Populating Next Right Pointers in Each Node

最新推荐文章于 2022-02-28 18:16:11 发布

原创最新推荐文章于 2022-02-28 18:16:11 发布 · 567 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种解决二叉树中填充每个节点的下一个右侧节点指针的问题的方法，该方法适用于任意二叉树，并且仅使用常数额外空间。通过递归方式实现，确保了树的每个层级的节点正确地连接起来。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Populating Next Right Pointers in Each Node II

Oct 28 '12

5299 / 13177

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.

For example,
Given the following binary tree,

/ \

2 3

/ \ \

4 5 7

After calling your function, the tree should look like:

1 -> NULL

/ \

2 -> 3 -> NULL

/ \ \

4-> 5 -> 7 -> NULL

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {


        if(root==null) return ;
        else{
            root.next = null;
            next(root);
        }
            
    }
    
    public void next(TreeLinkNode root){
        if(root==null) return ;
        TreeLinkNode p1 = root;
        TreeLinkNode p2=null,now;
        
        while(p1!=null){
            if(p1.left!=null){
                if(now ==null){
                    p2 = p1.left;
                    now = p1.left;
                }else{
                    now.next = p1.left;
                    now = now .next;
                }
            }
            if(p1.right!=null){
                if(now==null){
                    p2 = p1.right;
                    now = p1.right;
                }else{
                    now.next = p1.right;
                    now = now.next;
                    
                }
            }
            p1 = p1.next;
        }
        if(now!=null)
            now.next = null;
        next(p2);
    }
}