题目
July 10, 2021 · View on GitHub
给你二叉树的根节点 root ,返回它节点值的 前序 遍历。
示例 1:

输入:root = [1,null,2,3]
输出:[1,2,3]
示例 2:
输入:root = []
输出:[]
示例 3:
输入:root = [1]
输出:[1]
示例 4:

输入:root = [1,2]
输出:[1,2]
示例 5:

输入:root = [1,null,2]
输出:[1,2]
提示:
- 树中节点数目在范围 [0, 100] 内
- -100 <= Node.val <= 100
参考答案
class Solution {
public:
vector<int> preorderTraversal(TreeNode *root) {
vector<int> res;
if (root == nullptr) {
return res;
}
TreeNode *p1 = root, *p2 = nullptr;
while (p1 != nullptr) {
p2 = p1->left;
if (p2 != nullptr) {
while (p2->right != nullptr && p2->right != p1) {
p2 = p2->right;
}
if (p2->right == nullptr) {
res.emplace_back(p1->val);
p2->right = p1;
p1 = p1->left;
continue;
} else {
p2->right = nullptr;
}
} else {
res.emplace_back(p1->val);
}
p1 = p1->right;
}
return res;
}
};