递归

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

迭代

  
/**
 * 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) {
        vector<int> res;
        stack<TreeNode*> stk;
        while (root || stk.size()) {
            // 遍历所有左子树
            while(root) {
                res.push_back(root->val);  // 把当前结点值存入答案
                stk.push(root);  // 把所有root压入栈
                root = root->left;  // 左边
            }
            // 遍历完左子树之后直接变就可以遍历右子树
            root = stk.top()->right;  // 回溯到栈顶然后直接遍历右边 
            stk.pop();  // 弹出
        }
        return res;
    }
};