标签:for namespace 图片 else 面试 iostream 二叉树 let cout
删除结点1之后,下一个删除的就是3了。
程序:
1 #include <iostream> 2 #include "BTreeNode.h" 3 4 using namespace std; 5 using namespace DTLib; 6 7 template < typename T > 8 BTreeNode<T>* createTree() 9 { 10 static BTreeNode<int> ns[9]; 11 12 for(int i=0; i<9; i++) 13 { 14 ns[i].value = i; 15 ns[i].parent = NULL; 16 ns[i].left = NULL; 17 ns[i].right = NULL; 18 } 19 20 ns[0].left = &ns[1]; 21 ns[0].right = &ns[2]; 22 ns[1].parent = &ns[0]; 23 ns[2].parent = &ns[0]; 24 25 ns[1].left = &ns[3]; 26 ns[1].right = NULL; 27 ns[3].parent = &ns[1]; 28 29 ns[2].left = &ns[4]; 30 ns[2].right = &ns[5]; 31 ns[4].parent = &ns[2]; 32 ns[5].parent = &ns[2]; 33 34 ns[3].left = NULL; 35 ns[3].right = &ns[6]; 36 ns[6].parent = &ns[3]; 37 38 ns[4].left = &ns[7]; 39 ns[4].right = NULL; 40 ns[7].parent = &ns[4]; 41 42 ns[5].left = &ns[8]; 43 ns[5].right = NULL; 44 ns[8].parent = &ns[5]; 45 46 return ns; 47 } 48 49 template < typename T > 50 void printInOrder(BTreeNode<T>* node) 51 { 52 if( node != NULL ) 53 { 54 printInOrder(node->left); 55 56 cout << node->value <<" "; 57 58 printInOrder(node->right); 59 } 60 } 61 62 template < typename T > 63 void printDualList(BTreeNode<T>* node) 64 { 65 BTreeNode<T>* g = node; 66 67 cout << "head -> tail: " << endl; 68 69 while( node != NULL ) 70 { 71 cout << node->value << " "; 72 73 g = node; 74 75 node = node->right; 76 } 77 78 cout << endl; 79 80 cout << "tail -> head: " << endl; 81 82 while( g != NULL ) 83 { 84 cout << g->value << " "; 85 86 g = g->left; 87 } 88 89 cout << endl; 90 } 91 92 template < typename T> 93 BTreeNode<T>* delOdd1(BTreeNode<T>* node) 94 { 95 BTreeNode<T>* ret = NULL; 96 97 if( node != NULL ) 98 { 99 if( ((node->left != NULL) && (node->right == NULL)) || 100 ((node->left == NULL) && (node->right != NULL)) ) 101 { 102 BTreeNode<T>* parent = dynamic_cast<BTreeNode<T>*>(node->parent); 103 BTreeNode<T>* node_child = (node->left != NULL) ? node->left : node->right; 104 105 if( parent != NULL ) //根节点有可能是单度的 106 { 107 BTreeNode<T>*& parent_child = (parent->left == node) ? parent->left : parent->right; 108 109 parent_child = node_child; 110 node_child->parent = parent; 111 } 112 else 113 { 114 node_child->parent = NULL; 115 } 116 117 if( node->flag() ) 118 { 119 delete node; 120 } 121 122 ret = delOdd1(node_child); 123 } 124 else 125 { 126 delOdd1(node->left); 127 delOdd1(node->right); 128 129 ret = node; 130 } 131 } 132 133 return ret; 134 } 135 136 int main() 137 { 138 BTreeNode<int>* ns = createTree<int>(); 139 140 printInOrder(ns); 141 142 cout << endl; 143 144 ns = delOdd1(ns); 145 146 printInOrder(ns); 147 148 cout << endl; 149 150 int a[] = {6, 7, 8}; 151 152 for(int i = 0; i < 3; i++) 153 { 154 TreeNode<int>* n = ns + a[i]; 155 156 while( n != NULL ) 157 { 158 cout << n->value << " "; 159 160 n = n->parent; 161 } 162 163 cout << endl; 164 } 165 166 cout << endl; 167 168 return 0; 169 }
结果如下:
第一个要删除的结点是1,node是一个引用,指向1这个结点,也就是0结点中的孩子指针的引用,node = node->child执行完之后,0结点里面的孩子指针就指向了3,也就是node这个指针指向了3,这个node此时还是0结点中的孩子指针的引用。
树结点没有父节点的情况,删除单度结点的程序如下:
