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| 1 | +package cn.byhieg.algorithmtutorial; |
| 2 | + |
| 3 | +/** |
| 4 | + * Created by byhieg on 17/3/30. |
| 5 | + * Mail to byhieg@gmail.com |
| 6 | + */ |
| 7 | + |
| 8 | +import java.util.Stack; |
| 9 | + |
| 10 | +/** |
| 11 | + * 该类是二叉搜索树 |
| 12 | + * 该树在实现的时候,不考虑数组中有重复的数字。 |
| 13 | + * 节点的左子节点的值都小于这个节点的值,节点的右子节点的值都大于等于这个节点的值 |
| 14 | + */ |
| 15 | +public class BinarySearchTree { |
| 16 | + |
| 17 | + private Node root; |
| 18 | + |
| 19 | + |
| 20 | + public BinarySearchTree(){ |
| 21 | + |
| 22 | + } |
| 23 | + |
| 24 | + |
| 25 | + public BinarySearchTree(int[] nums) { |
| 26 | + Node[] nodes = new Node[nums.length]; |
| 27 | + for (int i = 0 ; i < nums.length;i++) { |
| 28 | + nodes[i] = new Node(nums[i]); |
| 29 | + insert(nodes[i]); |
| 30 | + } |
| 31 | + } |
| 32 | + |
| 33 | + /** |
| 34 | + * 查找指定的元素 |
| 35 | + * @param des |
| 36 | + * @return |
| 37 | + */ |
| 38 | + public Node find(int des){ |
| 39 | + if (root == null) { |
| 40 | + System.out.println("树是空的"); |
| 41 | + throw new RuntimeException(); |
| 42 | + } |
| 43 | + Node current = root; |
| 44 | + while (current.data != des) { |
| 45 | + if (current.data < des) { |
| 46 | + current = current.right; |
| 47 | + }else{ |
| 48 | + current = current.left; |
| 49 | + } |
| 50 | + if (current == null) return null; |
| 51 | + } |
| 52 | + return current; |
| 53 | + } |
| 54 | + |
| 55 | + /** |
| 56 | + * 对BST执行插入操作,采用非递归的形式 |
| 57 | + * 保证插入后,左节点的值小于根节点的值,根节点的值小于右节点的值 |
| 58 | + * @param node |
| 59 | + * @return |
| 60 | + */ |
| 61 | + public boolean insert(Node node){ |
| 62 | + if (root == null) { |
| 63 | + root = node; |
| 64 | + return true; |
| 65 | + } |
| 66 | + |
| 67 | + if (find(node.data) != null) { |
| 68 | + System.out.println("不允许插入相同data的数"); |
| 69 | + throw new RuntimeException(); |
| 70 | + } |
| 71 | + |
| 72 | + Node current = root; |
| 73 | + while (current != null) { |
| 74 | + if (current.data < node.data) { |
| 75 | + if (current.right == null) { |
| 76 | + current.right = node; |
| 77 | + return true; |
| 78 | + } |
| 79 | + current = current.right; |
| 80 | + }else{ |
| 81 | + if (current.left == null) { |
| 82 | + current.left = node; |
| 83 | + return true; |
| 84 | + } |
| 85 | + current = current.left; |
| 86 | + } |
| 87 | + } |
| 88 | + return false; |
| 89 | + |
| 90 | + } |
| 91 | + |
| 92 | + /** |
| 93 | + * 树的前序遍历,递归实现 |
| 94 | + */ |
| 95 | + public void preOrder(Node node) { |
| 96 | + System.out.print(node.data + "-->"); |
| 97 | + if (node.left != null) { |
| 98 | + preOrder(node.left); |
| 99 | + } |
| 100 | + if (node.right != null) { |
| 101 | + preOrder(node.right); |
| 102 | + } |
| 103 | + } |
| 104 | + |
| 105 | + /** |
| 106 | + * 树的中序遍历,递归实现 |
| 107 | + * 针对BST,该结果会从小到大输出树 |
| 108 | + * @param node |
| 109 | + */ |
| 110 | + public void inOrder(Node node){ |
| 111 | + if (node.left != null) { |
| 112 | + inOrder(node.left); |
| 113 | + } |
| 114 | + System.out.print(node.data + "-->"); |
| 115 | + if (node.right != null) { |
| 116 | + inOrder(node.right); |
| 117 | + } |
| 118 | + } |
| 119 | + |
| 120 | + /** |
| 121 | + * 树的后续遍历,递归实现 |
| 122 | + */ |
| 123 | + public void postOrder(Node node){ |
| 124 | + if (node.left != null) { |
| 125 | + postOrder(node.left); |
| 126 | + } |
| 127 | + if (node.right != null) { |
| 128 | + postOrder(node.right); |
| 129 | + } |
| 130 | + System.out.print(node.data + "-->"); |
| 131 | + } |
| 132 | + |
| 133 | + |
| 134 | + /** |
| 135 | + * 树的先续遍历,非递归实现 |
