leetcode：Populating Next Right Pointers in Each Node 【Java】

最新推荐文章于 2021-12-24 11:09:57 发布

原创最新推荐文章于 2021-12-24 11:09:57 发布 · 311 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

leetcode 专栏收录该内容

79 篇文章

订阅专栏

本文介绍了一种在完美二叉树中填充每个节点的next指针的方法，使得每个节点指向其右侧相邻节点。文章详细解释了递归算法的具体实现，并提供了完整的代码示例。

一、问题描述

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *next;
    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

You may only use constant extra space.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

二、问题分析

利用递归思想。

三、算法代码

/**
 * 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) {
        connect(root, null);
    }
    
    public void connect(TreeLinkNode root, TreeLinkNode sibling){
        if(root == null){
            return;
        }else{
            root.next = sibling;
        }
        connect(root.left, root.right);
        if(sibling != null){
            connect(root.right, sibling.left);
        }else{
            connect(root.right, null);
        }
    }
}