I like to read this solution as a small machine: keep the useful information, throw away the noise. For 117. Populating Next Right Pointers in Each Node II, the solution in this repository is mainly a two pointers solution.
Guide
What?
The first job is to translate the English prompt into state, transition, and stopping conditions. Instead of trying to be clever immediately, read the code as a sequence of questions:
- What state are we keeping?
- How do we move from one state to the next?
- When do we know the answer is already determined?
For this file, the main tools are: two pointers, sliding window.
The notes already sitting in the source point us in the right direction:
- BFS with queue
- time: O(N), space: O(N)
Guide
When?
This pattern shows up when the brute force version has too many repeated checks, too many possible branches, or too much bookkeeping to do by hand. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The solution is organized around the main LeetCode entry point and a few local helpers.
Guide
Why?
The code is doing bookkeeping so your brain does not have to keep the entire search space open at once.
- The queue gives the solution a level-by-level or frontier-style traversal.
- The final return is not magic; it is the invariant after the loops or recursion have finished doing their accounting.
Guide
How?
Walk through the solution in this order:
- Initialize the memory or helper structure.
- Process candidates in the order the invariant expects.
- Update the answer only when the current state is valid.
- Return the value that represents the fully processed input.
The most important competitive-programming habit here is to trust the invariant. Once the invariant is right, the loops become much less scary.
Guide
Complexity
- Time: O(N), space: O(1)
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
01//BFS with queue
02//Runtime: 20 ms, faster than 52.71% of C++ online submissions for Populating Next Right Pointers in Each Node II.
03//Memory Usage: 17.9 MB, less than 66.31% of C++ online submissions for Populating Next Right Pointers in Each Node II.
04//time: O(N), space: O(N)
05class Solution {
06public:
07 Node* connect(Node* root) {
08 if(!root) return root;
09
10 queue<Node*> q;
11
12 q.push(root);
13
14 while(!q.empty()){
15 int level_size = q.size();
16
17 Node* prev = nullptr;
18
19 while(level_size-- > 0){
20 Node* cur = q.front(); q.pop();
21
22 if(prev) prev->next = cur;
23
24 if(cur->left) q.push(cur->left);
25 if(cur->right) q.push(cur->right);
26
27 prev = cur;
28 }
29 }
30
31 return root;
32 }
33};
34
35//O(1) space iterative solution
36//https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/discuss/37811/Simple-solution-using-constant-space
37//Runtime: 12 ms, faster than 96.85% of C++ online submissions for Populating Next Right Pointers in Each Node II.
38//Memory Usage: 17.7 MB, less than 66.31% of C++ online submissions for Populating Next Right Pointers in Each Node II.
39//time: O(N), space: O(1)
40class Solution {
41public:
42 Node* connect(Node* root) {
43 //first node of next level
44 Node *nhead = nullptr;
45
46 Node *prev = nullptr, *cur = root;
47
48 while(cur){
49 //process cur's next level
50 while(cur){
51 if(cur->left){
52 if(prev){
53 prev->next = cur->left;
54 }else{
55 nhead = cur->left;
56 }
57 prev = cur->left;
58 }
59 //else
60 if(cur->right){
61 if(prev){
62 prev->next = cur->right;
63 }else{
64 nhead = cur->right;
65 }
66 prev = cur->right;
67 }
68 //go to next node in current level
69 cur = cur->next;
70 }
71
72 //go to next level
73 cur = nhead;
74 prev = nhead = nullptr;
75 }
76
77 return root;
78 }
79};
80
81//O(1) space(not consider recursion stack) recursive solution
82//https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/discuss/172861/Mostly-recursive-solution-O(n)-time-(beats-99.32)-and-O(1)-space-(without-considering-stack)
83//Runtime: 4 ms, faster than 100.00% of C++ online submissions for Populating Next Right Pointers in Each Node II.
84//Memory Usage: 17.6 MB, less than 66.31% of C++ online submissions for Populating Next Right Pointers in Each Node II.
85class Solution {
86public:
87 Node* findnext(Node* node){
88 //recursively find the next leftmost child
89 if(!node) return node;
90 if(node->left) return node->left;
91 if(node->right) return node->right;
92 //current node is leaf, go to its right neighbor
93 return findnext(node->next);
94 }
95
96 Node* connect(Node* root) {
97 if(!root) return root;
98
99 if(root->left){
100 Node* target;
101 if(root->right){
102 target = root->right;
103 }else{
104 target = findnext(root->next);
105 }
106 root->left->next = target;
107 }
108
109 if(root->right){
110 root->right->next = findnext(root->next);
111 }
112
113 //https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/discuss/172861/Mostly-recursive-solution-O(n)-time-(beats-99.32)-and-O(1)-space-(without-considering-stack)/260137
114 /*
115 in the function "findnext",
116 we will recursively go right,
117 and it requires that the right subtree is connected first,
118 otherwise, therewill be a gap btw left and right subtree
119
120 consider: [2,1,3,0,7,9,1,2,null,1,0,null,null,8,8,null,null,null,null,7]
121 */
122 connect(root->right);
123 connect(root->left);
124
125 return root;
126 }
127};Cost