题解 | #二叉树的前序遍历#

/**
 * struct TreeNode {
 *  int val;
 *  struct TreeNode *left;
 *  struct TreeNode *right;
 *  TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param root TreeNode类
     * @return int整型vector
     */
    vector<int> preorderTraversal(TreeNode* root) {
        // write code here
        vector<int> res;
        if(root == nullptr) return res;
        stack<TreeNode* > s;
        s.push(root);
        while (!s.empty()) {
            TreeNode* node = s.top();
            s.pop();
            res.push_back(node->val);
            if(node->right!=nullptr) s.push(node->right);
            if(node->left!=nullptr) s.push(node->left);
        }
        return res;
    }
};