leetcode -- 116. Populating Next Right Pointers in Each Node

最新推荐文章于 2026-01-09 18:00:00 发布

原创最新推荐文章于 2026-01-09 18:00:00 发布 · 206 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

算法专栏收录该内容

212 篇文章

订阅专栏

本文介绍了一种解决完美二叉树中填充Next指针的问题的方法，旨在让每个节点的Next指针指向其右侧相邻的节点。提供两种解决方案，一种使用队列进行层次遍历，另一种通过连接相邻节点实现。

题目描述

题目难度：Medium
You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

struct Node {
int val;
Node *left;
Node *right;
Node *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.
Example:
在这里插入图片描述

Note:

You may only use constant extra space.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.

AC代码1

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}

    public Node(int _val,Node _left,Node _right,Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/
class Solution {
    public Node connect(Node root) {
        if(root == null) return root;
        Queue<Node> queue = new LinkedList<Node>();
        queue.offer(root);
        Node pre = null;
        Node cur = null;
        while(!queue.isEmpty()){
            int size = queue.size();
            pre = queue.poll();
            if(pre.left != null) queue.offer(pre.left);
            if(pre.right != null) queue.offer(pre.right);
            for(int i = 0;i < size - 1;i++){
                cur = queue.poll();
                pre.next = cur;
                pre = cur;
                if(pre.left != null) queue.offer(pre.left);
                if(pre.right != null) queue.offer(pre.right);
            }
        }
        return root;
    }
}

AC代码2

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}

    public Node(int _val,Node _left,Node _right,Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/
class Solution {
    public Node connect(Node root) {
        if(root == null) return root;
        Node startNode = root;
        while(startNode != null){
            Node cur = startNode;
            startNode = cur.left;
            while(cur != null){
                if(cur.left != null) cur.left.next = cur.right;
                if(cur.right != null) cur.right.next = (cur.next == null) ? null : cur.next.left;
                cur = cur.next;
            }
        }
        return root;
    }
}