• 二叉树学习总结(Java实现)


    最近学习二叉树,做一些总结,内容包括:

    1.创建二叉树

    2.递归方法:先序 中序 后序

    3.非递归:先序 中序 后序

    4.宽度优先搜索

    5.寻找父节点

    如果有错误,还请及时指出,谢谢!

      1 import java.util.LinkedList;
      2 import java.util.Scanner;
      3 import java.util.Stack;
      4 
      5 public class BinaryTreeStudy {
      6 
      7     public static void main(String[] args) {
      8         BinTree tree = new BinTree();
      9         tree.create();
     10 
     11         tree.preOrderWithRecusion();
     12         tree.inOrderWithRecusion();
     13         tree.postOrderWithRecusion();
     14 
     15         tree.preOrderWithoutRecusion();
     16         tree.inOrderWithoutRecusion();
     17         tree.postOrderWithoutRecusion();
     18 
     19         tree.levelOrader();
     20 
     21         tree.searchParent(5);
     22 
     23     }
     24 }
     25 
     26 class Node {
     27     int data;
     28     Node leftChild;
     29     Node rightChild;
     30 }
     31 
     32 class Item {
     33     Node node;
     34     boolean tag = false;
     35 }
     36 
     37 class BinTree {
     38     Node root;
     39     Scanner scanner = new Scanner(System.in);
     40 
     41     public BinTree() {
     42         root = null;
     43     }
     44 
     45     public void create() {
     46         System.out.println("请输入根节点,-1表示节点为空");
     47         root = createTree();
     48     }
     49 
     50     public Node createTree() {
     51         Node subRoot;
     52         int data = scanner.nextInt();
     53         if (data == -1) {
     54             subRoot = null;
     55         } else {
     56             subRoot = new Node();
     57             subRoot.data = data;
     58             System.out.println("请输入" + data + "的左子树");
     59             subRoot.leftChild = createTree();
     60             System.out.println("请输入" + data + "的右子树");
     61             subRoot.rightChild = createTree();
     62         }
     63 
     64         return subRoot;
     65     }
     66 
     67     public void preOrderWithRecusion() {
     68         System.out.println("递归:先序输出二叉树");
     69         preOrderWithRecusion(root);
     70         System.out.println();
     71     }
     72 
     73     // 递归:先序
     74     private void preOrderWithRecusion(Node root) {
     75         if (root != null) {
     76             System.out.print(root.data);
     77             preOrderWithRecusion(root.leftChild);
     78             preOrderWithRecusion(root.rightChild);
     79         }
     80     }
     81 
     82     public void inOrderWithRecusion() {
     83         System.out.println("递归:中序输出二叉树");
     84         inOrderWithRecusion(root);
     85         System.out.println();
     86     }
     87 
     88     // 递归:中序
     89     private void inOrderWithRecusion(Node root) {
     90         if (root != null) {
     91             inOrderWithRecusion(root.leftChild);
     92             System.out.print(root.data);
     93             inOrderWithRecusion(root.rightChild);
     94         }
     95     }
     96 
     97     public void postOrderWithRecusion() {
     98         System.out.println("递归:后序输出二叉树");
     99         postOrderWithRecusion(root);
    100         System.out.println();
    101 
    102     }
    103 
    104     // 递归:后序
    105     private void postOrderWithRecusion(Node root) {
    106         if (root != null) {
    107             postOrderWithRecusion(root.leftChild);
    108             postOrderWithRecusion(root.rightChild);
    109             System.out.print(root.data);
    110         }
    111     }
    112 
    113     public void preOrderWithoutRecusion() {
    114         System.out.println("非递归:先序输出二叉树");
    115         preOrderWithoutRecusion(root);
    116         System.out.println();
    117     }
    118 
    119     // 非递归:先序
    120     private void preOrderWithoutRecusion(Node subRoot) {
    121         Stack<Node> stack = new Stack<Node>();
    122         Node pointer = subRoot;
    123 
    124         while (pointer != null) {
    125             System.out.print(pointer.data);
    126 
    127             if (pointer.rightChild != null) {
    128                 stack.push(pointer.rightChild);
    129             }
    130 
    131             if (pointer.leftChild != null) {
    132                 pointer = pointer.leftChild;
    133             } else {
    134                 if (!stack.isEmpty()) {
    135                     pointer = stack.pop();
    136                 } else {
    137                     pointer = null;
    138                 }
    139             }
    140 
    141         }
    142 
    143     }
    144 
    145     public void inOrderWithoutRecusion() {
    146         System.out.println("非递归:中序输出二叉树");
    147         inOrderWithoutRecusion(root);
    148         System.out.println();
    149     }
    150 
    151     // 非递归:中序
    152     private void inOrderWithoutRecusion(Node subRoot) {
    153         Stack<Node> stack = new Stack<Node>();
    154         Node pointer = subRoot;
    155 
    156         while (pointer != null || !stack.isEmpty()) {
    157             if (pointer != null) {
