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| 1 | +import 'LinkedListStatck.dart'; |
| 2 | +import 'LinkedListQueue.dart'; |
| 3 | +/** |
| 4 | + * 二分搜索树 |
| 5 | + */ |
| 6 | +class BST<T extends Comparable<T>> { |
| 7 | + _Node? root; |
| 8 | + |
| 9 | + int? size; |
| 10 | + |
| 11 | + BST() { |
| 12 | + root = null; |
| 13 | + size = 0; |
| 14 | + } |
| 15 | + |
| 16 | + int? getSize() { |
| 17 | + return size; |
| 18 | + } |
| 19 | + |
| 20 | + bool isEmpty() { |
| 21 | + return size == 0; |
| 22 | + } |
| 23 | + |
| 24 | + //添加元素 |
| 25 | +// add(T t) { |
| 26 | +// if (root == null) { |
| 27 | +// root = _Node(t); |
| 28 | +// size = size! + 1; |
| 29 | +// } else { |
| 30 | +// addNext(root, t); |
| 31 | +// } |
| 32 | +// } |
| 33 | + add(T e) { |
| 34 | + root = addNext(root, e); |
| 35 | + } |
| 36 | + |
| 37 | +// 向以node为根的二分搜索树中插入元素e,递归算法 |
| 38 | +// addNext(_Node? node, T? t) { |
| 39 | +// if (t! == node!.e) |
| 40 | +// return; |
| 41 | +// else if (t.compareTo(node.e) < 0 && node.left == null) { |
| 42 | +// node.left = _Node(t); |
| 43 | +// size = size! +1; |
| 44 | +// return; |
| 45 | +// } else if (t.compareTo(node.e) > 0 && node.right == null) { |
| 46 | +// node.right = _Node(t); |
| 47 | +// size = size! +1; |
| 48 | +// return; |
| 49 | +// } |
| 50 | +// if (t.compareTo(node.e) < 0) |
| 51 | +// addNext(node.left, t); |
| 52 | +// else //e.compareTo(node.e) > 0 |
| 53 | +// addNext(node.right, t); |
| 54 | +// } |
| 55 | + //返回新节点后二分搜索树的根 |
| 56 | + _Node addNext(_Node? node, T t) { |
| 57 | + if (node == null) { |
| 58 | + size = size! + 1; |
| 59 | + return _Node(t); |
| 60 | + } |
| 61 | + if (t.compareTo(node.e) < 0) { |
| 62 | + node.left = addNext(node.left, t); |
| 63 | + } else if (t.compareTo(node.e) > 0) { |
| 64 | + node.right = addNext(node.right, t); |
| 65 | + } |
| 66 | + return node; |
| 67 | + } |
| 68 | + |
| 69 | + // 向二分搜索树中添加新的元素e,非递归写法 |
| 70 | + add2(T e) { |
| 71 | + if (root == null) { |
| 72 | + root = _Node(e); |
| 73 | + size = size! + 1; |
| 74 | + return; |
| 75 | + } |
| 76 | + _Node? p = root; |
| 77 | + while (p != null) { |
| 78 | + if (e.compareTo(p.e) < 0) { |
| 79 | + if (p.left == null) { |
| 80 | + p.left = _Node(e); |
| 81 | + size = size! + 1; |
| 82 | + return; |
| 83 | + } |
| 84 | + p = p.left; |
| 85 | + } else if (e.compareTo(p.e) > 0) { |
| 86 | + if (p.right == null) { |
| 87 | + p.right = _Node(e); |
| 88 | + size = size! + 1; |
| 89 | + return; |
| 90 | + } |
| 91 | + p = p.right; |
| 92 | + } else |
| 93 | + return; |
| 94 | + } |
| 95 | + } |
| 96 | + |
| 97 | + //查询元素是否包含 |
| 98 | + bool contains(T t) { |
| 99 | + return _containsAll(root, t); |
| 100 | + } |
| 101 | + |
| 102 | + bool _containsAll(_Node? node, T t) { |
| 103 | + if (node == null) { |
| 104 | + return false; |
| 105 | + } |
| 106 | + if (t.compareTo(node.e) == 0) { |
| 107 | + return true; |
| 108 | + } else if (t.compareTo(node.e) < 0) { |
| 109 | + return _containsAll(node.left, t); |
| 110 | + } else { |
| 111 | + return _containsAll(node.right, t); |
| 112 | + } |
| 113 | + } |
| 114 | + |
| 115 | + // 二分搜索树的前序遍历 |
| 116 | + preOrder() { |
| 117 | + _preOrderDetail(root!); |
| 118 | + } |
| 119 | + |
| 120 | + _preOrderDetail(_Node? node) { |
