oj 二叉树相关
数据结构实验之二叉树一：树的同构
在没有new的时候，不会自动调用析构函数
1 #include<bits/stdc++.h> 2 using namespace std; 3 struct Node{ 4 char u; 5 int left; 6 int right; 7 Node(){ 8 u = 'A'; 9 left = 11; 10 right = 11; 11 } 12 }node1[20 + 3], node2[20 + 3]; 13 14 bool cal(int i, int j){ 15 if(node1[node1[i].left].u == node2[node2[j].left].u && node1[node1[i].right].u == node2[node2[j].right].u)return true; 16 if(node1[node1[i].left].u == node2[node2[j].right].u && node1[node1[i].right].u == node2[node2[j].left].u)return true; 17 return false; 18 } 19 bool judge(int n, int m){ 20 int i, j; 21 for(i = 0; i < n; i++){ 22 for(j = 0; j < m; j++){ 23 if(node1[i].u == node2[j].u){ 24 if(cal(i, j)){ 25 break; 26 } 27 else { 28 return false; 29 } 30 } 31 } 32 if(j == m){ 33 return false; 34 } 35 } 36 return true; 37 } 38 int main(){ 39 int n, m; 40 while(~scanf("%d", &n)){ 41 char str[10]; 42 int l, r, i; 43 44 for(i = 0; i < n; i++){ 45 scanf("%s", str); 46 node1[i].u = str[0]; 47 scanf("%s", str); 48 if(str[0] != '-'){ 49 node1[i].left = str[0] - '0'; 50 } 51 else node1[i].left = 11; 52 scanf("%s", str); 53 if(str[0] != '-'){ 54 node1[i].right = str[0] - '0'; 55 } 56 else 57 node1[i].right = 11; 58 } 59 60 scanf("%d", &m); 61 for(i = 0; i < m; i++){ 62 scanf("%s", str); 63 node2[i].u = str[0]; 64 scanf("%s", str); 65 if(str[0] != '-'){ 66 node2[i].left = str[0] - '0'; 67 } 68 else node2[i].left = 11; 69 scanf("%s", str); 70 if(str[0] != '-'){ 71 node2[i].right = str[0] - '0'; 72 } 73 else node2[i].right = 11; 74 } 75 if(judge(n, m)) 76 printf("Yes\n"); 77 else 78 printf("No\n"); 79 } 80 return 0; 81 }
数据结构实验之二叉树二：遍历二叉树
1 #include<bits/stdc++.h> 2 using namespace std; 3 struct Node 4 { 5 char u; 6 Node* left; 7 Node* right; 8 Node() 9 { 10 left = NULL; 11 right = NULL; 12 } 13 }; 14 char str[100]; 15 int i; 16 Node *create(int len) 17 { 18 if(i >= len) 19 return NULL; 20 if(str[i] == ',') 21 { 22 i++; 23 return NULL; 24 } 25 Node* tmp = new Node(); 26 tmp->u = str[i++]; 27 tmp->left = create(len); 28 tmp->right = create(len); 29 return tmp; 30 } 31 void ceng(Node* root) 32 { 33 if(root) 34 { 35 ceng(root->left); 36 printf("%c", root->u); 37 ceng(root->right); 38 } 39 } 40 void hou(Node* root) 41 { 42 if(root) 43 { 44 hou(root->left); 45 hou(root->right); 46 printf("%c", root->u); 47 } 48 } 49 int main() 50 { 51 52 while(scanf("%s", str) != EOF) 53 { 54 i = 0; 55 Node* root = new Node(); 56 int len = strlen(str); 57 root = create(len); 58 ceng(root); 59 printf("\n"); 60 hou(root); 61 printf("\n"); 62 } 63 return 0; 64 } 65 66 67 /*************************************************** 68 User name: ACM2017002信科1701黄庆祥 69 Result: Accepted 70 Take time: 0ms 71 Take Memory: 200KB 72 Submit time: 2019-09-27 21:27:21 73 ****************************************************/
前序和中序推层序遍历，中序遍历（层序遍历通过队列实现）（非递归版）
