117. 填充每个节点的下一个右侧节点指针 II 题目有点类似于层序线索化，就是在层序遍历的基础上魔改一下，使得可以获得当前遍历到第几层的信息。

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
/*
// Definition for a Node.
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) {}
};
*/

class Solution {
   
public:
    void bfs(Node *root) {
   
        if(root == NULL) return ;
        queue<Node*> q;
        q.push(root);
        //cnt:当前正在遍历的这一层已经遍历了几个节点
        //tail:当前正在遍历的这一层一共有几个节点
        //num:下一层一共有几个节点
        //pre:这一层上一个遍历到的节点
        int cnt = 0, tail = 1, num = 0;
        Node *pre = NULL;
        while(!q.empty()) {
   
            Node *u = q.front();
            q.pop();
            if(pre) pre->next = u;
            pre = u;
            ++ cnt;
            if(u->left) {
   
                q.push(u->left);
                ++ num;
            }
            if(u->right) {
   
                q.push(u->right);
                ++ num;
            }
            if(cnt == tail) {
   
                cnt = 0;
                tail = num;
                num = 0;
                u->next = NULL;
                pre = NULL;
            }
        }
    }
    Node* connect(Node* root) {
   
        bfs(root);
        return root;
    }
};

还有一种简单的写法，直接得出当前层节点个数

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
/*
// Definition for a Node.
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) {}
};
*/

class Solution {
   
public:
    void bfs(Node *root) {
   
        if(root == NULL) return ;
        queue<Node*> q;
        q.push(root);
        while(!q.empty()) {
   
            Node *pre = NULL;
            int len = q.size();	//为当前层的总节点数
            for(int i = 0; i < len; i ++) {
   	//遍历当前层节点
                Node *u = q.front();
                q.pop();
                if(pre) pre->next = u;
                pre = u;
                if(u->left) q.push(u->left);
                if(u->right) q.push(u->right);
            }
        }
    }
    Node* connect(Node* root) {
   
        bfs(root);
        return root;
    }
};