| 121 | + if (node == null) { |
| 122 | + return; |
| 123 | + } |
| 124 | + print(node.e); |
| 125 | + _preOrderDetail(node.left); |
| 126 | + _preOrderDetail(node.right); |
| 127 | + } |
| 128 | + |
| 129 | + inOrder() { |
| 130 | + _inOrderDetail(root); |
| 131 | + } |
| 132 | + |
| 133 | + _inOrderDetail(_Node? node) { |
| 134 | + if (node == null) { |
| 135 | + return; |
| 136 | + } |
| 137 | + _inOrderDetail(node.left); |
| 138 | + print(node.e); |
| 139 | + _inOrderDetail(node.right); |
| 140 | + } |
| 141 | + |
| 142 | + postOrder() { |
| 143 | + _postOrderDetail(root); |
| 144 | + } |
| 145 | + |
| 146 | + _postOrderDetail(_Node? node) { |
| 147 | + if (node == null) { |
| 148 | + return; |
| 149 | + } |
| 150 | + _postOrderDetail(node.left); |
| 151 | + |
| 152 | + _postOrderDetail(node.right); |
| 153 | + print(node.e); |
| 154 | + } |
| 155 | + |
| 156 | + // 二分搜索树的非递归前序遍历 |
| 157 | + preOrderNR() { |
| 158 | + if (root == null) return; |
| 159 | + LinkedListStack stack = new LinkedListStack(); |
| 160 | + stack.push(root); |
| 161 | + while (!stack.isEmpty()) { |
| 162 | + _Node cur = stack.pop(); |
| 163 | + print(cur.e); |
| 164 | + if (cur.right != null) { |
| 165 | + stack.push(cur.right); |
| 166 | + } |
| 167 | + if (cur.left != null){ |
| 168 | + stack.push(cur.left); |
| 169 | + } |
| 170 | + } |
| 171 | + } |
| 172 | + |
| 173 | + // 二分搜索树的层序遍历 |
| 174 | + levelOrder(){ |
| 175 | + if(root == null) |
| 176 | + return; |
| 177 | + LinkedListQueue q= new LinkedListQueue(); |
| 178 | + q.enqueue(root); |
| 179 | + while(!q.isEmpty()){ |
| 180 | + _Node cur = q.dequeue(); |
| 181 | + print(cur.e); |
| 182 | + if(cur.left != null){ |
| 183 | + q.enqueue(cur.left); |
| 184 | + } |
| 185 | + if(cur.right != null){ |
| 186 | + q.enqueue(cur.right); |
| 187 | + } |
| 188 | + } |
| 189 | + } |
| 190 | + |
| 191 | + // 寻找二分搜索树的最小元素 |
| 192 | + searchMinimum<T>(){ |
| 193 | + if(size == 0){ |
| 194 | + throw new Exception("BST is empty!"); |
| 195 | + } |
| 196 | + return _minimum(root!).e; |
| 197 | + } |
| 198 | + // 返回以node为根的二分搜索树的最小值所在的节点 |
| 199 | + _Node _minimum(_Node node){ |
| 200 | + if(node.left == null){ |
| 201 | + return node; |
| 202 | + } |
| 203 | + return _minimum(node.left!); |
| 204 | + } |
| 205 | + |
| 206 | + // 寻找二分搜索树的最大元素 |
| 207 | + searchMaximum<T>(){ |
| 208 | + if(size == 0){ |
| 209 | + throw new Exception("BST is empty"); |
| 210 | + } |
| 211 | + return _maximum(root!).e; |
| 212 | + } |
| 213 | + // 返回以node为根的二分搜索树的最大值所在的节点 |
| 214 | + _Node _maximum(_Node node){ |
| 215 | + if(node.right == null){ |
| 216 | + return node; |
| 217 | + } |
| 218 | + return _maximum(node.right!); |
| 219 | + } |
| 220 | + |
| 221 | + // 从二分搜索树中删除最小值所在节点, 返回最小值 |
| 222 | + removeMin<T>(){ |
| 223 | + T ret = searchMinimum(); |
| 224 | + root = _removeMinDetail(root!); |
| 225 | + return ret; |
| 226 | + } |
| 227 | + // 删除掉以node为根的二分搜索树中的最小节点 |
| 228 | + // 返回删除节点后新的二分搜索树的根 |
| 229 | + _Node _removeMinDetail(_Node node){ |
| 230 | + if(node.left == null){ |
| 231 | + _Node rightNode = node.right!; |
| 232 | + node.right = null; |
| 233 | + size = size! - 1; |
| 234 | + return rightNode; |
| 235 | + } |
| 236 | + node.left = _removeMinDetail(node.left!); |