| 136 | + * 1. 建立一个栈,现将头结点压入栈中。 |
| 137 | + * 2. 现将每出栈一个节点,打印他的值,然后都要先加入他的右节点,在加入他的左节点。因为栈的后进先出的特性,才能让左边先出。 |
| 138 | + * 3. 不断重复2,直到栈空 |
| 139 | + * |
| 140 | + */ |
| 141 | + public void preOrder2(Node node) { |
| 142 | + if (node != null) { |
| 143 | + Stack<Node> stack = new Stack<>(); |
| 144 | + stack.push(node); |
| 145 | + while (!stack.isEmpty()) { |
| 146 | + Node tmp = stack.pop(); |
| 147 | + System.out.print(tmp.data + "-->"); |
| 148 | + if (tmp.right != null) { |
| 149 | + stack.push(tmp.right); |
| 150 | + } |
| 151 | + if (tmp.left != null) { |
| 152 | + stack.push(tmp.left); |
| 153 | + } |
| 154 | + } |
| 155 | + } |
| 156 | + } |
| 157 | + |
| 158 | + /** |
| 159 | + * 树的中序遍历,非递归实现 |
| 160 | + * 1. 设定cur,初始化cur = root节点,不断遍历里cur.left,并将其压入栈中,直到null。 |
| 161 | + * 2. 出栈一个节点,打印他的值,然后cur = node.right,并不断重复第二步 |
| 162 | + * 3. 当栈为空,并且cur为空时,停止 |
| 163 | + */ |
| 164 | + public void inorder2(Node node){ |
| 165 | + if (node != null) { |
| 166 | + Stack<Node> stack = new Stack<>(); |
| 167 | + Node cur = node; |
| 168 | + while (!stack.isEmpty() || cur != null) { |
| 169 | + if (cur == null) { |
| 170 | + cur = stack.pop(); |
| 171 | + System.out.print(cur.data + "-->"); |
| 172 | + cur = cur.right; |
| 173 | + }else{ |
| 174 | + stack.push(cur); |
| 175 | + cur = cur.left; |
| 176 | + } |
| 177 | + } |
| 178 | + } |
| 179 | + } |
| 180 | + |
| 181 | + /** |
| 182 | + * 树的后续遍历,非递归实现 |
| 183 | + * 1. 树的先续遍历中,是栈存放顺序是根,右节点,左节点。 |
| 184 | + * 2. 我们可以将其反过来,用栈存放是根,左节点,右节点。然后出栈的时候,将出栈的结果存放到另一个栈里。 |
| 185 | + * 3. 第二栈里的顺序从上到下就是左节点,右节点,根的顺序。 |
| 186 | + * @param node |
| 187 | + */ |
| 188 | + |
| 189 | + public void postOrder2(Node node) { |
| 190 | + if (node != null) { |
| 191 | + Stack<Node> stack = new Stack<>(); |
| 192 | + Stack<Node> result = new Stack<>(); |
| 193 | + Node cur = node; |
| 194 | + stack.push(cur); |
| 195 | + while (!stack.isEmpty()){ |
| 196 | + Node tmp = stack.pop(); |
| 197 | + result.push(tmp); |
| 198 | + if (tmp.left != null) { |
| 199 | + stack.push(tmp.left); |
| 200 | + } |
| 201 | + if (tmp.right != null) { |
| 202 | + stack.push(tmp.right); |
| 203 | + } |
| 204 | + } |
| 205 | + |
| 206 | + while (!result.isEmpty()) { |
| 207 | + System.out.print(result.pop().data + "-->"); |
| 208 | + } |
| 209 | + } |
| 210 | + } |
| 211 | + |
| 212 | + |
| 213 | + /** |
| 214 | + * 得到树中最小的节点 |
| 215 | + * @return |
| 216 | + */ |
| 217 | + public Node getMinNode(){ |
| 218 | + if (root == null) { |
| 219 | + throw new RuntimeException("树为空"); |
| 220 | + } |
| 221 | + Node current = root; |
| 222 | + while (current.left != null) { |
| 223 | + current = current.left; |
| 224 | + } |
| 225 | + return current; |
| 226 | + |
| 227 | + } |
| 228 | + |
| 229 | + /** |
| 230 | + * 得到树中最大的节点 |
| 231 | + * @return |
| 232 | + */ |
| 233 | + public Node getMaxNode(){ |
| 234 | + if (root == null) { |
| 235 | + throw new RuntimeException("树为空"); |
| 236 | + } |
| 237 | + Node current = root; |
| 238 | + while (current.right != null) { |
| 239 | + current = current.right; |
| 240 | + } |
| 241 | + |
| 242 | + return current; |
| 243 | + } |
| 244 | + |
| 245 | + public static class Node{ |
| 246 | + public int data; |
| 247 | + public Node left; |
| 248 | + public Node right; |
| 249 | + |
| 250 | + public Node(int data){ |
| 251 | + this.data = data; |
| 252 | + } |
| 253 | + } |
| 254 | + |
| 255 | + public Node getRoot() { |
| 256 | + return root; |
| 257 | + } |
| 258 | +} |
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