## [144\. Binary Tree Preorder Traversal](https://leetcode.com/problems/binary-tree-preorder-traversal/)

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

**Example 1:**


**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`

**Follow up:** Recursive solution is trivial, could you do it iteratively?

recursion

/**
 * 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 {
private:
    vector<int> res;
public:
    vector<int> preorderTraversal(TreeNode* root) {
        dfs(root);
        return res;
    }
    void dfs(TreeNode* root) {
        if(!root) return;
        res.push_back(root->val);
        dfs(root->left);
        dfs(root->right);
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(N)$
:::

iteration

/**
 * 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) {
        // preorder: root -> left ->right
        vector<int> res;
        stack<TreeNode*> stk;
        stk.push(root);
        
        while(!stk.empty()) {
            TreeNode* node = stk.top(); stk.pop();
            if(!node) continue;
            res.push_back(node->val);
            // 為了先處理左側，先 push 右側到 stack
            if(node->right) {
                stk.push(node->right);
            }
            // 利用 stack 的特性，最後再 push 左側
            if(node->left) {
                stk.push(node->left);
            }
        }
        return res;
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(N)$
:::