• 在二叉树中找到一个节点的后继节点


    现在有一种新的二叉树节点类型如下:

    public class Node {

       public int value; public Node left;

      public Node right;

      public Node parent;

      public Node(int data) { this.value = data;

    } }

    在二 叉树的中序遍历的序列中, node的下一个节点叫作node的后继节点。该结构比普通二叉树节点结构多了一个指向父节点的parent指针。假 设有一 棵Node类型的节点组成的二叉树,树中每个节点的parent指针 都正确地指向 自己的父节点,头节点的parent指向null。只给一个在 二叉树中的某个节点 node,请实现返回node的后继节点的函数。

     1 package my_basic.class_4;
     2 
     3 public class Code_03_SuccessorNode {
     4     public static class Node {
     5         int value;
     6         Node parent;
     7         Node left;
     8         Node right;
     9         public Node(int value) {
    10             super();
    11             this.value = value;
    12         }
    13     }
    14     
    15     public static Node getSuccessNode(Node node) {
    16         if (node == null) {
    17             return null;
    18         }
    19         if (node.right != null) {
    20             return getMostLeft(node.right);
    21         }else {
    22             Node parent = node.parent;
    23             while (parent != null && parent.left != node) {
    24                 node = parent;
    25                 parent = node.parent;    
    26             }
    27             return parent;
    28         }
    29     }
    30 
    31     private static Node  getMostLeft(Node head) {
    32         if (head.left!=null) {
    33             head = head.left;
    34         }
    35         return head;
    36         
    37     }
    38     
    39     public static void main(String[] args) {
    40         Node head = new Node(6);
    41         head.parent = null;
    42         head.left = new Node(3);
    43         head.left.parent = head;
    44         head.left.left = new Node(1);
    45         head.left.left.parent = head.left;
    46         head.left.left.right = new Node(2);
    47         head.left.left.right.parent = head.left.left;
    48         head.left.right = new Node(4);
    49         head.left.right.parent = head.left;
    50         head.left.right.right = new Node(5);
    51         head.left.right.right.parent = head.left.right;
    52         head.right = new Node(9);
    53         head.right.parent = head;
    54         head.right.left = new Node(8);
    55         head.right.left.parent = head.right;
    56         head.right.left.left = new Node(7);
    57         head.right.left.left.parent = head.right.left;
    58         head.right.right = new Node(10);
    59         head.right.right.parent = head.right;
    60         
    61         Node test = head.left.left;
    62         System.out.println(test.value + " next: " + getSuccessNode(test).value);
    63         test = head.left.left.right;
    64         System.out.println(test.value + " next: " + getSuccessNode(test).value);
    65         test = head.left;
    66         System.out.println(test.value + " next: " + getSuccessNode(test).value);
    67         test = head.left.right;
    68         System.out.println(test.value + " next: " + getSuccessNode(test).value);
    69         test = head.left.right.right;
    70         System.out.println(test.value + " next: " + getSuccessNode(test).value);
    71         test = head;
    72         System.out.println(test.value + " next: " + getSuccessNode(test).value);
    73         test = head.right.left.left;
    74         System.out.println(test.value + " next: " + getSuccessNode(test).value);
    75         test = head.right.left;
    76         System.out.println(test.value + " next: " + getSuccessNode(test).value);
    77         test = head.right;
    78         System.out.println(test.value + " next: " + getSuccessNode(test).value);
    79         test = head.right.right; // 10's next is null
    80         System.out.println(test.value + " next: " + getSuccessNode(test));
    81     }
    82 }
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  • 原文地址:https://www.cnblogs.com/lihuazhu/p/10963714.html
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