    158                 stack.push(pointer);
    159                 pointer = pointer.leftChild;
    160             } else {
    161                 pointer = stack.pop();
    162                 System.out.print(pointer.data);
    163                 pointer = pointer.rightChild;
    164             }
    165         }
    166     }
    167 
    168     public void postOrderWithoutRecusion() {
    169         System.out.println("非递归:后序输出二叉树");
    170         postOrderWithoutRecusion(root);
    171         System.out.println();
    172     }
    173 
    174     // 非递归:后序
    175     private void postOrderWithoutRecusion(Node subRoot) {
    176         boolean LEFT = false;
    177         boolean RIGHT = true;
    178         Stack<Item> stack = new Stack<Item>();
    179         Node pointer = subRoot;
    180         while (pointer != null || !stack.isEmpty()) {
    181             Item item = null;
    182             while (pointer != null) {
    183                 item = new Item();
    184                 item.node = pointer;
    185                 item.tag = LEFT;
    186                 stack.push(item);
    187                 pointer = pointer.leftChild;
    188             }
    189 
    190             item = stack.pop();
    191             pointer = item.node;
    192 
    193             if (item.tag == LEFT) {
    194                 item.tag = RIGHT;
    195                 stack.push(item);
    196                 pointer = pointer.rightChild;
    197             } else {
    198                 System.out.print(pointer.data);
    199                 pointer = null;
    200             }
    201 
    202         }
    203     }
    204 
    205     public void levelOrader() {
    206         System.out.println("宽度优先:");
    207         levelOrader(root);
    208         System.out.println();
    209     }
    210 
    211     // 宽度优先搜索二叉树
    212     private void levelOrader(Node subRoot) {
    213         LinkedList<Node> nodes = new LinkedList<Node>();
    214 
    215         Node pointer = subRoot;
    216         if (pointer != null) {
    217             nodes.offer(pointer);
    218         }
    219 
    220         while (!nodes.isEmpty()) {
    221             pointer = nodes.poll();
    222             System.out.print(pointer.data);
    223 
    224             if (pointer.leftChild != null) {
    225                 nodes.offer(pointer.leftChild);
    226             }
    227 
    228             if (pointer.rightChild != null) {
    229                 nodes.offer(pointer.rightChild);
    230             }
    231 
    232         }
    233 
    234     }
    235 
    236     public void searchParent(int data) {
    237         System.out.println("搜索" + data + "的父节点:");
    238         Node parent = searchParent(root, data);
    239         System.out.println(parent.data);
    240     }
    241 
    242     // 查找父节点
    243     private Node searchParent(Node subRoot, int data) {
    244         if (subRoot == null) {
    245             return null;
    246         }
    247 
    248         if (subRoot.leftChild != null && subRoot.leftChild.data == data) {
    249             return subRoot;
    250         }
    251 
    252         if (subRoot.rightChild != null && subRoot.rightChild.data == data) {
    253             return subRoot;
    254         }
    255 
    256         Node pointer = null;
    257         if ((pointer = searchParent(subRoot.leftChild, data)) != null) {
    258             return pointer;
    259         }
    260 
    261         if ((pointer = searchParent(subRoot.rightChild, data)) != null) {
    262             return pointer;
    263         }
    264 
    265         return null;
    266     }
    267 
    268 }

     非递归搜索父节点:

     1    public void searchParentWithoutRecusion(int data) {
     2         System.out.println("非递归搜索" + data + "的父节点:");
     3         Node parent = searchParentWithoutRecusion(root, data);
     4         System.out.println(parent.data);
     5     }
     6 
     7     private Node searchParentWithoutRecusion(Node subRoot, int data) {
     8         Node pointer = subRoot;
     9         Stack<Node> nodes = new Stack<Node>();
    10         nodes.push(null);
    11         while (pointer != null) {
    12             if (pointer.leftChild != null && pointer.leftChild.data == data) {
    13                 return pointer;
    14             }
    15 
    16             if (pointer.rightChild != null && pointer.rightChild.data == data) {
    17                 return pointer;
    18             }
    19 
    20             if (pointer.rightChild != null) {
    21                 nodes.push(pointer.rightChild);
    22             }
    23 
    24             if (pointer.leftChild != null) {
    25                 pointer = pointer.leftChild;
    26             } else {
    27                 pointer = nodes.pop();
    28             }
    29         }
    30 
    31         return null;
    32     }
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  • 原文地址:https://www.cnblogs.com/androidstudy/p/3480543.html
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