时间轴

2025-10-26

init

题目：

P117 填充每个节点的下一个右侧节点指针 II

https://leetcode.cn/problems/populating-next-right-pointers-in-each-node-ii/description/?envType=study-plan-v2&envId=top-interview-150

层次遍历即可

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
/*
// Definition for a Node.
*/
#include <cstddef>
class Node {
    public:
	int val;
	Node *left;
	Node *right;
	Node *next;

	Node()
		: val(0)
		, left(NULL)
		, right(NULL)
		, next(NULL)
	{
	}

	Node(int _val)
		: val(_val)
		, left(NULL)
		, right(NULL)
		, next(NULL)
	{
	}

	Node(int _val, Node *_left, Node *_right, Node *_next)
		: val(_val)
		, left(_left)
		, right(_right)
		, next(_next)
	{
	}
};

#include <queue>
using std::queue;

class Solution {
    public:
	Node *connect(Node *root)
	{
		if (root == nullptr) {
			return root;
		}
		// 层次遍历
		int i, n;
		Node *curr, *last;
		queue<Node *> que;

		que.push(root);

		while (!que.empty()) {
			n = que.size();
			last = nullptr;
			for (i = 0; i < n; i++) {
				curr = que.front();
				que.pop();
				if (last != nullptr) {
					last->next = curr;
				}

				if (curr->left != nullptr) {
					que.push(curr->left);
				}
				if (curr->right != nullptr) {
					que.push(curr->right);
				}

				last = curr;
			}
			curr->next = nullptr;
		}
		return root;
	}
};