A good way into this one is to ask: what do we need to remember so we never redo work blindly? For 94. Binary Tree Inorder Traversal, the solution in this repository is mainly a two pointers solution.
Guide
What?
The code is easier to read if we treat it as a controlled search through possible states. 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, stack, sliding window.
The notes already sitting in the source point us in the right direction:
- Recursive
Guide
When?
Use this approach when the hard part is not syntax, but deciding what must stay true after every update. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are inorderTraversalR, inorderTraversal.
Guide
Why?
The code is doing bookkeeping so your brain does not have to keep the entire search space open at once.
- The stack stores unfinished context, which is usually the cleanest way to handle nested or monotonic structure.
- 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:
- Read the setup variables first.
- Follow the main loop or recursive helper next.
- Watch where invalid states get skipped.
- Check which value survives to the return statement.
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) to O(n log n), depending on the dominant loop or data structure operation
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
solution.cpp
01//Recursive
02//Runtime: 4 ms, faster than 56.80% of C++ online submissions for Binary Tree Inorder Traversal.
03//Memory Usage: 9.3 MB, less than 77.00% of C++ online submissions for Binary Tree Inorder Traversal.
04/**
05 * Definition for a binary tree node.
06 * struct TreeNode {
07 * int val;
08 * TreeNode *left;
09 * TreeNode *right;
10 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
11 * };
12 */
13class Solution {
14public:
15 vector<int> ans;
16
17 void inorderTraversalR(TreeNode* root){
18 if(!root) return;
19 if(root->left){
20 inorderTraversal(root->left);
21 }
22 ans.push_back(root->val);
23 if(root->right){
24 inorderTraversal(root->right);
25 }
26 }
27
28 vector<int> inorderTraversal(TreeNode* root) {
29 inorderTraversalR(root);
30 return ans;
31 }
32};
33
34//Runtime: 0 ms, faster than 100.00% of C++ online submissions for Binary Tree Inorder Traversal.
35//Memory Usage: 8 MB, less than 100.00% of C++ online submissions for Binary Tree Inorder Traversal.
36/**
37 * Definition for a binary tree node.
38 * struct TreeNode {
39 * int val;
40 * TreeNode *left;
41 * TreeNode *right;
42 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
43 * };
44 */
45/**
46 * Definition for a binary tree node.
47 * struct TreeNode {
48 * int val;
49 * TreeNode *left;
50 * TreeNode *right;
51 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
52 * };
53 */
54class Solution {
55public:
56 vector<int> inorderTraversal(TreeNode* root) {
57 stack<TreeNode*> stk;
58 TreeNode* cur = root;
59 vector<int> ans;
60
61 stk.push(cur);
62
63 do{
64 cur = stk.top();
65
66 //go to its leftmost child
67 TreeNode* tmp = cur ? cur->left : NULL;
68 while(tmp){
69 stk.push(tmp);
70 tmp = tmp->left;
71 }
72
73 //if it has left child, process leftmost child first
74 cur = stk.top(); stk.pop();
75 if(cur) ans.push_back(cur->val);
76
77 //cut current node and its parent,
78 //so current node won't be traversed again
79 if(stk.size() > 0) stk.top()->left = NULL;
80
81 //deal with right child
82 if(cur && cur->right) stk.push(cur->right);
83 }while(!stk.empty());
84
85 return ans;
86 }
87};
88
89//iterative, another version
90//Runtime: 4 ms, faster than 56.80% of C++ online submissions for Binary Tree Inorder Traversal.
91//Memory Usage: 8 MB, less than 100.00% of C++ online submissions for Binary Tree Inorder Traversal.
92/**
93 * Definition for a binary tree node.
94 * struct TreeNode {
95 * int val;
96 * TreeNode *left;
97 * TreeNode *right;
98 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
99 * };
100 */
101class Solution {
102public:
103 vector<int> inorderTraversal(TreeNode* root) {
104 vector<int> ans;
105 stack<TreeNode*> stk;
106 TreeNode* cur;
107
108 stk.push(root);
109
110 while(!stk.empty()){
111 cur = stk.top();
112 while(cur && cur->left){
113 stk.push(cur->left);
114 cur = cur->left;
115 }
116 cur = stk.top(); stk.pop();
117 if(cur) ans.push_back(cur->val);
118 if(stk.size() > 0) stk.top()->left = nullptr;
119 if(cur && cur->right) stk.push(cur->right);
120 }
121
122 return ans;
123 }
124};
125
126//Approach 2: Iterating method using Stack(official sol)
127//https://leetcode.com/problems/binary-tree-postorder-traversal/discuss/45551/Preorder-Inorder-and-Postorder-Iteratively-Summarization/188240
128//time: O(N), space: O(N)
129//Runtime: 0 ms, faster than 100.00% of C++ online submissions for Binary Tree Inorder Traversal.
130//Memory Usage: 8.5 MB, less than 100.00% of C++ online submissions for Binary Tree Inorder Traversal.
131/**
132 * Definition for a binary tree node.
133 * struct TreeNode {
134 * int val;
135 * TreeNode *left;
136 * TreeNode *right;
137 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
138 * };
139 */
140class Solution {
141public:
142 vector<int> inorderTraversal(TreeNode* root) {
143 vector<int> ans;
144 stack<TreeNode*> stk;
145 TreeNode* cur = root;
146 while(cur || !stk.empty()){
147 //if cur is nullptr, it's ok, because it will later be stk.top()
148 while(cur){
149 stk.push(cur);
150 cur = cur->left;
151 }
152 cur = stk.top(); stk.pop();
153 ans.push_back(cur->val);
154 //it's fine if cur->right is nullptr
155 cur = cur->right;
156 }
157
158 return ans;
159 }
160};
161
162//Approach 3: Morris Traversal
163//time: O(n), space: O(n)
164//Runtime: 4 ms, faster than 56.80% of C++ online submissions for Binary Tree Inorder Traversal.
165//Memory Usage: 8 MB, less than 100.00% of C++ online submissions for Binary Tree Inorder Traversal.
166/**
167 * Definition for a binary tree node.
168 * struct TreeNode {
169 * int val;
170 * TreeNode *left;
171 * TreeNode *right;
172 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
173 * };
174 */
175class Solution {
176public:
177 vector<int> inorderTraversal(TreeNode* root) {
178 vector<int> ans;
179 TreeNode *cur = root, *pre;
180
181 while(cur){
182 if(cur->left){
183 pre = cur->left;
184 while(pre->right){
185 pre = pre->right;
186 }
187 //pre->right is now cur's rightmost child
188 //it doesn't have right child
189 //pre is the parent of cur's rightmost child
190 pre->right = cur;
191 TreeNode* tmp = cur;
192 //current root becomes its left child
193 cur = cur->left;
194 //cur original root's left subtree
195 tmp->left = nullptr;
196 }else{
197 ans.push_back(cur->val);
198 cur = cur->right;
199 }
200 }
201
202 return ans;
203 }
204};Cost
Complexity
Time
O(n) to O(n log n), depending on the dominant loop or data structure operation
Dominated by the main traversal, recursion, or data-structure operations in the code.
Space
O(n) in the usual case for auxiliary containers or recursion
Auxiliary state plus the answer structure where the problem requires one.