134 lines
2.6 KiB
Markdown
134 lines
2.6 KiB
Markdown
# Leetcode Binary-Tree-Inorder-Traversal
|
|
|
|
#### 2022-07-04 15:42
|
|
|
|
> ##### Algorithms:
|
|
> #algorithm #DFS #DFS_inorder
|
|
> ##### Data structures:
|
|
> #DS #binary_tree
|
|
> ##### Difficulty:
|
|
> #coding_problem #difficulty-easy
|
|
> ##### Additional tags:
|
|
> #leetcode
|
|
> ##### Revisions:
|
|
> N/A
|
|
|
|
##### Related topics:
|
|
```expander
|
|
tag:#DFS
|
|
```
|
|
|
|
- [[Leetcode Binary-Tree-Postorder-Traversal]]
|
|
- [[Leetcode Binary-Tree-Preorder-Traversal]]
|
|
|
|
|
|
##### Links:
|
|
- [Link to problem](https://leetcode.com/problems/binary-tree-inorder-traversal/)
|
|
___
|
|
### Problem
|
|
Given the `root` of a binary tree, return _the inorder traversal of its nodes' values_.
|
|
|
|
#### Examples
|
|
**Example 1:**
|
|
|
|
|
|
|
|
**Input:** root = [1,null,2,3]
|
|
**Output:** [1,3,2]
|
|
|
|
**Example 2:**
|
|
|
|
**Input:** root = []
|
|
**Output:** []
|
|
|
|
**Example 3:**
|
|
|
|
**Input:** root = [1]
|
|
**Output:** [1]
|
|
|
|
#### Constraints
|
|
- The number of nodes in the tree is in the range `[0, 100]`.
|
|
- `-100 <= Node.val <= 100`
|
|
|
|
### Thoughts
|
|
|
|
> [!summary]
|
|
> This is a #DFS traversal problem
|
|
|
|
Many of them are same to [[Leetcode Binary-Tree-Preorder-Traversal]]
|
|
### Solution
|
|
|
|
Recursion
|
|
```cpp
|
|
/**
|
|
* Definition for a binary tree node.
|
|
* struct TreeNode {
|
|
* int val;
|
|
* TreeNode *left;
|
|
* TreeNode *right;
|
|
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
|
|
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
|
|
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
|
|
* right(right) {}
|
|
* };
|
|
*/
|
|
class Solution {
|
|
void inorder(TreeNode *root, vector<int> &answer) {
|
|
if (!root) {
|
|
return;
|
|
}
|
|
|
|
inorder(root->left, answer);
|
|
answer.push_back(root->val);
|
|
inorder(root->right, answer);
|
|
}
|
|
|
|
public:
|
|
vector<int> inorderTraversal(TreeNode *root) {
|
|
// Recursion.
|
|
vector<int> answer;
|
|
inorder(root, answer);
|
|
return answer;
|
|
}
|
|
};
|
|
```
|
|
|
|
Iteration
|
|
```cpp
|
|
/**
|
|
* Definition for a binary tree node.
|
|
* struct TreeNode {
|
|
* int val;
|
|
* TreeNode *left;
|
|
* TreeNode *right;
|
|
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
|
|
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
|
|
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
|
|
* right(right) {}
|
|
* };
|
|
*/
|
|
class Solution {
|
|
public:
|
|
vector<int> inorderTraversal(TreeNode *root) {
|
|
// Iteration
|
|
stack<TreeNode *> pending;
|
|
vector<int> answer;
|
|
|
|
while (root || !pending.empty()) {
|
|
// traverse left first.
|
|
while (root) {
|
|
pending.push(root);
|
|
root = root->left;
|
|
}
|
|
|
|
root = pending.top();
|
|
pending.pop();
|
|
answer.push_back(root->val);
|
|
root = root->right;
|
|
}
|
|
|
|
return answer;
|
|
}
|
|
};
|
|
|
|
``` |