1 #include <iostream> 2 #include "BTreeNode.h" 3 4 using namespace std; 5 using namespace DTLib; 6 7 template < typename T > 8 BTreeNode<T>* createTree() 9 { 10 static BTreeNode<int> ns[9]; 11 12 for(int i=0; i<9; i++) 13 { 14 ns[i].value = i; 15 ns[i].parent = NULL; 16 ns[i].left = NULL; 17 ns[i].right = NULL; 18 } 19 20 ns[0].left = &ns[1]; 21 ns[0].right = &ns[2]; 22 ns[1].parent = &ns[0]; 23 ns[2].parent = &ns[0]; 24 25 ns[1].left = &ns[3]; 26 ns[1].right = NULL; 27 ns[3].parent = &ns[1]; 28 29 ns[2].left = &ns[4]; 30 ns[2].right = &ns[5]; 31 ns[4].parent = &ns[2]; 32 ns[5].parent = &ns[2]; 33 34 ns[3].left = NULL; 35 ns[3].right = &ns[6]; 36 ns[6].parent = &ns[3]; 37 38 ns[4].left = &ns[7]; 39 ns[4].right = NULL; 40 ns[7].parent = &ns[4]; 41 42 ns[5].left = &ns[8]; 43 ns[5].right = NULL; 44 ns[8].parent = &ns[5]; 45 46 return ns; 47 } 48 49 template < typename T > 50 void printInOrder(BTreeNode<T>* node) 51 { 52 if( node != NULL ) 53 { 54 printInOrder(node->left); 55 56 cout << node->value <<" "; 57 58 printInOrder(node->right); 59 } 60 } 61 62 template < typename T > 63 void printDualList(BTreeNode<T>* node) 64 { 65 BTreeNode<T>* g = node; 66 67 cout << "head -> tail: " << endl; 68 69 while( node != NULL ) 70 { 71 cout << node->value << " "; 72 73 g = node; 74 75 node = node->right; 76 } 77 78 cout << endl; 79 80 cout << "tail -> head: " << endl; 81 82 while( g != NULL ) 83 { 84 cout << g->value << " "; 85 86 g = g->left; 87 } 88 89 cout << endl; 90 } 91 92 template < typename T> 93 BTreeNode<T>* delOdd1(BTreeNode<T>* node) 94 { 95 BTreeNode<T>* ret = NULL; 96 97 if( node != NULL ) 98 { 99 if( ((node->left != NULL) && (node->right == NULL)) || 100 ((node->left == NULL) && (node->right != NULL)) ) 101 { 102 BTreeNode<T>* parent = dynamic_cast<BTreeNode<T>*>(node->parent); 103 BTreeNode<T>* node_child = (node->left != NULL) ? node->left : node->right; 104 105 if( parent != NULL ) //根节点有可能是单度的 106 { 107 BTreeNode<T>*& parent_child = (parent->left == node) ? parent->left : parent->right; 108 109 parent_child = node_child; 110 node_child->parent = parent; 111 } 112 else 113 { 114 node_child->parent = NULL; 115 } 116 117 if( node->flag() ) 118 { 119 delete node; 120 } 121 122 ret = delOdd1(node_child); 123 } 124 else 125 { 126 delOdd1(node->left); 127 delOdd1(node->right); 128 129 ret = node; 130 } 131 } 132 133 return ret; 134 } 135 136 template < typename T> 137 void delOdd2(BTreeNode<T>*& node) 138 { 139 if( node != NULL ) 140 { 141 if( ((node->left != NULL) && (node->right == NULL)) || 142 ((node->left == NULL) && (node->right != NULL)) ) 143 { 144 BTreeNode<T>* node_child = (node->left != NULL) ? node->left : node->right; 145 146 if( node->flag() ) 147 { 148 delete node; 149 } 150 151 node = node_child; 152 153 delOdd2(node); 154 } 155 else 156 { 157 delOdd2(node->left); 158 delOdd2(node->right); 159 } 160 } 161 } 162 163 int main() 164 { 165 BTreeNode<int>* ns = createTree<int>(); 166 167 printInOrder(ns); 168 169 cout << endl; 170 171 delOdd2(ns); 172 173 printInOrder(ns); 174 175 cout << endl; 176 177 return 0; 178 }
结果如下:
程序如下:
1 #include <iostream> 2 #include "BTreeNode.h" 3 4 using namespace std; 5 using namespace DTLib; 6 7 template < typename T > 8 BTreeNode<T>* createTree() 9 { 10 static BTreeNode<int> ns[9]; 11 12 for(int i=0; i<9; i++) 13 { 14 ns[i].value = i; 15 ns[i].parent = NULL; 16 ns[i].left = NULL; 17 ns[i].right = NULL; 18 } 19 20 ns[0].left = &ns[1]; 21 ns[0].right = &ns[2]; 22 ns[1].parent = &ns[0]; 23 ns[2].parent = &ns[0]; 24 25 ns[1].left = &ns[3]; 26 ns[1].right = NULL; 27 ns[3].parent = &ns[1]; 28 29 ns[2].left = &ns[4]; 30 ns[2].right = &ns[5]; 31 ns[4].parent = &ns[2]; 32 ns[5].parent = &ns[2]; 33 34 ns[3].left = NULL; 35 ns[3].right = &ns[6]; 36 ns[6].parent = &ns[3]; 37 38 ns[4].left = &ns[7]; 39 ns[4].right = NULL; 40 ns[7].parent = &ns[4]; 41 42 ns[5].left = &ns[8]; 43 ns[5].right = NULL; 44 ns[8].parent = &ns[5]; 45 46 return ns; 47 } 48 49 template < typename T > 50 void printInOrder(BTreeNode<T>* node) 51 { 52 if( node != NULL ) 53 { 54 printInOrder(node->left); 55 56 cout << node->value <<" "; 57 58 printInOrder(node->right); 59 } 60 } 61 62 template < typename T > 63 void printDualList(BTreeNode<T>* node) 64 { 65 BTreeNode<T>* g = node; 66 67 cout << "head -> tail: " << endl; 68 69 while( node != NULL ) 70 { 71 cout << node->value << " "; 72 73 g = node; 74 75 node = node->right; 76 } 77 78 cout << endl; 79 80 cout << "tail -> head: " << endl; 81 82 while( g != NULL ) 83 { 84 cout << g->value << " "; 85 86 g = g->left; 87 } 88 89 cout << endl; 90 } 91 92 template < typename T> 93 BTreeNode<T>* delOdd1(BTreeNode<T>* node) 94 { 95 BTreeNode<T>* ret = NULL; 96 97 if( node != NULL ) 98 { 99 if( ((node->left != NULL) && (node->right == NULL)) || 100 ((node->left == NULL) && (node->right != NULL)) ) 101 { 102 BTreeNode<T>* parent = dynamic_cast<BTreeNode<T>*>(node->parent); 103 BTreeNode<T>* node_child = (node->left != NULL) ? node->left : node->right; 104 105 if( parent != NULL ) //根节点有可能是单度的 106 { 107 BTreeNode<T>*& parent_child = (parent->left == node) ? parent->left : parent->right; 108 109 parent_child = node_child; 110 node_child->parent = parent; 111 } 112 else 113 { 114 node_child->parent = NULL; 115 } 116 117 if( node->flag() ) 118 { 119 delete node; 120 } 121 122 ret = delOdd1(node_child); 123 } 124 else 125 { 126 delOdd1(node->left); 127 delOdd1(node->right); 128 129 ret = node; 130 } 131 } 132 133 return ret; 134 } 135 136 template < typename T> 137 void delOdd2(BTreeNode<T>*& node) 138 { 139 if( node != NULL ) 140 { 141 if( ((node->left != NULL) && (node->right == NULL)) || 142 ((node->left == NULL) && (node->right != NULL)) ) 143 { 144 BTreeNode<T>* node_child = (node->left != NULL) ? node->left : node->right; 145 146 if( node->flag() ) 147 { 148 delete node; 149 } 150 151 node = node_child; 152 153 delOdd2(node); 154 } 155 else 156 { 157 delOdd2(node->left); 158 delOdd2(node->right); 159 } 160 } 161 } 162 163 template < typename T > 164 void inOrderThread(BTreeNode<T>* node, BTreeNode<T>*& pre) 165 { 166 if( node != NULL ) 167 { 168 inOrderThread(node->left, pre); 169 170 node->left = pre; 171 172 if( pre != NULL ) 173 { 174 pre->right = node; 175 } 176 177 pre = node; 178 179 inOrderThread(node->right, pre); 180 } 181 } 182 183 184 template < typename T > 185 BTreeNode<T>* inOrderThread1(BTreeNode<T>* node) 186 { 187 BTreeNode<T>* pre = NULL; 188 189 inOrderThread(node, pre); 190 191 while( (node != NULL) && (node->left != NULL) ) 192 { 193 node = node->left; 194 } 195 196 return node; 197 } 198 199 int main() 200 { 201 BTreeNode<int>* ns = createTree<int>(); 202 203 printInOrder(ns); 204 205 cout << endl; 206 207 delOdd2(ns); 208 209 printInOrder(ns); 210 211 cout << endl; 212 213 ns = inOrderThread1(ns); 214 215 printDualList(ns); 216 217 return 0; 218 }
结果如下:
标签:for namespace 图片 else 面试 iostream 二叉树 let cout
原文地址:https://www.cnblogs.com/wanmeishenghuo/p/9696042.html