=0;i<n;++i) { int a; scanf("%d",&a); valtoid[a] = i; } for(int i=n-1;i>=0;--i) root = Insert(root,post[i]); int m; scanf("%d",&m); getchar(); for(int i=0;i<m;++i) { string str = mygets(); bool bRight = isRight(str); printf("%s\n",bRight?"Yes":"No"); } return 0; }View Code
題目在下面：
7-4 Structure of a Binary Tree （30 分） Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, a binary tree can be uniquely determined. Now given a sequence of statements about the structure of the resulting tree, you are supposed to tell if they are correct or not. A statment is one of the following: A is the root A and B are siblings A is the parent of B A is the left child of B A is the right child of B A and B are on the same level It is a full tree Note: Two nodes are on the same level, means that they have the same depth. A full binary tree is a tree in which every node other than the leaves has two children. Input Specification: Each input file contains one test case. For each case, the first line gives a positive integer N (≤30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are no more than 10 ?3 ?? and are separated by a space. Then another positive integer M (≤30) is given, followed by M lines of statements. It is guaranteed that both A and B in the statements are in the tree.
Output Specification:
For each statement, print in a line Yes if it is correct, or No if not.
Sample Input: 9 16 7 11 32 28 2 23 8 15 16 23 7 32 11 2 28 15 8 7 15 is the root 8 and 2 are siblings 32 is the parent of 11 23 is the left child of 16 28 is the right child of 2 7 and 11 are on the same level It is a full tree Sample Output: Yes No Yes No Yes Yes Yes
2019春季考試第4題 7-4 Structure of a Binary Tree （30 分）