1. 递归解法

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 

 
class Solution {
public:
    vector<int> path;
    void preorder(TreeNode *root)
    {
        if (root == NULL)
            return;
        path.push_back(root->val);
        preorder(root->left);
        preorder(root->right);
    }
    vector<int> preorderTraversal(TreeNode *root) {
        
        preorder(root);
        
        return path;
    }
};

2. 非递归解法一（空间复杂度O(n))

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode *root) {
        vector<int> path;
        path.clear();
        stack<TreeNode *> st;
        if (root == NULL)
            return path;
        st.push(root);
        while (!st.empty())
        {
            TreeNode *node = st.top();
            st.pop();
            path.push_back(node->val);
            if (node->right != NULL)
                st.push(node->right);
            if (node->left != NULL)
                st.push(node->left);
        }
        return path;
    }
};

3. 非递归解法（空间复杂度O(1))

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode *root) {
        vector<int> path;
        if (root == NULL)
            return path;
            
        TreeNode *cur = root;
        TreeNode *pre = NULL;
        
        while (cur != NULL)
        {
            if (cur->left == NULL)
            {
                path.push_back(cur->val);
                cur = cur->right;
            }
            else
            {
                pre = cur->left;
                while (pre->right != NULL && pre->right != cur)
                    pre = pre->right;
                if (pre->right == NULL)
                {
                    pre->right = cur;
                    path.push_back(cur->val);
                    cur = cur->left;
                }
                else
                {
                    pre->right = NULL;
                    cur = cur->right;
                }
            }
        }
        return path;
    }
};