# Leetcode Binary-Tree-Preorder-Traversal

#### 2022-07-04 14:51

> ##### Algorithms:
>
> #algorithm #DFS #DFS_preorder #Morris_traversal
>
> ##### Data structures:
>
> #DS #binary_tree
>
> ##### Difficulty:
>
> #coding_problem #difficulty-easy
>
> ##### Additional tags:
>
> #leetcode #CS_list_need_understanding
>
> ##### Revisions:
>
> N/A

##### Related topics:

##### Links:

- [Link to problem](https://leetcode.com/problems/binary-tree-preorder-traversal/)

---

### Problem

Given the `root` of a binary tree, return _the preorder traversal of its nodes' values_.

#### Examples

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/09/15/inorder_1.jpg)

**Input:** root = [1,null,2,3]
**Output:** [1,2,3]

**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 #binary_tree #DFS_preorder problem.

Preorder, means root is at the "Pre" position, so the order is:

- Search root node
- Search left sub-tree
- Search right sub-tree

The recursion version is easy to implement.

Use Stacks for the iteration method, because stacks are FILO, the **subtree will get traversed first.**
And remember to push **right subtree** first, so that the left one will be traversed first.

The Morris traversal needs more understandings.

### Solution

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> preorderTraversal(TreeNode *root) {
    if (!root) {
      return {};
    }

    // Using stacks, since FILO, the latest gets served first.
    vector<int> answer;
    stack<TreeNode *> pending;

    do {
      if (root != nullptr) {
        answer.push_back(root->val);
        // nothing gets pushed if is nullptr;
        pending.push(root->right);
        pending.push(root->left);
      }
      root = pending.top();
      pending.pop();
    } while (root || !pending.empty());

    return answer;
  }
};

```

Recursion, using private funcs:

```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> preorderTraversal(TreeNode *root) {
    // Use two functions:
    vector<int> answer;
    preorder(root, answer);
    return answer;
  }

private:
  void preorder(TreeNode *root, vector<int> &answer) {
    if (root == nullptr) {
      return;
    }

    answer.push_back(root->val);
    preorder(root->left, answer);
    preorder(root->right, answer);
  }
};
```

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 {
public:
  vector<int> preorderTraversal(TreeNode *root) {
    // If the node is empty
    if (root == nullptr) {
      return {};
    }
    // First check root
    vector<int> answer;

    answer.push_back(root->val);

    vector<int> left = preorderTraversal(root->left);
    answer.insert(answer.end(), left.begin(), left.end());

    vector<int> right = preorderTraversal(root->right);
    answer.insert(answer.end(), right.begin(), right.end());

    return answer;
  }
};

```