1 #include<bits/stdc++.h> 2 using namespace std; 3 struct Node{ 4 char u; 5 Node* left; 6 Node* right; 7 Node(){ 8 left = NULL; 9 right = NULL; 10 } 11 }; 12 char str1[100], str2[100]; 13 int cnt; 14 Node* build(Node* root, int L, int R) 15 { 16 int i; 17 char t = str1[cnt++]; 18 for(i = L; i < R; i++){ 19 if(str2[i] == t){ 20 break; 21 } 22 } 23 root = new Node(); 24 root->u = t; 25 if(i > L && i < R){ 26 root->left = build(root->left, L, i); 27 } 28 if(i + 1 > L && i + 1 < R){ 29 root->right = build(root->right, i + 1, R); 30 } 31 return root; 32 } 33 Node* hou(Node* root){ 34 if(root){ 35 hou(root->left); 36 hou(root->right); 37 printf("%c",root->u); 38 } 39 } 40 void ceng(Node* root){ 41 queue<Node*>q; 42 if(root) 43 q.push(root); 44 while(!q.empty()){ 45 Node* tmp = new Node(); 46 tmp = q.front(); 47 q.pop(); 48 printf("%c",tmp->u); 49 if(tmp->left) 50 q.push(tmp->left); 51 if(tmp->right) 52 q.push(tmp->right); 53 } 54 return ; 55 } 56 int main(){ 57 int T; 58 scanf("%d", &T); 59 while(T--){ 60 cnt = 0; 61 Node* root = new Node(); 62 scanf("%s", str1); 63 scanf("%s", str2); 64 int len = strlen(str2); 65 root = build(root, 0, len); 66 hou(root); 67 printf("\n"); 68 ceng(root); 69 printf("\n"); 70 } 71 return 0; 72 }
非递归版：
1 Node* build(Node* root, int L, int R) 2 { 3 // cout << str1 << " " << str2 << endl; 4 stack<Node*>q; 5 root = new Node(); 6 root->u = str1[0]; 7 q.push(root); 8 int i, j; 9 for(i = 1, j = 0; i < R; i++){ 10 Node* node = new Node(); 11 node->u = str1[i]; 12 Node *tmp = NULL; 13 while(!q.empty() && q.top()->u == str2[j]){ 14 tmp = q.top(); 15 q.pop(); 16 j++; 17 } 18 if(tmp) 19 tmp->right = node; 20 else 21 q.top()->left = node; 22 q.push(node); 23 } 24 return root; 25 }
知道中序和后序求树
通过后序最后一个来判断
1 #include<bits/stdc++.h> 2 using namespace std; 3 struct Node{ 4 char u; 5 Node* left; 6 Node* right; 7 Node(){ 8 left = NULL; 9 right = NULL; 10 } 11 }; 12 char str1[100], str2[100]; 13 int cnt; 14 int flag = 0; 15 Node* build(Node* root, int l, int r){ 16 if(flag) 17 return NULL; 18 root = new Node(); 19 root->u = str2[--cnt]; 20 int i, k = 0; 21 for(i = l; i < r; i++){ 22 if(str1[i] == root->u){ 23 k = 1; 24 break; 25 } 26 } 27 if(!k) 28 return NULL; 29 if(i + 1 < r) 30 root->right = build(root->right, i + 1, r); 31 if(i > l) 32 root->left = build(root->left, l, i); 33 return root; 34 } 35 void ceng(Node* root){ 36 queue<Node*>q; 37 if(root) 38 q.push(root); 39 while(!q.empty()){ 40 Node* tmp = new Node(); 41 tmp = q.front(); 42 q.pop(); 43 printf("%c",tmp->u); 44 if(tmp->left) 45 q.push(tmp->left); 46 if(tmp->right) 47 q.push(tmp->right); 48 } 49 return ; 50 } 51 52 int main(){ 53 scanf("%s", str1); 54 scanf("%s", str2); 55 int len = strlen(str1); 56 cnt = len; 57 Node* root = new Node(); 58 root = build(root, 0, len); 59 ceng(root); 60 return 0; 61 }
非递归版
Node* build(int L, int R) { // cout << str1 << " " << str2 << endl; stack<Node*>q; Node* root = new Node(); Node* tmp = new Node(); root->u = str2[R - 1]; q.push(root); for(int i = R - 2, j = R - 1; i >= 0; i--){ Node* node = new Node(); tmp = NULL; node->u = str2[i]; while(!q.empty() && str1[j] == q.top()->u){ tmp = q.top(); q.pop(); j--; } if(tmp == NULL) q.top()->right = node; else tmp->left = node; q.push(node); } return root; }