| 237 | + return node; |
| 238 | + } |
| 239 | + |
| 240 | + |
| 241 | + // 从二分搜索树中删除最大值所在节点 |
| 242 | + removeMax<T>(){ |
| 243 | + T ret = searchMaximum(); |
| 244 | + root = _removeMaxDetail(root!); |
| 245 | + return ret; |
| 246 | + } |
| 247 | + // 删除掉以node为根的二分搜索树中的最大节点 |
| 248 | + // 返回删除节点后新的二分搜索树的根 |
| 249 | + _Node _removeMaxDetail(_Node node){ |
| 250 | + if(node.right == null){ |
| 251 | + _Node leftNode = node.left!; |
| 252 | + node.left = null; |
| 253 | + size = size! -1 ; |
| 254 | + return leftNode; |
| 255 | + } |
| 256 | + node.right = _removeMaxDetail(node.right!); |
| 257 | + return node; |
| 258 | + } |
| 259 | + |
| 260 | + // 从二分搜索树中删除元素为e的节点 |
| 261 | + remove(T e){ |
| 262 | + root = _removeNode(root!, e); |
| 263 | + } |
| 264 | + |
| 265 | + _removeNode(_Node node, T e){ |
| 266 | + if( node == null ) |
| 267 | + return null; |
| 268 | + |
| 269 | + if( e!.compareTo(node.e!) < 0 ){ |
| 270 | + node.left = _removeNode(node.left! , e); |
| 271 | + return node; |
| 272 | + } |
| 273 | + else if(e.compareTo(node.e) > 0 ){ |
| 274 | + node.right = _removeNode(node.right!, e); |
| 275 | + return node; |
| 276 | + } |
| 277 | + else{ // e.compareTo(node.e) == 0 |
| 278 | + // 待删除节点左子树为空的情况 |
| 279 | + if(node.left == null){ |
| 280 | + _Node rightNode = node.right!; |
| 281 | + node.right = null; |
| 282 | + size = size! - 1 ; |
| 283 | + return rightNode; |
| 284 | + } |
| 285 | + // 待删除节点右子树为空的情况 |
| 286 | + if(node.right == null){ |
| 287 | + _Node leftNode = node.left!; |
| 288 | + node.left = null; |
| 289 | + size = size! -1; |
| 290 | + return leftNode; |
| 291 | + } |
| 292 | + // 待删除节点左右子树均不为空的情况 |
| 293 | + // 找到比待删除节点大的最小节点, 即待删除节点右子树的最小节点 |
| 294 | + // 用这个节点顶替待删除节点的位置 |
| 295 | + _Node successor = _minimum(node.right!); |
| 296 | + successor.right = _removeMinDetail(node.right!); |
| 297 | + successor.left = node.left; |
| 298 | + node.left = node.right = null; |
| 299 | + return successor; |
| 300 | + } |
| 301 | + } |
| 302 | + |
| 303 | + |
| 304 | + @override |
| 305 | + String toString() { |
| 306 | + StringBuffer res = new StringBuffer(); |
| 307 | + _generateBSTString(root!, 0, res); |
| 308 | + return res.toString(); |
| 309 | + } |
| 310 | + |
| 311 | + _generateBSTString(_Node? node, int depth, StringBuffer res) { |
| 312 | + if (node == null) { |
| 313 | + res.write("${_generateBSTStringDepth(depth)} null \n"); |
| 314 | + return; |
| 315 | + } |
| 316 | + res.write("${_generateBSTStringDepth(depth)} ${node.e} \n"); |
| 317 | + _generateBSTString(node.left, depth + 1, res); |
| 318 | + _generateBSTString(node.right, depth + 1, res); |
| 319 | + } |
| 320 | + |
| 321 | + _generateBSTStringDepth(int depth) { |
| 322 | + StringBuffer res = new StringBuffer(); |
| 323 | + for (int i = 0; i < depth; i++) { |
| 324 | + res.write(" - -"); |
| 325 | + } |
| 326 | + return res.toString(); |
| 327 | + } |
| 328 | +} |
| 329 | + |
| 330 | +class _Node<T> { |
| 331 | + T? e; |
| 332 | + |
| 333 | + _Node? left; |
| 334 | + |
| 335 | + _Node? right; |
| 336 | + |
| 337 | + _Node(T t) { |
| 338 | + this.e = t; |
| 339 | + this.left = null; |
| 340 | + this.right = null; |
| 341 | + } |
| 342 | +} |
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