链表问题----复杂链表的复制+二叉搜索树与双向链表（Java）

本文链接：https://blog.youkuaiyun.com/zlhzlh11/article/details/49495229

本文介绍两种数据结构的转换方法：一是复杂链表的复制，通过哈希表记录随机指针对应关系及将复制节点插入原链表中间的方式实现；二是二叉搜索树转换为双向链表的方法，包括栈、递归和神级遍历法。

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复杂链表的复制

题目描述：输入一个复杂链表（每个节点中有节点值，以及两个指针，一个指向下一个节点，另一个特殊指针指向任意一个节点）。
（1）利用hash表记录随机指针对应关系；
import 
java.util.HashMap;
/*
public class RandomListNode {
    int label;
    RandomListNode next = null;
    RandomListNode random = null;
 
    RandomListNode(int label) {
        this.label = label;
    }
}
*/
public 
class Solution {
    publicRandomListNode Clone(RandomListNode pHead)
 
        {
           if(pHead==null){
                returnpHead;
            }
            RandomListNode head=pHead;
            RandomListNode result=newRandomListNode(pHead.label);
            RandomListNode temp=result;
            HashMap<RandomListNode,RandomListNode> map=newHashMap<RandomListNode,RandomListNode>();
            map.put(pHead, temp);
            pHead=pHead.next;
            while(pHead!=null){
                temp.next=newRandomListNode(pHead.label);
                map.put(pHead, temp.next);
                temp=temp.next;
                pHead=pHead.next;
            }
            pHead=head;
            temp=result;
            while(pHead!=null){
                if(pHead.random!=null){
                    temp.random=map.get(pHead.random);
                }
                pHead=pHead.next;
                temp=temp.next;
            }
            returnresult;
        }
     
}
（2）将每个结点复制并链接到原链表中间，再拆分。
/*
public class RandomListNode {
    int label;
    RandomListNode next = null;
    RandomListNode random = null;
 
    RandomListNode(int label) {
        this.label = label;
    }
}
*/
public 
class Solution {
    publicRandomListNode Clone(RandomListNode pHead)
    {
        if(pHead==null){
            returnpHead;
        }
        RandomListNode head=pHead;
        RandomListNode result;
        while(pHead!=null){
            RandomListNode temp=newRandomListNode(pHead.label);
            temp.next=pHead.next;
            pHead.next=temp;
            pHead=temp.next;
        }
        pHead=head;
        while(pHead!=null){
            if(pHead.random!=null){
                pHead.next.random=pHead.random.next;
            }
            pHead=pHead.next.next;
        }
        result=pHead=head.next;
        while(pHead.next!=null){
            head.next=pHead.next;
            head=head.next;
            pHead.next=head.next;
            pHead=pHead.next;
        }
        head.next=null;
        returnresult;
    }
}


二叉搜索树与双向链表

题目描述：输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点，只能调整树中结点指针的指向。


（1）利用栈
import 
java.util.Stack;
/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;
 
    public TreeNode(int val) {
        this.val = val;
 
    }
 
}
*/
public 
class Solution {
    publicTreeNode Convert(TreeNode pRootOfTree) {
        if(pRootOfTree==null||pRootOfTree.left==null&&pRootOfTree.right==null){
            returnpRootOfTree;
        }
        Stack<TreeNode> stack=newStack<TreeNode>();
        TreeNode p=pRootOfTree;
        TreeNode pre=null;
        TreeNode result=null;
        while(p!=null||!stack.empty()){
            while(p!=null){
                stack.push(p);
                p=p.left;
            }
            p=stack.pop();
            if(pre!=null){
                if(pre.left==null){
                    result=pre;
                }
                pre.right=p;
                p.left=pre;
            }
            pre=p;
            p=p.right;
        }
        returnresult;
    }
}


（2）利用递归
/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;
 
    public TreeNode(int val) {
        this.val = val;
 
    }
 
}
*/
public 
class Solution {
    publicTreeNode Convert(TreeNode pRootOfTree) {
        if(pRootOfTree==null||pRootOfTree.left==null&&pRootOfTree.right==null){
            returnpRootOfTree;
        }
        if(pRootOfTree.left!=null){
            TreeNode temp=Convert(pRootOfTree.left);
            while(temp.right!=null){
                temp=temp.right;
            }
            pRootOfTree.left=temp;
            temp.right=pRootOfTree;
        }
        if(pRootOfTree.right!=null){
            TreeNode temp=Convert(pRootOfTree.right);
            pRootOfTree.right=temp;
            temp.left=pRootOfTree;
        }
        while(pRootOfTree.left!=null){
            pRootOfTree=pRootOfTree.left;
        }
        returnpRootOfTree;
    }
}

（3）使用二叉树的神级遍历发，时间复杂度O（n），额外空间复杂度O（1）

/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;
 
    public TreeNode(int val) {
        this.val = val;
 
    }
 
}
*/
public 
class Solution {
    public
TreeNode Convert(TreeNode pRootOfTree) {
        if(pRootOfTree==null||pRootOfTree.left==null&&pRootOfTree.right==null){
            return
pRootOfTree;
        }
        TreeNode cur1=pRootOfTree;
        TreeNode cur2,pre=null;
        TreeNode result=null;
        while(cur1!=null){
            cur2=cur1.left;
            if(cur2==null||cur2.right==cur1){
                if(result==null){
                    result=cur1;
                }
                else{
                    pre.right=cur1;
                }
                cur1.left=pre;
                pre=cur1;
                cur1=cur1.right;
            }
            else{
                while(cur2.right!=null){
                    cur2=cur2.right;
                }
                cur2.right=cur1;
                cur1=cur1.left;
                cur2.right.left=cur2;
            }
        }
        return
result;
    }
}