leetcode 116. Populating Next Right Pointers in Each Node

最新推荐文章于 2025-12-03 17:03:32 发布

原创最新推荐文章于 2025-12-03 17:03:32 发布 · 102 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#算法

leetcode 专栏收录该内容

490 篇文章

订阅专栏

本文介绍了一种在完美二叉树中连接节点next指针的方法，通过BFS和递归两种方式实现，确保每层节点间的正确链接。

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.

在这里插入图片描述

给一棵二叉树，让把每层的左边节点的next指针指向右边节点

思路
(1) BFS
每层节点直到最后一个都是next指向下一节点，每层的最后一个节点next指向null

//1ms
	//BFS..
    public Node connect(Node root) {
        
        if (root == null) {
            return null;
        }
        
        List<Node> current = new ArrayList<>();
        List<Node> next = new ArrayList<>();
        
        current.add(root);
        root.next = null;
        
        while (!current.isEmpty()) {
            
            for (int i = 0; i < current.size() - 1; i++) {
                Node node = current.get(i);
                node.next = current.get(i + 1);
                
                if (node.left != null) {
                    next.add(node.left);
                }
                
                if (node.right != null) {
                    next.add(node.right);
                }
            }
            
            Node last = current.get(current.size() - 1);
            last.next = null;
            if (last.left != null) {
                next.add(last.left);
            }
                
            if (last.right != null) {
                next.add(last.right);
            }
            
            current.clear();
            List<Node> tmp = current;
            current = next;
            next = tmp;
        }
        
        return root;
    }

(2) 递归

//0ms
    public Node connect(Node root) {
        if (root == null) {
            return null;
        }
       
        root.next = null;
        helper(root.left, root.right);
        
        return root;
    }
    
    public void helper(Node root1, Node root2) {
        if (root1 == null && root2 == null) {
            return;
        }
        
        root1.next = root2;
        
        helper(root1.left, root1.right);
        helper(root1.right, root2.left);
        helper(root2.left, root2.right);
    }